chapter 3 · 2018. 9. 18. · chapter 3 applications of derivatives 3.1 extrema of functions on...

Chapter 3

Applications of Derivatives

3.1 Extrema of Functions on Intervals

• Maximum and Minimum Values of a Function • Relative Extrema and Crit-ical Numbers • Finding the Extrema on a Closed Interval

What you have been obliged

to discover by yourself leaves a

path in your mind which you

can use again when the need

arises. - G.C. Lichtenberg

Maximum and Minimum Values of a Function

1 2 3 4 x1

5

y

Figure 1aIn [1, 4], the maximumvalue is f(4) = 5 and theminimum value is f(2) = 1.

1 2 3 4 x1

5

y

Figure 1b

In [1, 4), f has

no maximum value and the

minimum value is f(2) = 1.

In Chapter 2, we studied several rules for finding the derivative of a function. In thissection, we use derivatives to find the maximum and minimum values of a differentiablefunction. Such values have important consequences. The information of knowing how tomaximize returns or minimize costs is important to know.

Definition 1 Maximum and Minimum Values of a Function

Let f be a function that is defined on an interval I containing c.

1. f(c) is the minimum value of f on I if f(c) ≤ f(x) for all x in I.

2. f(c) is the maximum value of f on I if f(c) ≥ f(x) for all x in I.

The maximum and minimum values of f on I are called the extreme values or extremaof f on I.

In the next example, we use the graph to identify any extrema. It is possible for afunction not to have a maximum or minimum value.

Example 1 Finding the Extrema of a Function

Find the extrema of f(x) = (x− 2)2 + 1 in the indicated interval.

a) [1, 4] b) [1, 4)

Solution

a) The graph of y = f(x) in [1, 4] is given in Figure 1a. The ‘highest point’ is (4, 5)and the ‘lowest point’ is (2, 1). Then the maximum value of f is f(4) = 5 and theminimum value is f(2) = 1.

b) In Figure 1b, we see the graph of y = f(x) in [1, 4). The point (4, 5) is not the‘highest point’ since (4, 5) does not belong to the graph of y = f(x) in [1, 4).

137

138 CHAPTER 3. APPLICATIONS OF DERIVATIVES

There is no highest point in [1, 4) because given any point we can always find ahigher point. In other words, none of the following values of f(x)

f(3.9) = (3.9− 2)2 + 1 = 4.61

f(3.99) = (3.99− 2)2 + 1 = 4.9601

...

f(x) = (x− 2)2 + 1, if x < 4

is the maximum value of f(x) on [1, 4). Then f has no maximum value on [1, 4).However, the point (2, 1) is the ‘lowest point’ on the graph. Hence, the minimumvalue of f on [1, 4) is f(2) = 1.

✷

�1 1 x

2

y

Figure 1c

The graph of f(x) = x2+ 1.

Try This 1Find the extrema of f(x) = x2 + 1 in the indicated interval, see Figure 1c.

a) [−1, 0] b) (−1, 1) c) (0, 1) d) (0, 1]

1 2 3 4 x1

3

5

y

Figure 2

The graph has no minimum

value in [1, 4], but the

maximum value is 5.

Example 2 Finding the Extrema of a Function

Find the extrema of

g(x) =

�(x− 2)2 + 1 if x �= 23 if x = 2

in the interval [1, 4]

Solution In Figure 2, the point (4, 5) is the ‘highest point’ on the graph of y = g(x).Then the maximum value of g on [1, 4] is g(4) = 5.

Note, the point (2, 1) does not lie on the graph of y = g(x) and cannot be the ‘lowestpoint ” on the graph. We cannot claim that a certain point on the graph is the lowestpoint since there will be another point on the graph that is ‘lower’. Hence, g(x) has nominimum value on [1, 4].

✷

Try This 2Find the extrema of g(x) = (x − 3)2 + 4 where x lies in [1, 4] and x �= 3. In particular,g(3) is undefined.

The next theorem states a condition when a function is guaranteed to have extremevalues. A proof of the theorem is omitted since it is beyond the scope of this book.

Theorem 3.1 The Extreme Value Theorem

A continuous function on a closed interval [a, b] is guaranteed to have a maximum valueand a minimum value on [a, b].

3.1. EXTREMA OF FUNCTIONS ON INTERVALS 139

y� f �x��1 1 2 3 4 x

5

�27

y

Figure 3

Relative minimum values

are f(0) = 0 and

f(3) = −27, and the relative

maximum value is f(1) = 5.

Relative Extrema and Critical NumbersA function f has a relative maximum value at point P (c, f(c)) if P is ‘higher than nearbypoints’, or geometrically P is at the ‘top of a hill’. Similarly, f has a relative minimumvalue at P if P is ‘lower than nearby points’, or P is at the ‘bottom of a valley’.

As seen in Figure 3, the graph of y = f(x) has a relative maximum value at thepoint (1, 5), or f(1) = 5 is a relative maximum value of f . Similarly, f has relativeminimum values at (0, 0) and (3,−27). That is, f(0) = 0 and f(3) = −27 are relativeminimum values of f .

Definition 2 Definition of Relative Extrema

Let f be a function.

1. f(c) is a relative maximum value of f if there is an open interval I containing c forwhich f(c) is the maximum value of f on I.

2. f(c) is a relative minimum value of f if there is an open interval I containing c forwhich f(c) is the minimum value of f on I.

The relative maximum values and relative minimum values of f are called the relativeextreme values, or relative extrema of f . The relative maximum values and relativeminimum values are also called relative maxima and relative minima, respectively.

f �x��x3�3x2�2

�1

2

1 2 3 4 x

�2

y

Figure 4af(0) = 2 is the relativemaximum value, andf(2) = −2 is the relativeminimum value of f .

f �x��1�x2�3

�1 1 x

1

y

Figure 4b

f(0) = 1 is the

maximum value of f .

Clearly, an extreme value of a function is a relative extreme value. However, a relativeextreme value is not necessarily an extreme value. In Figure 3, f(0) = 0 and f(1) = 5 arerelative extreme values of f but they are not the minimum and maximum values of f .

In Example 3, we will see a particular case of a general phenomena. That is, if f(c) isa relative extreme value then f �(c) is either zero or undefined.

Example 3 The Derivative at a Relative Extrema

Determine the relative extreme values f(c) of f , and evaluate f �(c).

a) f(x) = x3 − 3x2 + 2 b) f(x) = 1− x2/3

See the graphs in Figures 4a and 4b.

Solution

a) In Figure 4a, we see that the point (0, 2) is a ‘relative maximum point’. Thenf(0) = 2 is a relative maximum value of f . Similarly, the point (2,−2) is a ‘localminimum point’. Thus, f(2) = −2 is a relative minimum value of f .

The derivative of f isf �(x) = 3x2 − 6x = 3x(x− 2).

Substituting x = 0 and x = 2 into f �(x), we obtain

f �(0) = 0 and f �(2) = 0.

b) Note, the derivative of f is

f �(x) = −23x−1/3 = − 2

3 3√x.

Using Figure 4b, we see that f(0) = 1 is the maximum value of f .

Then f �(0) is undefined since it evaluates to the expression − 2

3 3√0.

Finally, from the graph we see that there is no relative minimum value as seen inFigure 4b.

✷


f �x��4x�x2

2 x

4

y

Figure 5af(x) = 4x− x2

f �x��1�x 2�3�3�2

�1 1 x

1

y

Figure 5b

f(x) =�1− x2/3

�3/2

Try This 3For each relative extreme value f(c), evaluate f �(c). See Figures 5a and 5b.

a) f(x) = 4x− x2 b) f(x) =�1− x2/3

�3/2

In Example 3, we have seen that the values of the derivative at the relative extrema areeither zero or undefined. These x-values are called critical numbers.

Definition 3 The Definition of a Critical Number

Let f be a function which is defined at a number c. If f �(c) = 0 or f �(c) does not exist,then c is a critical number of f .

Example 4 Finding the Critical Numbers

Find the critical numbers of each function.

a) f(x) = 2x3 − 3x2 − 36x+ 1 b) g(x) = x2/3(x+ 4)

Solution

a) Evaluate and factor the derivative f � as follows:

f �(x) = 6x2 − 6x− 36

= 6(x2 − x− 6)

= 6(x− 3)(x+ 2).

Then f �(x) = 0 exactly when x = −2 or x = 3. Thus, the critical numbers arex = −2, 3; see local extrema in Figure 6a.

f �x�� 2x3�3x2�36x�145

�2 3 x

�80

y

Figure 6a: The critical numbers of f are −2 and 3.

b) Apply the product rule and factor g�(x).

g(x) = x2/3(x+ 4)

g�(x) =23x−1/3(x+ 4) + x2/3

g�(x) =x−1/3

3[2(x+ 4) + 3x]

g�(x) =5x+ 8

3 3√x


Note, g�(x) = 0 if and only if5x+ 8 = 0.

Then x = − 85 is a critical number.

Moreover, the derivative g�(0) is undefined since it evaluates to an undefined expres-sion 8

3 3√0. Thus, x = 0 is also a critical number.

Hence, the critical numbers are − 85 and 0; see the relative extrema in Figure 6b.

g�x��x2�3�x�4�

� 85x

4

y

Figure 6b: The critical numbers of g are 0 and −8/5.

✷

Try This 4Find the critical numbers of each function.

a) f(x) = x3 + 3x2 − 24x b) f(x) =√x c) f(x) = 3x+ 1

A first step in finding the extreme values of a differentiable function is to find thecritical numbers. To illustrate this point, in Figure 7 we see that f has relative minimumand maximum values at C1 and C2, respectively. By Theorem 3.2, the abcissas of C1 andC2 are critical numbers of f . Thus, the relative extrema can be found from the list ofcritical numbers. However, we caution the student that there are critical numbers that donot provide extreme values. In any case, the set of critical numbers is a short list that wecould use to find the relative extrema.

y� f �x�C1

C2

x

y

Figure 7 The graph of f hasrelative extreme values at thecritical points C1 and C2.


Theorem 3.2 The Relative Extrema Occur Exactly at Critical Points

Let f be a function. If f(c) is a relative extreme value, then c is a critical number of f .

Proof If f �(c) does not exist, then the theorem is true and there would be nothing elseto prove. So, we assume f �(c) exists and we have to prove f �(c) = 0. If f �(c) �= 0, theneither f �(c) > 0 or f �(c) < 0. Suppose

f �(c) = limh→0

f(c+ h)− f(c)h

> 0.

By Exercise 47 in Chapter 1, there exists an open interval (−δ, δ) with δ > 0 such that

f(c+ h)− f(c)h

> 0

whenever −δ < h < δ with h �= 0. If we multiply both sides of the previous inequalitywith a positive h that is less than δ, then we obtain

h · f(c+ h)− f(c)h

> h · 0

f(c+ h)− f(c) > 0

f(c+ h) > f(c).

This implies f(c) is not a relative maximum value of f . On the other hand if we multiplythe same inequality with a negative h with −δ < h < 0, then we reverse the direction ofthe inequality and find

h · f(c+ h)− f(c)h

< h · 0

f(c+ h)− f(c) < 0

f(c+ h) < f(c).

This shows f(c) is not a relative minimum value of f . Since f(c) is a relative extremevalue of f , we have a contradiction.

Similarly, if f �(c) < 0 then we would reach the same contradiction. Hence, we concludef �(c) = 0.

✷

Finding the Extrema on a Closed IntervalRecall, the extrema of a continuous function f defined on a closed interval [a, b] are guar-anteed to exist because of Theorem 3.1. As a consequence of Theorem 3.2, the followingguidelines can be used to find the extrema of f on [a, b].

Guidelines for Finding the Extrema of aContinuous Function f on a Closed Interval [a, b]

1. Find the critical numbers c of f in (a, b).

2. Evaluate f(c) at the critical numbers c in (a, b).

3. Evaluate f(a) and f(b).

4. The largest value in steps 2-3 is the maximum value of f on [a, b].

The smallest value in steps 2-3 is the minimum value of f on [a, b].


y�x3�3x�12

3 x

6

�12�14

y

Figure 8On [0, 3], the maximum,value of f is f(3) = 6,and the minimum valueis f(1) = −14.

Example 5 Finding the Extrema on a Closed Interval

Find the extrema of f(x) = x3 − 3x− 12 on [0, 3], as in Figure 8

Solution The derivative of f isf �(x) = 3x2 − 3.

The critical numbers of f are found as follows:

3x2 − 3 = 0

3x2 = 3

x2 = 1

x = ±1.

The only critical number in the open interval (0, 3) is x = 1, and

f(1) = 13 − 3(1)− 12 = −14.

The values of f at the endpoints of [0, 3] are

f(0) = −12 and f(3) = 33 − 3(3)− 12 = 6.

From the list belowx f(x)

1, critical number −140, endpoint −123, endpoint 6

we conclude the maximum value of f is 6, and the minimum value is −14.✷

Try This 5Find the extrema of the given function on [0, 3].

f(x) = x3 + 3x2 − 24x+ 18.

Example 6 Find the Extrema on [a, b]

Find the extrema of g(x) = x2 − x2/3 on the interval [−2, 1], see Figure 9.

y�x2�x2�3�2 1 x

y

Figure 9On [−2, 1], themaximum value of g isachieved at x = −2, andthe minimum value at thecritical numbers ±1/ 4

√27.

Solution The derivative of g is

g�(x) = 2x− 2

3 3√x

=6x4/3 − 2

3 3√x

.

Note, x = 0 is a critical number of g. If g�(x) = 0, then

6x4/3 − 2 = 0

x4/3 =13

x = ± 14√27

≈ ±0.44

Then x = ±1/ 4√27 are critical numbers on (−2, 1). Moreover, we find

g

�± 1

4√27

�=

�1

4√27

�2

−�

14√27

�2/3

=1

3√3− 1√

3= − 2

3√3.


In addition, at the endpoints of [−2, 1] we have

g(−2) = 4− 3√4 and g(1) = 0.

Now, identify the largest and smallest values of g in the table

x g(x)0, critical number 0±1/ 4

√27, critical numbers −2/(3

√3) ≈ −0.38

−2, endpoint 4− 3√4 ≈ 2.4

1, endpoint 0

.

Hence, the maximum value of g is

g(−2) = 4− 3√4

and the minimum value is

g(±1/ 4√27) = − 2

3√3= −2

√3

9.

✷

Try This 6

Find the extrema of g(x) = x− 3x1/3 on [0, 8].

y�2cos�x��sin�2x�

�Π Π� Π2Π6

5 Π6

x

3 32

� 3 32

y

Figure 10The maximum value ish (π/6) = 3

√3/2 and

the minimum value ish (5π/6) = −3

√3/2.

Example 7 Finding the Extrema on a Closed Interval

Find the extrema of h(x) = 2 cosx+ sin 2x on [−π,π], see Figure 10.

Solution The derivative of h is

h�(x) = −2 sinx+ 2 cos 2x

= −2 sinx+ 2(1− 2 sin2 x) Since cos 2x = 1− 2 sin2 x

= −2(2 sin2 x+ sinx− 1)

h�(x) = −2(2 sinx− 1)(sinx+ 1)

If h�(x) = 0 for x on the interval (−π,π), then

sinx =12

or sinx = −1

x =π6,5π6

or x = −π2.

Moreover, we find

h�π6

�= 2 cos

π6+ sin

π3=

√3 +

√32

=3√3

2,

h

�5π6

�= 2 cos

5π6

+ sin5π3

= −√3−

√32

= −3√3

2, and

h�−π2

�= 2 cos

�−π2

�+ sin (−π) = 0.

The values of h at the endpoints of [−π,π] are

h (−π) = 2 cos (−π) + sin (−2π) = −2 + 0 = −2

h (π) = 2 cosπ + sin 2π = −2 + 0 = −2.


The values of h at the critical numbers and endpoints are listed below:

x h(x)π/6, critical number 3

√3/2 ≈ 2.6

5π/6, critical number −3√3/2 ≈ −2.6

−π/2, critical number 0−π, endpoint −2π, endpoint −2

Identify the largest and smallest values of h in the table. Hence, the maximum value of h

is h(π/6) = 3√3

2 , and the minimum value is h(5π/6) = − 3√3

2 .

✷

Try This 7Find the extrema of f(x) = cos 2x+ 2 sinx on [0,π].

3.1 Check-It Out

1. Find the critical numbers of f(x) = 2√x− x

2. Find the critical numbers of g(t) = 2 sin t− t on the open interval (0, 2π).

3. Find the extrema of f(x) = x4 − 4x on [0, 2].

True or False. If false, explain or show an example that shows it is false.

1. The number f(1) is the minimum value of f(x) = x2 − 2x.

2. If f(c) is a relative extreme value of f , then f �(c) = 0.

3. If c is a critical number of a function f , then f(c) is a relative extreme value of f .

4. The critical numbers of f(x) = x3 − x are x = ±√3.

5. The critical number of f(x) = (x− a)(x− b) is x = (a+ b)/2.

6. The critical numbers of y = sin(2πx) on the open interval (0, 1) are x =14,34.

7. g(x) =1√x

has a critical number.

8. The extrema of f(x) = x3 − 12x on [0, 3] are 0 and −16.

9. The extrema of f(x) = 2 sinx− x on [0,π] are√3− π/3 and −π.

10. The critical numbers of f(t) =t2

t+ 1are t = 0,−2.

Exercises for Section 3.1

In Exercises 1-4, use the graph to find the extrema of the function in the indicated interval.

1. f(x) = −(x− 3)2 + 5, [1, 4] 2. f(x) = −(x− 3)2 + 5, (1, 4)

1 3 4 x1

5

y

1 3 4 x1

5

y

For No. 1 For No. 2


3. g(x) =

�−(x− 2)2 + 5 if x �= 2

3 if x = 2, [0, 3]

2 3 x1

345

y

2 3 4 x

1

2

3

4

y

For No. 3 For No. 4

4. g(x) =

�(x− 3)3 + 2 if x is in (2, 3) ∪ (3, 4)4 if x = 3

, [2, 4]

In Exercises 5-8, find the value of the derivative at each relative extremum.

5. f(x) = x3 − 12x 6. f(x) = 2x3 − 9x2 + 12x− 2

2�2 x

16

�16

y

1 2 x

23

y

For No. 5 For No. 6

7. f(x) = 2− 3x2/3 8. f(x) = 18x2/3 − x4/3

1�1 x

2

y

27�27 x

y

For No. 7 For No. 8

In Exercises 9-18, find the critical numbers of the function.

9. f(x) = x3 − 48x+ 2 10. f(x) = x3 + 3x2 − 72x+ 4

11. g(x) = x5/3 − x2/3 12. g(x) = x4/3(x− 3)

13. f(x) =x2

x+ 114. f(x) =

x2 + 81− x

15. g(x) = 4 cosx+ 2x− 1, 0 < x < 2π

16. g(x) = 3x− 6 sinx+ 2, 0 < x < 2π

17. f(x) = 2 sinx− cos 2x, 0 < x < 2π

18. f(x) =12sin 2x− sinx, 0 < x < 2π


In Exercises 19-34, find the extrema of the function in the indicated closed interval.

19. f(x) = 2x(x− 4), [1, 4] 20. f(x) = −2x(5 + x), [−3, 0]

21. f(x) = x3 − 27x+ 5, [−4, 4] 22. f(x) = 4x3 − 3x+ 2, [−1, 1]

23. g(x) = x3 + 6x2 − 15x+ 10, [−6, 2] 24. g(x) = x3 + x2 − x+ 1, [−1.5, 0.5]

25. g(x) = (x− 1)2(x+ 1)2, [−1/2, 2] 26. g(x) = (x+ 1)3(2x− 1), [−1, 1/4]

27. f(x) = (x− 2)√x, [0, 1] 28. f(x) = x

√1− x, [−1, 1]

29. f(x) = cosx, [0, 2π] 30. f(x) = 1− sin 2πx, [0, 1]

31. f(t) =t2 − 1t2 + 1

, [−1, 1] 32. h(t) =t

t2 + 4, [0, 3]

33. k(t) = 6 sin t− 3t, [0,π] 34. M(s) = 2 sin s+ cos 2s, [0,π/3]

Applications

35. The Path of Least Cost A plumbing project involves installing pvc pipes from Ato C to B, see figure below. Along the horizontal, the installation costs $4 per foot.Along the diagonal, the cost is $12 per foot due to extra labor. Find the minimumcost of the plumbing project.

C B

A

40 x

30

y

P 100 x

50

75

y

For No. 35 For No. 36

36. Staking Two Antennas Two antennas are 100 feet apart, and their heights fromthe ground are 50 feet and 75 feet. Suppose a cable is connected from point P tothe top of each antenna, see above figure. Where should P be located so that theleast amount of cable is used?

37. Generating a Right Circular Cylinder The perimeter of a rectangle is 6 feet. Ifthe rectangle is revolved about one of it sides, a right circular cylinder is generated.Find the maximum volume of the cylinder. See figure below.

A�0,1�B�x,0�

C�2,�1�2 x

y


38. Bring out a Calculator A line segment has endpoints A(0, 1) and B(x, 0) where0 ≤ x ≤ 2. A second line segment has endpoints B(x, 0) and C(2,−1). Find x iftwo times the length of the first segment plus the length of the second segment isthe minimum. See figure above.


3.2 The Mean Value Theorem

• Rolle’s Theorem • The Mean Value Theorem

Rolle’s TheoremMichel Rolle (1652-1719),

a French mathematician,

published Rolle’s Theorem

in 1691. His theorem plays

an important role in the proofs

of several calculus theorems.

The above-named theorem is a basic result that makes possible the application of deriva-tives to finding the extreme values of a differentiable function. Also, Rolle’s Theoremestablishes a connection between critical numbers and the functional values of a differen-tiable function, see Figure 1.

�c1, f �c1��

�c2, f �c2��

a b x

y

Figure 1Rolle’s Theorem: If f(a) = f(b),then f �(c) = 0 for some c in (a, b).

Theorem 3.3 Rolle’s Theorem

Let f be a function that is continuous on [a, b] and differentiable on (a, b). If f(a) = f(b),then there exists a number c in (a, b) satisfying f �(c) = 0.

Proof We consider three cases.Case 1 If f(x) = f(a) for all x in [a, b], then f is a constant function. Thus, f �(c) = 0for all c in (a, b).

Case 2 Suppose f(x0) > f(a) for some x0 in (a, b). Recall, the Extreme Value Theoremassures that f has a maximum value f(c) where c is some number c in [a, b]. Sincef(x0) > f(a) = f(b), we find c �= a, b. Then f(c) is a relative extreme value of f . ByTheorem 3.2, c is a critical number of f in (a, b). Since f is differentiable in (a, b), we havef �(c) = 0.

Case 3 Suppose f(x0) < f(a) for some x0 in (a, b). We consider the minimum value f(c)of f . Similarly, as in Case 2 we conclude that c lies in (a, b) and f �(c) = 0.

✷

Example 1 Illustrating Rolle’s Theorem

Show Rolle’s Theorem applies to the function on the indicated interval:

f(x) = x1/3 − x4/3, [0, 1]

Then find all the numbers c that satisfy Rolle’s Theorem.

Solution To show Rolle’s Theorem applies we have to verify the following: a) f iscontinuous on [0, 1], b) f is differentiable on (0, 1), and c) f(0) = f(1). We establish thevalidity of statements a), b), and c) as follows.

3.2. THE MEAN VALUE THEOREM 149

14 1

x

0.47

y

Figure 2The values of f at theendpoints of [0, 1] aref(0) = f(1) = 0. Rolle’sTheorem guarantees theexistence of c = 1/4 in theinterval (0, 1) such thatf �(c) = 0.

a) The radical function y = 3√x and the composite function y = ( 3

√x)4 are continuous

everywhere, see page 44 and Theorem 1.10. Then the difference function f(x) =x1/3 − x4/3 is continuous on [0, 1].

b) Applying the power rule, we find

f �(x) =13x−2/3 − 4

3x1/3

=1

3x2/3(1− 4x) (1)

Since f �(x) is undefined only when x = 0, f is differentiable on (0, 1).

c) Clearly, f(1) = 11/3 − 14/3 = 0 and f(0) = 0.

Thus, Rolle’s Theorem applies. Using (1), we find that f �(c) = 0 implies

1− 4c = 0.

Hence, c = 1/4 is the only number in (0, 1) that satisfies Rolle’s Theorem, see Figure 2.

✷

Try This 1Find all numbers c satisfying Rolle’s Theorem.

a) f(x) = 9x− x3, [−3, 0] b) f(x) = sin 2x, [π/4, 5π/4]

Many functions satisfy the hypothesis of Rolle’s Theorem. These include polynomial,rational, and trigonometric functions provided these functions are defined on the interval[a, b] in Rolle’s Theorem. Recall, polynomial, rational, and trigonometric functions aredifferentiable in their domains of definition, see Sections 1.3

Example 2 Applying Rolle’s Theorem

Apply Rolle’s Theorem and the Intermediate Value Theorem to show that the graph of

f(x) = x5 + 2x+ 2

has exactly one x-intercept.

1�1 x

2

y

Figure 3

A sketch of

the graph of

f(x) = x5+ 2x+ 2.

Solution We find two functional values of f with opposite signs:

f(1) = 15 + 2(1) + 2 = 5 and f(−1) = (−1)5 + 2(−1) + 2 = −1

By the Intermediate Value Theorem (see page 45), we can find a number x0 in (−1, 1)such that f(x0) = 0. Then (x0, 0) is an x-intercept of the graph of f .

Suppose (x1, 0) is another x-intercept of the graph f and x1 �= x0. Then f(x1) =f(x0) = 0. Applying Rolle’s Theorem, there exists a real number c between x1 and x0

such that f �(c) = 0. But this is impossible for the derivative has no zero, i.e.,

f �(x) = 5x4 + 2 �= 0 for all x.

Hence, the graph of f has exactly one x-intercept, namely, (x0, 0).✷

Try This 2Show the graph of

f(x) = 4− 9x− x3

has exactly one x-intercept using Rolle’s Theorem and the Intermediate Value Theorem.


The Mean Value Theorem was

first proved by Joseph-Louis

Lagrange (1736-1813). At the

young age of 19, Lagrange be-

came a professor in Turin.

�a, f �a��b, f �b��

a bc x

y

Figure 4

Mean Value Theorem:

The line joining

(a, f(a)) to (b, f(b))

is parallel to a

tangent line at some

point (c, f(c)) where

a < c < b.

The Mean Value TheoremThe following theorem is a generalization of Rolle’s Theorem. The Mean Value Theoremcan be applied to classify ‘increasing’ or ‘decreasing’ functions in terms of the derivative,as we will see in Section 4.3.

Theorem 3.4 The Mean Value Theorem

Let f be a function that is continuous on [a, b] and differentiable on (a, b). Then thereexists a number c in (a, b) satisfying

f �(c) =f(b)− f(a)

b− a.

Proof The slope of the secant line joining A(a, f(a)) to B(b, f(b)) is

m =f(b)− f(a)

b− a.

An equation of the secant line is

s(x) = m(x− a) + f(a).

Let g(x) be the difference f(x) and s(x) as defined below:

g(x) = f(x)− (m(x− a) + f(a)) .

Then g(a) = 0 and g(b) = 0 because of the definition of m. Observe,

g�(x) =ddx

[f(x)]− ddx

[m(x− a) + f(a)]

= f �(x)−m.

Applying Rolle’s Theorem, there exists a number c in (a, b) satisfying

g�(c) = 0

f �(c)−m = 0.

Hence, we obtain

f �(c) = m =f(b)− f(a)

b− a.

This completes the proof of the Mean value Theorem. ✷

In other words, the Mean Value Theorem implies that the average rate of change of afunction f over [a, b] is equal to the rate of change of f at some number c in (a, b).Geometrically, the line joining the end points (a, f(a)) and (b, f(b)) is parallel to a tangentline at some point (c, f(c)) where a < c < b, see Figure 4.

Example 3 Illustrating the Mean Value Theorem

Find the values of c that satisfy the Mean Value Theorem for the given function in theindicated interval.

f(x) = x3 − 2x2 − 3x+ 4, [−3, 5]


�5,64�

��3,�32�� 53 3�3 5 x

y

Figure 5The tangent lines atthe points where x = −5/3and x = 3 are parallelto the secant line thatcontains the endpoints(−3,−32) and (5, 64).

Solution If c satisfies the Mean Value Theorem, then

f �(c) =f(5)− f(−3)

5 + 3

=64− (−32)

8

=968

f �(c) = 12.

Next, solve the equation f �(x) = 12.

3x2 − 4x− 3 = 12

3x2 − 4x− 15 = 0

(3x+ 5)(x− 3) = 0

x = −53, 3

Note, c must lie in the open interval (−3, 5). Thus, the values of c satisfying the MeanValue Theorem are c = 3 and c = −5/3, as shown in Figure 5.

✷

Try This 3Find the values of c that satisfy the Mean Value Theorem for the function

h(x) = 2√x+ x

in the interval [0, 1].

Note, the values of c in the Mean Value Theorem belong to the open interval (a, b). Thetheorem does not indicate how many such c’s there are. The theorem simply states theexistence of such a number.

Example 4 An Application of the Mean Value Theorem

A police officer clocks the speed of a certain car at 65 mph. Five minutes later anotherpolice officer finds the same car going at 60 mph. If the police officers are eight milesapart, explain why the car was speeding at 96 mph at some point between the two policeofficers.

Solution Let t be a fraction of an hour after the first officer clocked the car’s speed.Denote by s(t) the corresponding distance in miles between the first officer and the car attime t.

Note, five minutes is equivalent to 1/12 of an hour. Then

s(0) = 0 and s

�112

�= 8.

Suppose s(t) is a differentiable function of t. Then the Mean Value Theorem applies tos(t) on [0, 1

12 ]. Thus, there exists a number c in (0, 112 ) such that

s�(c) =s(1/12)− s(0)

1/12− 0=

8− 01/12

= 96 mph.

Hence, at some point between the two police officers the car’s speed was exactly 96 mph.✷


Try This 4Johnny made a 300-mile trip in 4 hours. Show that at some point in the trip Johnny wasdriving at 75 mph.

3.2 Check-It Out

1. Find the values of c that satisfy Rolle’s Theorem for f(x) = x3 − 3x+ 2 in [−1, 2].

2. Find the values of c that satisfy the Mean Value Theorem for f(x) =√x in [0, 4].

3. Use Rolle’s Theorem to explain why f(x) = cosx+2x− 1 has only one x-intercept.


1. For the function f(x) = x3−6x2+9x−2 in [0, 3], the value of c that satisfies Rolle’sTheorem is c = 1.

2. Rolle’s Theorem applies to f(x) = 3(x− 1)2/3 in the interval [0, 2].

3. For the function h(t) =√t− 4t in [1, 4], the value of c that satisfies the Mean Value

Theorem is c = 3/2.

4. If y = f(x) is a differentiable function and f(a) = f(b), then there is exactly one cin (a, b) satisfying f �(c) = 0.

5. Suppose a differentiable function y = f(x) has x-intercepts (a, 0) and (b, 0), a �= b.Then y = f(x) has a critical number c in (a, b).

6. A police officer spotted a car traveling at 55 mph. After 3 minutes, a state trooperclocked the same car at 60 mph. If the officer and trooper are 5 miles apart, thenthe car’s speed at some point in between was exactly 100 mph.

7. The function f(x) = sin(2πx) + 2x2 − x satisfies Rolle’s Theorem in the interval[0, 1/2].

8. For f(x) = 1/x2 in [−2, 1], there exists a number c in (−2, 1) such that

f �(c) =f(1)− f(−2)

1− (−2).

9. For f(x) = 1/x in [−1, 1], there exists a number c in (−1, 1) such that

f �(c) =f(1)− f(−1)

1− (−1).

10. If a car can accelerate from 60 mph to 70 mph in 1 minute, then the car’s accelerationat some instant is exactly 600 mph per hour.


In Exercises 1-8, determine if Rolle’s theorem applies to the function in the indicatedinterval. If it does, find the values of c that satisfy Rolle’s Theorem.

1. f(x) = 2 + 6x− x2, [0, 6] 2. f(x) = 3x2 − x− 2, [0, 1/3]

3. s(t) = t3 − t2 + 4, [0, 1] 4. g(t) = t3 − 6t2 + 11t− 2, [1, 2]

5. T (θ) = 2 sin θ − 1, [π/6, 5π/6] 6. T (θ) = tan θ + cot θ, [π/6,π/3]

7. p(w) = sinw2, [−2, 2] 8. h(α) = cos[(α2 − 4α+ 5)π], [1, 3]


In Exercises 9-16, find the values of c that satisfy the Mean Value Theorem.

9. g(t) = t3 − t2 + t− 1, [0, 1] 10. f(x) = 2x2 − x3 − 3x+ 1, [−3, 0]

11. R(s) = s2 − s4 + 3s, [−1, 1] 12. C(w) = w(w − 3)2 + w, [0, 2]

13. f(x) =x+ 1x− 1

, [2, 3] 14. f(x) =x− 2x+ 2

, [0, 3]

15. A(h) = h−√h, [0, 4] 16. p(t) = 2t+

√t, [0, 1]

Applications of the Mean Value Theorem

17. Let y = f(x) be a function such that f �(a) �= 1 for any a. Prove that the equationf(w) = w has at most one solution w.

18. Let y = g(t) be a function such that for some nonzero constant k we have g�(t) �= kfor any t. Prove g(x) = kx has at most one solution x.

19. Arithmetic Mean Verify that the value of c that satisfies the Mean Value Theoremfor f(x) = x2 on the interval [a, b] is c = (a+ b)/2.

20. Geometric Mean Let a, b > 0 be positive numbers. Show that the value of c thatsatisfies the Mean Value Theorem for r(x) = 1/x on the interval [a, b] is c =

√ab.

21. Let f be a differentiable function such that f(1) = 4 and 0 ≤ f �(x) ≤ 2 for all x.Find a maximum possible value for f(6).

22. Let p be a differentiable function such that p(10) = 2 and p�(x) ≥ 3 for all x. Finda minimum possible value for p(12).

23. If a �= b, show that ��sin b− sin a

b− a

�� ≤ 1.

24. If x �= y, show that ��tanx− tan y

x− y

�� ≥ 1.

25. Let y = f(x) be differentiable function that satisfies f(1) = 1 and f(2) = 2. Showthat there exists a number c in the open interval (1, 2) such that the tangent line tothe graph of f at the point (c, f(c)) passes through the origin.

26. Show that y = x3 − Ax2 + 1 has three distinct zeros if A >3√3

2, and has exactly

one zero if A <3√3

2.

27. Driving in an interstate Jimmie went from one exit to another exit in 80 seconds.If the exits are 2 miles apart, explain why Jimmie’s speed was exactly 90 mph atsome point between the two exits.

28. Let y = f(x) be continuous on [a, b] and differentiable on (a, b). If f �(x) = 0 for allx in (a, b), then f(s) = f(t) for all s, t in [a, b].


3.3 Increasing and Decreasing Functions

• Increasing and Decreasing Functions • The First Derivative Test

a b x

y

Figure 1aThe function f isincreasing on theopen interval (a, b).

cb x

y

Figure 1bThe function f isdecreasing on (b, c).

Increasing and Decreasing FunctionsIn this section, we will use the derivative to identify the relative extrema of a differentiablefunction. First, we need a few preliminaries.

Definition 4 Increasing, Decreasing, and Constant Functions

Let f be a function on an interval I.

1. f is increasing on I if

f(x1) < f(x2) whenever x1 < x2 and x1, x2 belong to I.

2. f is decreasing on I if

f(x1) > f(x2) whenever x1 < x2 and x1, x2 belong to I.

3. f is constant on I if f(x1) = f(x2) for all x1, x2 in I.

Geometrically, a function f is increasing on an interval I if its graph is rising as xmoves to the right in I. In Figure 1a, f is increasing on an open interval (a, b). Likewise,f is decreasing on I if its graph is falling as x moves to the right in I. In Figure 1b, f isdecreasing on (b, c).

The derivative can tell us if a differentiable function is increasing or decreasing. InFigure 1c, we see f is increasing on (a, b), the tangent lines have positive slopes for x in(a, b), and consequently f �(x) > 0. Similarly, f is decreasing on (b, c), the tangent lineshave negative slopes, and f �(x) < 0 for x in (b, c). The next theorem summarizes theseresults.

a cb d x

y

Figure 1cA function is increasing, decreasing, or constantaccording to whether its derivative is positive,negative, or zero, respectively.

Theorem 3.5 Increasing and Decreasing Test

Suppose f is a continuous function on [a, b], and differentiable on (a, b).

1. If f �(x) > 0 for all x in (a, b), then f is increasing on [a, b].

2. If f �(x) < 0 for all x in (a, b), then f is decreasing on [a, b].

3 If f �(x) = 0 for all x in (a, b), then f is constant on [a, b].

3.3. INCREASING AND DECREASING FUNCTIONS 155

Proof Suppose f �(x) > 0 for all x in (a, b). If a ≤ x1 < x2 ≤ b, then by the Mean ValueTheorem there exists a number c such that x1 < c < x2 and

f �(c) =f(x2)− f(x1)

x2 − x1.

Since f �(c) and x2 − x1 are positive, f(x2)− f(x1) is also positive. Thus,

f(x2) > f(x1)

and consequently f is increasing on [a, b]. The remaining two cases are proved similarly,see Exercises 59 and 61 at the end of the section.

✷

The above theorem implies that we have to solve inequalities such as f �(x) > orf �(x) < 0 to determine where a function is increasing or decreasing. In the next example,we use a standard method to solve such inequalities. The method involves finding thecritical numbers and determining the signs of f �(x) off the critical numbers. This methodis described in more details after Try This 1.

Example 1 Identify Where f is Increasing or Decreasing

Find the open intervals where

f(x) = 4x3 − 11x2 + 6x+ 15

is increasing or decreasing.

Solution Since the derivative is given by

f �(x) = 12x2 − 22x+ 6 = 2(3x− 1)(2x− 3)

the critical numbers are x = 13 ,

32 .

�3�2, 51�4��1�3, 430�27�

13

32

x

y

Figure 2f is increasing on(−∞, 1

3 ) and ( 32 ,∞),and decreasing on ( 13 ,

32 ).

If we delete the critical numbers from the number line R, we obtain a union of openintervals:

R−�13,32

�=

�−∞,

13

�∪�13,32

�∪�32,∞

�.

Choose a number in I1 =�−∞, 1

3

�, say x1 = −1. Determine the sign of f �(x1); in fact

f �(−1) > 0. Then f �(x) > 0 for all x in I1.1 Thus, f is increasing on I1 by Theorem 3.5.

Repeat this procedure and choose test values x2 and x3 from

I2 =

�13,32

�and I3 =

�32,∞

�.

Then determine the sign of f �(xi) which is the sign of f � on the test open interval Ii. Thefollowing table summarizes the results.

Test Open Interval�−∞, 1

3

� �13 ,

32

� �32 ,∞

�

Test Value xi x1 = −1 x2 = 1 x3 = 2Sign of f �(xi) f �(−1) > 0 f �(1) < 0 f �(2) > 0Apply Theorem 3.5 Increasing Decreasing Increasing

Hence, f is increasing on (−∞, 13 ), increasing on ( 32 ,∞), and decreasing on ( 13 ,

32 ). The

graph of f is shown in Figure 2.

✷

1The sign of f �

(−1) will be the same as the sign of f �(x) for any other test value x in I1. For

if f �(x1) and f �

(x2) have opposite signs with x1 and x2 in I1, then by the Intermediuate Value

Theorem there exist a number x0 in I1 for which f �(x0) = 0; but this is a contradiction since I1

does not contain any critical number of f .


Try This 1Find the open intervals on which the function is increasing or decreasing.

a) s(t) = −16t2 + 96t+ 10 b) p(x) = x3 − 6x2 + 4

The guidelines below are for students to help them develop a strategy for finding openintervals on which a function is increasing or decreasing.

Guidelines for finding open intervalswhere a function is increasing or decreasing

Let f be a differentiable function on an open interval (a, b).

1. Find the critical numbers xi in (a, b) and arrange in ascending order:

a < x1 < x2 < · · · < xn < b.

2. Form the open intervals I1 = (a, x1), I2 = (x1, x2), ..., In+1 = (xn, b)

3. Select any test value xi in Ii.

4. Apply Theorem 3.5:

• If f �(xi) > 0 then f is increasing on Ii.

• If f �(xi) < 0 then f is decreasing on Ii.

In the above, we allow for a = −∞ or b = ∞.

The First Derivative TestThe sign of the slope of a tangent line is an indicator of whether a function is locallyincreasing or decreasing. This observation is the basis of the First Derivative Test.However, we need to introduce some terminologies. For our purposes, think of the functiong in the next theorem as the first or second derivative of a function f .

�3�2, 51�4��1�3, 430�27�

13

32

x

y

Figure 3The derivative changessign at a local extrema.

Definition 5 A Function Changing its Sign at a Number

Let g be a function, and let c be a number.

1. g changes from positive to negative at c if there are open intervals (b, c) and (c, d)such that g is positive on (b, c) and g is negative on (c, d).

2. g changes from negative to positive at c if there are open intervals (b, c) and (c, d)such that g is negative on (b, c) and g is positive on (c, d).

3. The sign of g stays constant about c if there are open intervals (b, c) and (c, d)such that either g is negative on (b, c) and (c, d), or g is positive on (b, c) and (c, d).

For instance, see Figure 3, the sign of the derivative f � changes from positive to negativeat x = 1

3 . Also, f � changes its sign from negative to positive at x = 32 . Moreover, we see

that f has relative maxima f( 13 ) and relative minima f( 32 ). The First Derivative Teststates that sign changes of f � are indicators of the local extrema of a function.


Theorem 3.6 The First Derivative Test

Let f be a differentiable function, and let c be a critical number of f .

1. If f � changes from positive to negative at c, then f(c) is a relative maximum of f .

2. If f � changes from negative to positive at c, then f(c) is a relative minimum of f .

3. If the sign of f � stays constant about c, then f(c) is not a relative extremum of f .

Proof Suppose f � changes from positive to negative at c. Choose b and d such that

f �(x) > 0 for all x in (b, c)

andf �(x) < 0 for all x in (c, d).

Then f is increasing on (b, c) and f is decreasing on (c, d) by Theorem 3.5. Since f iscontinuous at c, f is increasing on (b, c] and f is decreasing on [c, d). Thus, f(c) is themaximum value of f on the (b, d). Hence, f(c) is a relative maximum value of f .

The proof of the second part is similar. We leave the proof as an exercise for thestudent to prove, see Exercise 62.

If the sign of f � stays constant at c, then there exist numbers c1, c2 such that either f �

is positive on (c1, c) and (c, c2), or f� is negative on (c1, c) and (c, c2). Applying Theorem

3.5 and the continuity of f at c, we obtain that f is either increasing or decreasing on(c1, c2). Hence, f(c) is neither a relative maximum nor relative minimum value of f .

✷

Example 2 Applying the First Derivative Test

Find the relative extrema of g(x) = x1/3 − x2/3.

�1�8,1�4�Rel. max.

1 x

1

y

Figure 4g has a relative maxima atx = 1

8 since g� changes frompositive to negative at x = 1

8 .

Solution The derivative of g is

g�(x) =13x−2/3 − 2

3x−1/3 =

1− 2x1/3

3x2/3.

Then x = 0 is a critical number of g since g�(0) is undefined and g(0) is defined. To findanother critical number, suppose g�(x) = 0. Then

1− 2x1/3 = 0

x =

�12

�3

=18.

Thus, the critical numbers are x = 0, 18 . Consider the open intervals

I1 = (−∞, 0) , I2 =

�0,

18

�, I3 =

�18,∞

�

following the guidelines after Example 1. Find test values xi in Ii, and evaluate g�(xi) asfollows:

Test Open Interval I1 = (−∞, 0) I2 =�0, 1

8

�I3 =

�18 ,∞

�

Test Value xi x1 = −1 x2 = 1/9 x3 = 1Sign of g�(xi) g�(−1) > 0 g�

�19

�> 0 g�(1) < 0

Positive Positive Negative

Note, g� changes from positive to negative at 18 . Applying the First Derivative Test, g has

a relative maximum value at x = 18 . Since the sign of g� stays constant at x = 0, g does

not have a relative extreme value at x = 0. Hence, the relative maximum value of g is

g

�18

�=

�18

�1/3

−�18

�2/3

=12− 1

4=

14.

The graph of g is shown in Figure 4. ✷


Try This 2

Find the relative extrema of g(t) =14t4/3 − 1

5t5/3.

Example 3 Using the First Derivative Test

Find the relative extrema ofh(x) =

√3 sinx+ cosx

in the open interval (0, 2π), see Figure 5.

Π3

4 Π3

x

1

�2

y

Figure 5The relative extremaof h are y = ±2.

Solution The derivative is

h�(x) =√3 cosx− sinx.

To find the critical numbers, we find the zeros of the derivative.√3 cosx− sinx = 0√

3 cosx = sinx√3 = tanx

x =π3,4π3.

Then we find test values xi in the open intervals Ii where

I1 =�0,

π3

�, I2 =

�π3,4π3

�, I3 =

�4π3, 2π

�.

following the guidelines after Example 1.

Test Open Interval�0,

π3

� �π3,4π3

� �4π3, 2π

�

Test Value xi x1 =π6

x2 =π2

x3 =3π2

Sign of h�(xi) h� (x1) = 1 h� (x2) = −1 h� (x3) = 1Positive Negative Positive

Hence, by the First Derivative Test, the relative maximum value of h is

h�π3

�=

√3 sin

π3+ cos

π3= 2

and the relative minimum value of h is

h

�4π3

�=

√3 sin

4π3

+ cos4π3

= −2.

✷

Try This 3Find the relative extrema of k(x) = 2 cosx− sin 2x where 0 < x < 2π.


Example 4 Using the First Derivative Test

Find the relative extrema of m(x) = 23x

3 + 32x .

�2,64�3��2,�64�3� 2�2 x

20

y

Figure 6Relative maxima andrelative minima

Solution Evaluate m�(x) as follows:

m�(x) = 2x2 − 32x2

=2(x4 − 16)

x2

=2(x− 2)(x+ 2)(x2 + 4)

x2Factoring

The critical numbers of m are x = ±2. Note, x = 0 is not a critical number for m(0) isundefined. Form the open intervals using the critical numbers:

I1 = (−∞,−2) , I2 = (−2, 0) , I3 = (0, 2) , I4 = (2,∞).

Select certain test values xi in Ii and evaluate m�(xi).

Test Open Interval (−∞,−2) (−2, 0) (0, 2) (2,∞)Test Value xi x1 = −3 x2 = −1 x3 = 1 x4 = 3Value of m�(xi) m� (x1) > 0 m� (x2) < 0 m� (x3) < 0 h� (x4) > 0

Positive Negative Negative Positive

We apply the First Derivative Test. Since m� changes from positive to negative atx = −2, the relative maximum value of m is

m (−2) =23(−2)3 +

32−2

= −163

− 16 = −643

Similarly, m� changes from negative to positive at x = 2. Hence, the relative minimumvalue of m is

m (2) =163

+ 16 =643.

The graph of m is shown in Figure 6. ✷

Try This 4Find the relative extrema of R(x) = x2 + 25

x2+4.

L1

L24

4

Α

Α

Figure 7L1 and L2 are thehypotenuse of righttriangles with angle α.

Example 5 The Longest Rod Through a Corner

Find the length of the longest rod that can be carried horizontally from one hall onto theother hall, as in Figure 7. Suppose the halls are 4 feet wide and they meet at a right angle.

Solution The length of the longest rod that can be carried horizontally through the hallsis the minimum value of L1 + L2, as seen Figure 7. Using right triangle trigonometry, weobtain

L = L1 + L2 = 4 secα+ 4 cscα, 0 < α <π2.

The derivative of L with respect to α is

L�(α) = 4 secα tanα− 4 cscα cotα

= 4

�sinαcos2 α

− cosα

sin2 α

�

= 4

�sin3 α− cos3 α

sin2 α cos2 α

�.


L1

L23 3 1

Α

Α

Figure 8Find the minimum sum ofL1 + L2 given the widthsof 1 yd and 3

√3 yd.

Then the critical numbers of L in (0, π2 ) must satisfy

sin3 α− cos3 α = 0 or tanα = 1.

Thus, the critical number is α = π4 . Consider the open intervals

I1 =�0,

π4

�, I2 =

�π4,π2

�.

The sign of L�(α) in the above open intervals are described below.

Test Open Interval I1 =�0, π

4

�I2 =

�π4 ,

π2

�

Test Value αi α1 = π6 α2 = π

3

Sign of L�(αi) L� �π6

�≈ −11.2 L� �π

3

�≈ 11.2

Negative Positive

Note, L is decreasing on (0,π/4) and L is increasing on (π/4,π/2) byTheorem 3.5. Thus, the minimum value of L is

L�π4

�= 4 sec

�π4

�+ 4 csc

�π4

�= 8

√2 ≈ 11.3 feet.

Hence, the length of the longest possible rod is 8√2 feet.

✷

Try This 5As in Example 5 but suppose the widths of the halls are 1 yard and 3

√3 yards, see Figure

8. Find the length of the longest rod that can be carried horizontally from one hall ontothe other hall.

3.3 Check-It Out

1. Find the open intervals on which f(x) = 12x− x3 + 10 is increasing or decreasing.

2. Find the relative extrema of s(t) = 2t3 +3t2 − 12t+7 by using the First DerivativeTest.

3. Find the relative extrema of y = sin(x) + cos(x) where 0 < x < 2π.

4. Find the maximum product of two positive numbers x and y where x+ y = 1.


1. If f �(x) > 0 for x in (−10, 10), then f is increasing on (−10, 10).

2. If f �(x) < 0 for x in (−1, 1), then f is decreasing on [−1, 1].

3. If f(x) = (x− 2)5, then the sign of f �(x) stays constant about 2.

4. If y = f(x) is continuous on (0, 6) and the sign of f �(x) changes from negative topositive at 4, then f(4) is a relative minimum of f .

5. If f �(x) > 0 for x in (−∞, 0) and f �(x) < 0 for x in (0,∞), then f(0) is a relativemaximum of f .

6. If f �(x) = (x− 1)2(x+ 1)3, then f is increasing on (−1,∞).

7. If f �(x) = −3(x− 2)5, then f �(x) > 0 for x in (2,∞).

8. If g(t) =√t(t− 1), then g(1/3) is a relative minimum of g.

9. The function s(t) = 2 cos(t) + t− 15 has a relative maximum at t = π/6.

10. The minimum sum x+ y of two positive numbers x and y for which xy = 1 is two.


Exercises for Section 3.3In Exercises 1-20, determine the open intervals on which the function is increasing ordecreasing.

1. f(x) = 5(4− x)(x+ 2) 2. g(x) = −3x(x+ 2)

3. y = 3x(x− 4)− x(4− x) 4. y = (x− 1)(x+ 5)− 10(x+ 5)

5. p(x) = 4x3 − x2 − 2x− 5 6. r(x) = 8x3 − 3x2 − 9x+ 4

7. S(t) = −4t3 − t2 + 2t− 6 8. g(t) = −4t3 + 11t2 − 6t+ 7

9. y = t3 + 6t2 + 12t− 3 10. y = 6t2 − 3t3 − 4t+ 2

11. R(x) =x+ 12x− 3

12. P (x) =3x− 1x+ 4

13. m(x) =x2 + 1x2 − 9

14. N(x) =1− x2

x2 − 4

15. s(θ) = sin2(θ), 0 < θ < 2π 16. t(θ) = cos2(θ), 0 < θ < 2π

17. y = sin(x)− cos(x), 0 < x < 2π 18. y = cos(x) + sin(x), 0 < x < 2π

19. f(x) = 2 sin(x)− cos(2x), 0 < x < 2π 20. g(x) = cos(x)− sin(2x)2

, 0 < x < 2π

In Exercises 21-46, determine all the relative extreme values of the function. Apply theFirst Derivative Test.

21. f(x) = 4x(x− 4) 22. g(x) = 3(x− 1)(x+ 3)

23. h(x) = x3 + 3x2 − 9x+ 15 24. y(t) = (t+ 1)2(t− 3)

25. y(t) = (3t− 1)2(t+ 1) 26. y = 27x− 4x3 − 7

27. f(x) = 3x4 − 14x3 + 9x2 + 2 28. M(x) =x2 + 1x2 − 4

29. L(x) =x2 − 9x2 + 1

30. N(x) =x

x3 + 4

31. k(x) =x

8− x332. f(x) = x1/3 + x−1/3

33. f(x) =√x(2− x)2 34. f(x) = x4 − 18x2

35. f(x) =

�2 + x if x ≤ 4

22− x2 if x > 436. f(x) =

�1− 2x if x ≤ −2x2 + 1 if x > −2

37. f(x) = (x+ 1)2/3 − 2x 38. f(x) =13x3 +

16x

39. f(x) = x2/3(3− x)1/3 40. f(x) =√x(x− 1)3

41. g(t) = 4 + sin2(t), 0 < t < 2π 42. g(t) = cos2(t)− 3, −π < t < π

43. v(θ) = sin(θ) +√3 cos(θ), 0 < θ < 2π 44. A(w) = 2 sin(w) + w − π

3, −π < w < π

45. f(α) = cos2(α)− sin(α), 0 < α < 2π 46. g(β) = sin2(β) + cos(β), 0 < β < 2π

Applications

47. The difference between two numbers is one. Find the minimum product of two suchnumbers.

48. What is the minimum sum of two positive numbers whose product is one?

49. Find the maximum area of a rectangle that has a perimeter of 12 feet. What if theperimeter is p feet?


50. Find the minimum perimeter of a rectangle that has an area of 16 square inches.What if the area is A square inches?

51. Let f(x) be the square of the distance between the point (0, 1) and a point (x, 4−x2)on the parabola y = 4− x2.

a) Find open intervals on which y = f(x) is increasing or decreasing.

b) Which points on the parabola are closest to (0, 1)?

4x x

6

y

y

Figure for No.53

�1,2�x x

y

y

Figure for No.54

52. Let D(x) be the square of the distance between the point (1, 0) and a point (x,√x)

on the curve y =√x.

a) Find open intervals on which y = D(x) is increasing or decreasing.

b) Which point on the graph of y =√x is nearest to (1, 0)?

53. A rectangle is bounded by the x-and y-axis and the line 3x + 2y = 12, see figurebelow. Find the dimensions of the rectangle that has the maximum area.

54. A right triangle is bounded by the x- and y-axis and a line that passes through thepoint (1, 2). Find the dimensions of such a triangle that has the minimum area.

Theory and Proofs

55. Find and sketch the graph of a function y = p(x) that satisfies

a) p(2) = 1, p�(2) = 0, p�(x) > 0 if x > 2, and p�(x) < 0 when x < 2

b) p(1) = −3, p�(1) = 0, p�(x) > 0 if x �= 1

56. Sketch the graph of a differentiable function m = v(t) that satisfies

a) v(2) = 5, v(−2) = −5, v�(−2) = v�(2) = 0, v�(t) > 0 if |t| < 2,

and v�(t) < 0 when |t| > 2

b) v(1) = v(−1) = 1, v�(1) = v�(−1) = 0, v�(t) > 0 if t < −1, v�(t) < 0

if −1 < t < 0, v�(t) > 0 if 0 < t < 1, and v�(t) < 0 if t > 1

57. Let f(x) = x(x− a)(x− b). Prove y = f(x) is decreasing on the open

interval�a+b−c

3 , a+b+c3

�where c =

√a2 − ab+ b2

58. Let p(x) = (x− a)2(x− b)2, a < b. Prove y = p(x) is increasing

on�a, a+b

2

�and (b,∞).

59. Let f be continuous on [a, b] and differentiable on (a, b). Suppose f �(x) < 0 for allx in (a, b). Prove f is decreasing on [a, b]. Hint: Mean Value Theorem.

60. Suppose f is continuous on [a, b]. If f �(x) > 0 for x in (a, b), prove f is increasingon [a, b].

61. Let f be continuous on [a, b] and differentiable on (a, b). If f �(x) = 0 for all x in(a, b), prove f is a constant function on [a, b].

62. Let c be a critical number of a differentiable function f . If f � changes from negativeto positive at c, prove f(c) is a local minimum of f .

Odd Ball Problems

63. Find a 3rd degree polynomial that has a relative maximum point at (−1, 2) and arelative minimum point at (1, 1).

64. Find a 3rd degree polynomial that has a relative maximum point at (4, 5) and arelative minimum point at (1, 3).

65. If 0 ≤ x <π2, prove tan2

�x2

�≤ tan2(x).

66. If 0 ≤ x ≤ π2, prove 3

√cosx ≤ cos

�x3

�.

3.4. CONCAVITY OF THE GRAPHS OF FUNCTIONS 163

3.4 Concavity of the Graphs of Functions

• Concavity • Points of Inflection • The Second Derivative Test

ConcavityIn Section 3.3, we discussed how the sign of the derivative f � determines whether f isincreasing or decreasing. Also, we studied the First Derivative Test which is a criterion forfinding the relative extrema of f . In this section, we will see how the sign of the secondderivative f �� determines whether the graph of f is curving upward or curving downward.In addition, we discuss the Second Derivative Test which is another criterion for findingthe relative extrema of f .

Definition 6 Concavity of a Graph

Let f be a differentiable function on an open interval I. The graph of f is concaveupward if the derivative f � is an increasing function on I. Likewise, the graph of f isconcave downward if f � is decreasing on I.

For simplicity, we say f is concave upward or downward on an interval I if the graphof f is concave or downward on I, respectively.

Consider the graphs of f and p in Figures 1a and 2a, respectively. Note, f � and p� areincreasing since the slopes of the tangent lines to the graphs of f and p are increasing. Bydefinition, f and p are concave upward. Similarly, we see the graphs of g and q in Figures1b and 2b. Since g� and q� are decreasing functions, g and q are concave downward.

The graph of f isconcave upward

x

yThe graph of g is

concave downward

x

y

Figure 1a. f � is increasing. Figure 1b. g� is decreasing.

The graph of p is

concave upward

x

y

The graph of q is

concave downward

x

y

Figure 2a. p� is increasing. Figure 2b. q� is decreasing.

Moreover, we claim f is concave upward if and only if the graph of f lies above all itstangent lines. Also, f is concave downward if and only if the graph of f lies below all itstangent lines. For a proof, see Theorem A.6, page 413.


concavedownward

concaveupward

f �x� � x3�3x�3

x

y

f '' is positive

f '' is negative

f ''�x��6x

x

y

Figure 3The concavity of thegraph of f is determinedby the signs of f ��.

The next theorem is a straightforward way to determine concavity. The proof followsdirectly from Definition 6 and Theorem 3.5 in Section 3.3.

Theorem 3.7 Concavity Test

Let f be twice differentiable on an open interval I.

1. If f ��(x) > 0 for all x in I, then the graph of f is concave upward on I.

2. If f ��(x) < 0 for all x in I, the graph of f is concave downward on I.

In Figure 3, we see the graphs of f(x) = x3 − 2x+3 and f ��(x) = 6x. Since f ��(x) > 0on (0,∞), f is concave upward on (0,∞) by Theorem 3.7. Likewise, since f ��(x) < 0 on(−∞, 0), f is concave downward on (−∞, 0).

Example 1 Determining Concavity

Find the open intervals where the graph of

f(x) = x4 − 2x3 + 2

is concave upward or concave downward.

1 x

3

y

Figure 4A sketch of the graph of

f(x) = x4 − 2x3 + 2

Solution We obtain the derivatives as follows:

f �(x) = 4x3 − 6x2

f ��(x) = 12x2 − 12x = 12x(x− 1).

Then x = 0, 1 are the solutions to f ��(x) = 0. We test the signs of f ��(x) on the openintervals I0 = (−∞, 0), I1 = (0, 1), and I2 = (1,∞).

Test open interval (−∞, 0) (0, 1) (1,∞)Test value xi x1 = −1 x2 = 1

2 x3 = 2Value of f ��(xi) f ��(x1) = 24 f ��(x2) = −3 f ��(x3) = 24Sign of f �� on Ii Positive Negative Positive

Applying Theorem 3.7, f is concave upward on (−∞, 0) and (1,∞), and concave downwardon (0, 1). The graph of f is shown in Figure 4.

✷

Try This 1

Find the open intervals on which the graph of y = x4 + 2x3 − 1 is concave upward orconcave downward.

Example 2 Determining Concavity on Open Intervals

Discuss the concavity of the graph of f(x) =24

x2 + 12.

Solution Applying the General Power Rule,we obtain

f �(x) = −24(x2 + 12)−2(2x)

=−48x

(x2 + 12)2.


�2,1.5��2,1.5�

2�2 x

y

Figure 5.The graph of f is concaveupward on (−∞,−2) and(2,∞), and concave down-ward on (−2, 2).

Then by the Quotient Rule, we find

f ��(x) =(x2 + 12)2(−48) + 48x(2(x2 + 12)(2x))

(x2 + 12)4

=48(x2 + 12)

�−(x2 + 12) + 4x2

�

(x2 + 12)4Factor out 48(x2 + 12)

=48(x2 + 12)

�3x2 − 12

�

(x2 + 12)4

f ��(x) =144(x− 2)(x+ 2)

(x2 + 12)3. Simplify

Note, f ��(x) = 0 precisely when x = ±2. We test the signs of f ��(x) on (−∞,−2), (−2, 2),and (2,∞) as follows:

Test open interval (−∞,−2) (−2, 2) (2,∞)Test value xi x0 = −3 x0 = 0 x0 = 3Value of f ��(xi) f ��(x1) ≈ 0.1 f ��(x2) = −1/3 f ��(x3) ≈ 0.1Sign of f �� on testopen interval

Positive Negative Positive

By Theorem 3.7, f is concave upward (−∞,−2) and (2,∞), and concave downward on(−2, 2). The graph of f is shown in Figure 5.

✷

Try This 2

Find the open intervals on which the graph of y =1

x2 + 3is concave upward or concave

downward.

Example 3 Determining Concavity on Open Intervals

Discuss the concavity of the graph of g(x) =x

x2 − 1, as shown in Figure 6.

1�1 x

3

�3

y

Figure 6Apply the Concavity Test todetermine the concavity ofthe graph of g.

Solution The first derivative is

g�(x) =(x2 − 1)− x(2x)

(x2 − 1)2Quotient rule

= − 1 + x2

(x2 − 1)2. Simplify

Then

g��(x) = − (x2 − 1)2(2x)− (1 + x2)[2(x2 − 1)(2x)](x2 − 1)4

Quotient rule

= −2x(x2 − 1)

�(x2 − 1)− 2(1 + x2)

�

(x2 − 1)4Factor out 2x(x2 − 1)

=2x(x2 + 3)(x2 − 1)3

. Simplify

Then x = 0 is the only solution to g��(x) = 0, Note, g��(x) is undefined when x = ±1.Next, we obtain the sign of g��(x) on the open intervals determined by x = 0,±1.

Test open interval (−∞,−1) (−1, 0) (0, 1) (1,∞)Test value xi x1 = −2 x2 = −0.5 x3 = 0.5 x4 = 2Value of g��(xi) g��(x1) ≈ −1.0 g��(x2) ≈ 7.7 g��(x3) ≈ −7.7 g��(x4) ≈ 1.0Sign of g��(x) Negative Positive Negative Positive


Hence, by Theorem 3.7, g is concave upward on (−1, 0) and (1,∞), and concave downwardon (−∞,−1) and (0, 1).

✷

Try This 3Find the open intervals on which the graph of

y =1

1− x

is concave upward or concave downward.

Points of InflectionA point on a graph where concavity changes is called a point of inflection. In Figure 7,we see three points of inflection P , Q, and R. Observe, the concavity to the nearby leftside is different from the concavity to the nearby right side of a point of inflection.

P

Q

R

1 2 3 x

1

2

3

y

Figure 7Points of Inflection at P,Q, and R.

Definition 7 Point of Inflection

Let f be continuous at c. Then (c, f(c)) is a point of inflection of the graph of f if thereare open intervals (a, c) and (c, b) such that either

a) f is concave upward on (a, c) and concave downward on (c, b), or

b) f is concave downward on (a, c) and concave upward on (c, b).

For brevity, we may write inflection point of f . A point of inflection of f is notnecessarily a critical point of f . However, an inflection point is a critical point of thederivative f � as the next theorem shows.


Theorem 3.8 Necessary Condition for a Point of Inflection

If (c, f(c)) is a point of inflection of f , then f ��(c) = 0 or f ��(c) does not exist.

Proof If f ��(c) does not exist, then the theorem is true. Suppose f ��(c) exists. Then weapply the continuity of f � at c and Definition 7. That is, there exist intervals (a, c] and[c, b) such that

a) f � is increasing on (a, c] and decreasing on [c, b), or

b) f � is decreasing on (a, c] and increasing on [c, b).

Thus, f �(c) is a relative extreme value of f �. Hence, f ��(c) = 0 by Theorem 3.2 in page142.

✷

Example 4 Finding Points of Inflection

Find the points of inflection and discuss the concavity of the graph of

f(x) = x4 + 2x3 − 1.

�1 4�4 x

�4

y

Figure 8A sketch of the graph of

f(x) = x4 + 2x3 − 1.

Solution The first two derivatives of f are given by

f �(x) = 4x3 + 6x2

f ��(x) = 12x2 + 12x = 12x(x+ 1).

Then f ��(x) = 0 only if x = 0 or x = −1. We determine the signs of f �� on

I1 = (−∞,−1) , I2 = (−1, 0) , I3 = (0,∞) .

Test open interval I1 = (−∞,−1) I2 = (−1, 0) I3 = (0,∞)Test value xi x1 = −2 x2 = −1/2 x3 = 1Value of f ��(xi) f ��(−2) = 24 f ��(−1/2) = −3 f ��(1) = 24Sign of f �� Positive Negative Positive

Applying Theorem 3.7, f is concave upward on (−∞,−1) and (0,∞), and concave down-ward on (−1, 0). Hence, by Definition 7 the points of inflection of f are (0, f(−1)) = (0,−2)and (−1, f(−1)) = (−1,−2). See the graph of f in Figure 8.

✷

Try This 4Find the points of inflection and discuss the concavity of y = 2x3 − x4 − 2.

Example 5 Finding Points of Inflection

Discuss the concavity and find the points of inflection of

f(x) = 4x5/3 − x8/3 + 1.


�0,1��1,4�1 2 3 x

y

Figure 9At a point of inflection(c, f(c)), either f ��(c) = 0or f ��(c) is undefined.

Solution The first two derivatives are

f �(x) =203x2/3 − 8

3x5/3

f ��(x) =409x−1/3 − 40

9x2/3

=40(1− x)

9 3√x

.

Note, f ��(x) = 0 if x = 1, and f ��(x) is undefined if x = 0. Next, determine the signs off ��(x) on the open intervals

I1 = (−∞, 0) , I2 = (0, 1) , I3 = (1,∞) .

The results are as follows:

Test open interval I1 = (−∞, 0) I2 = (0, 1) I3 = (1,∞)Test value xi x1 = −1 x2 = 0.5 x3 = 2Value of f ��(xi) f ��(x1) ≈ −8.8 f ��(x2) ≈ 2.8 f ��(x3) ≈ −3.5Sign of f �� Negative Positive Negative

By Theorem 3.7, f is concave upward on (0, 1), and concave downward on (−∞, 0) and(1,∞). Since the direction of concavity changes at x = 0 and x = 1, the points of inflectionof f are (0, f(0)) = (0, 1) and (1, f(1)) = (1, 4). See the graph of f in Figure 9.

✷

Concave Upward

c x

y

Figure 10aIf f �(c) = 0 and f ��(c) > 0,then f(c) is a relativeminimum value of f .

Concave Downward

c x

y

Figure 10bIf f �(c) = 0 and f ��(c) < 0,then f(c) is a relativemaximum value of f .

Try This 5Find the points of inflection of the graph of

y = x5/3 − 5x2/3.

Include a discussion of the concavity of the graph.

Observe, if f ��(c) = 0 or f ��(c) is undefined, we cannot conclude that (c, f(c)) is apoint of inflection. For example, let g(x) = x4 and h(x) = x4/3. Since g�(x) = 4x3

and h�(x) = 43

3√x are increasing functions, the graphs of g and h are concave upward on

(−∞,∞) by Definition 6. Thus, g and h do not have points of inflections even thoughg��(0) = 0 and h��(0) is undefined for g��(x) = 12x2 and h��(x) = 4

93√x2

.

The Second Derivative TestThe above named-test uses the second derivative to identify certain relative extreme valuesof a function. We describe a geometric interpretation of the test. If f �(c) = 0 and f isconcave upward on an open interval I containing c as in Figure 10a, then f(c) is a relativeminima of f . Similarly, if f is concave downward on I, then f(c) is a relative maxima off , see Figure 10b.

Theorem 3.9 The Second Derivative Test

Let f be a function and suppose f �(c) = 0.

a) If f ��(c) > 0, then f(c) is a relative minimum value of f .

b) If f ��(c) < 0, then f(c) is a relative maximum value of f .


Proof Suppose f ��(c) > 0.

f ��(c) = limx→c

f �(x)− f �(c)x− c

= limx→c

f �(x)x− c

> 0

Then there exists an open interval (a, b) containing c such that

f �(x)x− c

> 0 whenever x is in (a, b) and x �= c.

Thus, f �(x) < 0 for all x in (a, c), and f �(x) > 0 for all x in (c, b). Note, f is continuousat c since f �(c) exists. Then f is decreasing on (a, c] and decreasing on [c, b). Hence, f(c)is a relative minimum of f by the First Derivative Test in page 154. The proof of part b)is left as an exercise.

✷

Note, the Second Derivative Test is inconclusive if f ��(c) = 0. In such a case, analternative is to apply the First Derivative Test to identify any relative extrema.

Example 6 Applying the Second Derivative Test

Find the relative extrema of

f(x) = x3 − 320

x5 − 1.

�2 2 x

�3

3

y

Figure 11

The point (2, 2.2) is a

relative maximum, and

(−2,−4.2) is a relative

minimum of f .

Solution First, we determine the derivative of f .

f �(x) = 3x2 − 34x4

=3x2

4(4− x2) (2)

Then the solutions to f �(x) = 0 are x = 0,±2. The second derivative of f is

f ��(x) =ddx

�3x2 − 3

4x4

�

= 6x− 3x3

f ��(x) = 3x(2− x2).

We apply the Second Derivative Test with c = 0,±2.

c −2 0 2Sign of f ��(c) f ��(−2) > 0 f ��(0) = 0 f ��(2) < 0Conclusion Relative minimum Test fails Relative maximum

Hence, a local maximum value of f is

f(2) = 8− 320

(32)− 1 = 2.2

and a local minimum value is

f(−2) = −8− 320

(−32)− 1 = −4.2

Note, the Second Derivative Test fails when x = 0. However, using (2) we claim that thesign of f �(x) stays constant at x = 0. Thus, f(0) is not a relative extrema of f by theFirst Derivative Test, see page 157. The graph of f is given in Figure 11

✷

Try This 6Apply the Second Derivative Test to find the relative extrema of

s(t) = 8t2 − t4 + 5.


Example 7 Applying the Second Derivative Test

Find the relative extrema of

g(x) = 2 sinx+ cos(2x), 0 < x < 2π.

Π6

5 Π6

Π2

3 Π2 2Π

x

�3

32

y

Figure 12

The graph of

g(x) = 2 sinx+ cos(2x)

in 0 < x < 2π.

Solution The derivative is

g�(x) = 2 cosx− 2 sin(2x)

= 2 cosx− 4 sinx cosx Since sin 2x = 2 sinx cosx

= 2 cosx(1− 2 sinx).

Then the critical numbers x in (0, 2π) satisfy either

cosx = 0 or sinx =12.

Thus, the critical numbers are

x =π2,3π2,π6,5π6.

The second derivative is

g��(x) =ddx

[2 cosx− 2 sin(2x)]

= −2 sinx− 4 cos(2x).

We apply the Second Derivative Test as follows:

c π6

π2

5π6

3π2

Value of g��(c) g��(π6 ) = −3 g��(π2 ) = 2 g��( 5π6 ) = −3 g��( 3π2 ) = 6Conclusion Rel max Rel min Rel max Rel min

Hence, the relative maximum value of g(x) is

g�π6

�=

32= g

�5π6

�.

and the relative minimum values are

g�π2

�= 1 and g

�3π2

�= −3.

The graph of g is in Figure 12.✷

Try This 7Use the Second Derivative Test to find the relative extrema of

f(x) = cos(2x)− 2 cosx,π4< x <

5π4.


3.4 Check-It Out

1. Find the open intervals on which the graph of f(x) = x4 + 4x3 + 12 is concave upward

or concave downward.

2. Find the points of inflection of the graph of p(x) = 3x5 − 10x4 + 10x3 + 2.

3. Use the Second Derivative Test to find the relative extrema of s(t) = 14 t

4 − 2t2.


1. The graph of p(x) = x3 − 6x2 + 9x+ 1 is concave upward on (−∞, 2).

2. The graph of T (x) = sinx is concave downward on (0,π).

3. A point of inflection of the graph of y = x4 is (0, 0).

4. A point of inflection of the graph of y = 3√x is (0, 0).

5. If (c, f(c)) is a point of inflection of y = f(x), then f ��(c) = 0 or f ��(c) is undefined.

6. If g��(c) = 0, then (c, g(c)) is a point of inflection of y = g(x).

7. If R�(c) = 0 and R��(c) > 0, then R(c) is relative minimum value of y = R(x).

8. If y = M(x) is concave upward on (a, b) and concave downward on (b, c),

then (b,M(b)) is a point of inflection of the graph of M .

9. If y = p(x) is a polynomial such that p��(x) < 0 on (a, b) and p��(x) > 0 on (b, c),

then (b, p(b)) is a point of inflection of y = p(x).

10. It is possible that a point of inflection of y = f(x) is also a relative extrema of f .


In Exercises 1-4, find the open intervals where the graph is concave upward or concave downward.

Also, identify the inflection points.

1. y = x2 − 4x 2. y = −x2 − 2x

1 2 3 x

�4

�2

1

y

1�1�2 x

�2

1

�1

y

Figure for No. 1 Figure for No. 2


2. p(t) = t3 − 3t2 + 3t+ 1 3. y = 2 sin t, 0 ≤ t ≤ 2π

1 3 t

2

1

y

2ΠΠΠ2

3 Π2

t

2

�2

y


In Exercises 5-22, find the open intervals where the graph of the function is concave upward or downward.

Then determine the points of inflection.

5. p(x) = x4 − 8x3 + 18x2 + 3 6. p(x) = x5 − 5x+ 2

7. g(x) = x4/3 − 4x1/3 8. h(x) = x4/3 + 2x1/3

9. f(x) = x5/3 − 5x4/3 10. g(x) = x7/3 − 14x4/3

11. y = 1 + cos2(x), 0 < x < π 12. y = x+ sin2(x), 0 < x < π

13. y = sin(x) + cos(x), 0 < x < π 14. y = cos(x)− sin(x), 0 < x < 2π

15. f(x) =64

x2 + 1216. f(x) =

50x2 + 75

17. P (t) =32t

3t2 + 1618. R(w) =

196w21w2 + 343

19. m(x) =x√

x2 + 120. n(x) =

x√x+ 1

21. s(t) =t√1− t

22. T (w) =w√

1− w2

In Exercises 23-38, find the relative extrema by using the Second Derivative Test when applicable.

23. f(x) = x3 − 3x2 + 5 24. f(x) = 3x2 − 2x3 + 4

25. g(x) = x3 + x2 − 8x+ 20 26. f(x) = 2x3 + 5x2 − 4x+ 15

27. C(w) = (w2 + 64)(100− w2) 28. A(w) = (w2 + 25)(169− w2)

29. g(x) = 3 sin(2x), 0 ≤ x ≤ π 30. f(x) = 4 cos(3x) + 1, π6 ≤ x ≤ 5π

6

31. L(α) = tan(α)− 4α, −π2 < α < π

2 32. R(β) = 4β + 3 cot(β), 0 < β < π

33. y =8t

t2 + 1634. Q(p) =

−10pp2 + 25

35. f(x) =x2 + x− 2

x− 236. g(x) =

x3

9− x2

37. K(x) = 4x1/3 − x4/3 38. M(c) = 8c2/3 − c4/3


In Exercises 39-42, sketch a graph of a function with the given properties. There are several possible answers.Then find a formula y = f(x) for such a function.

39. f(1) = 2, f �(1) = 0, and f ��(x) > 0 for −∞ < x < ∞

40. f(3) = −1, f �(3) = 0, and f ��(x) < 0 for −∞ < x < ∞

41. f(0) = 2, f �(0) = 0, f ��(x) < 0 for x < 1, and f ��(x) > 0 for x > 1

42. f(1) = 5, f(−1) = −5, f �(±1) = 0, f ��(x) > 0 for x < 0, and f ��(x) < 0 for x > 0

Applications

43. Find the points on the graph of y = 1x that are closest to the origin.

44. Find the shortest distance between the origin and the graph of y = 1x2 .

45. Find the shortest distance between the parabola y = x2 and the point (3, 0).

46. Find the point on the graph of y = sinx that is nearest to the point (π2 , 0).

47. A rectangle is bounded by the x and y-axes and the graph of y = 1x3 in the first quadrant,

see figure below. Find the smallest possible perimeter of such a rectangle.

y�1�x3x

y

y�1�xx

y


48. Find the minimum length of the line segment that is tangent to the graph of y = 1x

and whose endpoints lie on the positive x- and y-axes, see figure above.

49. The diagonals of a rectangle contain the origin and two vertices of the rectangle lie on the

branches of the graph of y = 1x3 . Find the minimum perimeter of such a rectangle.

50. Find the points on the graph of y = 1x3 that are nearest the origin.


Prove, or disprove by finding a counter example

51. If f ��(x) is a twice differentiable concave upward function, prove or disprove that f ��(x) > 0 for all x.

52. If f(x) is a concave upward function, prove or disprove that y = (f(x))2 is concave upward.

53. If f � and f �� are positive functions on the interval (0,∞), prove that y = f(x2) is concave upward on (0,∞).

Is the converse true?

54. Find conditions on f and g so that y = f(g(x)) is concave upward.

3.5. LIMITS AT INFINITY 175

3.5 Limits at Infinity

• Limits at Infinity • Horizontal Asymptotes • Infinite Limits at Infinity

Limits at Infinity

�4 4 x1

�1

y

Figure 1As x → ∞, we findf(x) approaches 1.As x → −∞, we seef(x) approaches −1.

In the section, we study the ‘end behavior’ of a function. Specifically, we analyze the limitof a function f(x) as x increases or decreases without bound. To illustrate, we list severalvalues of a function for large |x|:

f(x) =

√4 + x2

x

x −1000 −100 −10 10 100 1000f(x) −1.00001 −1.0002 −1.0198 1.0198 1.0002 1.00001

The above values suggest that f(x) approaches 1 as x increases without bound, andwe write

limx→∞

f(x) = 1. Limit at Positive Infinity

Similarly, as x decreases without bound the table of values indicate f(x) approaches −1,and we write

limx→−∞

f(x) = −1. Limit at Negative Infinity

In Figure 1, we see the graph of f is getting closer to the horizontal line y = 1 as x → ∞.Also, the graph of f is getting nearer to the line y = −1 as x → −∞. For clarity, we pointout that ∞ does not represent a number.

Example 1 Finding a Limit at Infinity

Complete the table below where f(x) =2x

x+ 1.

x −1000 −100 −10 10 100 1000f(x)

Then estimate the limits at ∞ and −∞:

a) limx→∞

2xx+ 1

b) limx→−∞

2xx+ 1

Solution Using a calculator, we find

x −1000 −100 −10 10 100 1000f(x) 2.002 2.020 2.222 1.818 1.980 1.998

Observe, f(x) approaches 2 as x → ∞ or x → −∞. Then we find

limx→∞

f(x) = 2 and limx→−∞

f(x) = 2.

✷

Try This 1Estimate the limits

a) limx→∞

2xx2 + 1

b) limx→−∞

2xx2 + 1

by constructing a table of values for f(x) = 2xx2+1

.


f �x� approaches Lx approaches�

M x

LL�Ε

L�Ε

y

Figure 2aFor x > M , the graphof y = f(x) liesbetween the linesy = L+ � and y = L− �.

f �x� approaches Lx approaches ��

N x

L

L�Ε

L�Ε

y

Figure 2bFor x < N , the graphof y = f(x) liesbetween the linesy = L+ � and y = L− �.

�Π�2Π Π 2Π x

1

�1

y

Figure 2clim

x→∞sinx does not exist

limx→−∞

sinx does not exist

For exactness, we will need the definition of a limit at infinity.

Definition 8 Limits at Infinity

Let L be a real number and let f be a function.

1. The statement limx→∞

f(x) = L means that for each ε > 0 there exists a positive number

M > 0 such that|f(x)− L| < ε whenever x > M.

2. The statement limx→−∞

f(x) = L means that for each ε > 0 there exists a negative

number N < 0 such that

|f(x)− L| < ε whenever x < N.

In words, the definition of a limit at infinity implies that f(x) is within ε units of Lfor all x > M (where M depends on ε). Geometrically, the graph of y = f(x) for x > Mlies between the horizontal lines y = L+ ε and y = L− ε, as shown in Figure 2a. Similarstatements can be made for a limit at minus infinity (or negative infinity), see Figure 2b.

Not all functions have limits at ±∞. The graph of y = sinx oscillates infinitely manytimes between the lines y = −1 and y = 1, see Figure 2c. Then it is impossible to finda positive number M > 0 such that the graph of y = sinx for x > M lies between thehorizontal lines Y = L± 1

2 . Thus, the infinite limit limx→∞

sinx does not exist. Similarly,

limx→−∞

sinx does not exist.

Example 2 Proving a Limit at Infinity

Prove the following limit by using Definition 8:

limx→∞

4x2

= 0

Solution Let f(x) = 4x2 , ε > 0, and L = 0. Applying the notation in Definition 8, we

analyze

|f(x)− L| =��4x2

− 0

�� =4x2

< ε

The above inequality is equivalent to

2√ε< |x|.

Let M = 2√ε. Thus, |f(x)− L| < ε whenever x > M .

Hence, by Definition 8 we have proved

limx→∞

4x2

= 0.

✷

Try This 2

Use Definition 8 to prove limx→∞

1x

= 0.


Several properties of limits extend to limits at infinity. Such properties of limits arediscussed in Section 1.3. For instance, if lim

x→∞f(x) and lim

x→∞g(x) both exist then

1. limx→∞

(f(x) + g(x)) = limx→∞

f(x) + limx→∞

g(x)

2. limx→∞

(f(x)− g(x)) = limx→∞

f(x)− limx→∞

g(x)

3. limx→∞

(f(x)g(x)) = limx→∞

f(x) limx→∞

g(x)

4. limx→∞

f(x)g(x)

=lim

x→∞f(x)

limx→∞

g(x)provided lim

x→∞g(x) �= 0, and

5. limx→∞

(cf(x)) = c limx→∞

f(x) where c is any real number.

The above formulas remain true when x → ∞ is replaced by x → −∞. The nexttheorem is useful for evaluating limits at infinity of rational functions. A proof of thetheorem is given in Appendix A.1, page 414.

Theorem 3.10 Basic Rules for Limits at Infinity

If r > 0 is a positive rational number and c is a real number, then

1. limx→∞

cxr

= 0, and

2. limx→−∞

cxr

= 0 provided xr is well-defined for all x < 0.

Example 3 Finding Limits at Infinity

Evaluate the limits:

a) limx→∞

�3− 10

x2

�b) lim

x→−∞

�4 +

5x

�

Solution

a) Applying the limit of a difference, we find

limx→∞

�3− 10

x2

�= lim

x→∞3− lim

x→∞

10x2

= 3− 0 By Theorem 3.10

= 3.

b) Likewise, applying the limit of a sum we obtain

limx→−∞

�4 +

5x

�= lim

x→−∞4 + lim

x→−∞

5x

= 4 + 0 Theorem 3.10

= 4.

✷

Try This 3Evaluate the limits.

a) limx→∞

10√x

b) limx→−∞

120x2

c) limx→−∞

1√x


Horizontal AsymptotesWhen they exist, the limits at infinity are useful in describing the ‘end behavior’ of thegraph of a function.

Definition 9 Horizontal Asymptote

A line y = L is a horizontal asymptote of the graph of a function y = f(x) if either

limx→∞

f(x) = L or limx→−∞

f(x) = L.

Since the limits in Definition 9 involve x → ∞ or x → −∞, a function may have atmost two distinct horizontal asymptotes.

f �x��3 x� 52 x� 9

x32

y

Figure 3

The line y =32is a

horizontal asymptote.

Example 4 Finding the Horizontal Asymptotes

Evaluate the limits

a) limx→∞

3x− 52x+ 9

b) limx→∞

3x− 42x2 + 1

Then find the horizontal asymptotes of the graph of f(x) =3x− 52x+ 9

.

Solution Evaluate the limit as follows:

limx→∞

3x− 52x+ 9

= limx→∞

3x− 52x+ 9

·1x1x

= limx→∞

3− 5x

2 + 9x

=3− 02 + 0

=32

By Theorem 3.10

Similarly, as x → −∞ we obtain

limx→−∞

3x− 52x+ 9

= limx→−∞

3− 5x

2 + 9x

=32.

Thus, the line y = 32 is the horizontal asymptote of the graph of f as shown in Figure 3.

Finally, we have

limx→∞

3x− 42x2 + 1

= limx→∞

3x − 4

x2

2 + 1x2

= 0.

✷

Try This 4Evaluate the limits

a) limx→∞

4x3− 8x

b) limx→−∞

4x5− 8x2

Then find the horizontal asymptotes of y =4x

3− 8x.


For the rational functions, the limit at infinity is determined by the leading coefficientsin the numerator and denominator, as discussed below.

Summary for Finding Limits at Infinity of Rational Functions

Let p(x) and q(x) be polynomials of degree n and m, respectively. Suppose a and b arethe leading coefficients of p(x) and q(x), respectively.

1. If n = m, then

limx→∞

p(x)q(x)

=ab= lim

x→−∞

p(x)q(x)

.

2. If n < m, then

limx→∞

p(x)q(x)

= 0 = limx→−∞

p(x)q(x)

.

3. If n > m, the limit does not exist as x → ∞ or x → −∞.

In any case, a rational function has at most one horizontal asymptote.

The proof of the above summary is left as an exercise at the end of this section. Forexample, the polynomials in the numerator and denominator in the rational function

R(x) =3x2 + 10x5x2 + 12

are of both of degree two. Thus, as x → ∞ or x → −∞, we obtain that R(x) → 3/5,which is the quotient of the leading coefficients. In Figure 4, the line y = 3

5

r�x�� 3 x2 � 10 x5 x2 � 12

x

35

y

Figure 4: R(x) =3x2 + 10x5x2 + 12

is the horizontal asymptote of the graph of R. However, for a non-rational function itsgraph may possibly have two distinct horizontal asymptotes.

Example 5 A Graph With Two Horizontal Asymptotes

Evaluate the limits:

a) limx→∞

4x√x2 + 9

b) limx→−∞

4x√x2 + 9


y�4 x

x2 � 9

x

�4

4

y

Figure 5The lines y = ±4 arehorizontal asymptotes.

Solution

a) We rewrite as follows:

4x√x2 + 9

·1x1x

=4√x2+9√x2

=4�

1 + 9x2

Applying Theorem 3.10, we obtain

limx→∞

4x√x2 + 9

= limx→∞

4�1 + 9

x2

=4√1 + 0

= 4.

b) Since x = −√x2 whenever x < 0, we rewrite

4x√x2 + 9

·1x1x

=4√x2+9

−√x2

=4

−�

1 +9x2

Thus, by Theorem 3.10 we obtain

limx→−∞

4x√x2 + 9

= limx→−∞

4

−�

1 + 9x2

=4

−√1 + 0

= −4.

In particular, the lines y = ±4 are the horizontal asymptotes of the graph of y = 4x√x2+9

as seen in Figure 5.✷

Try This 5Evaluate each limit.

a) limx→∞

x√x2 + 1

b) limx→−∞

x√9x2 + 4

In evaluating limits at infinity involving trigonometric functions, the Squeeze Theoremas explained in Exercise 54 will be helpful to know.

Example 6 Limits at Infinity with Trigonometric Functions

Evaluate each limit:

a) limx→∞

x+ cosxx

b) limx→−∞

x+ cosxx

Solution For all x, we recall−1 ≤ cosx ≤ 1.

Adding x, we obtainx− 1 ≤ x+ cosx ≤ x+ 1. (3)

If we divide by x > 0, then

x− 1x

≤ x+ cosxx

≤ x+ 1x

.

However, if x < 0 thenx− 1x

≥ x+ cosxx

≥ x+ 1x

.


y� x� cos �x�x

�20 x

1

y

Figure 6The graph of

y =x+ cosx

x

and the horizontalasymptote y = 1.

Using the leading coefficients, we find

limx→∞

x− 1x

= 1 = limx→∞

x+ 1x

and

limx→−∞

x− 1x

= 1 = limx→−∞

x+ 1x

.

Hence, by the Squeeze Theorem we obtain

limx→∞

x+ cosxx

= 1 = limx→−∞

x+ cosxx

.

As seen in Figure 6, the line y = 1 is the horizontal asymptote of y = x+cos xx .

✷

Try This 6Evaluate the following limits.

a) limx→∞

2x+ sinx√x2 + 1

b) limx→−∞

2x+ sinx√x2 + 1

N�t��5000�30000 t2100� t2

24 48 72 t

N�t�

Figure 7As t increases withoutbound, N(t) approaches35,000.

Example 7 The Limit of a Population of Bacteria

A nutrient is needed to help grow a population of bacteria. The number of bacteria ismodeled by the function

N(t) = 5000 +30000t2

100 + t2

where t ≥ 0 is in hours. Determine the number of bacteria when t = 24, t = 48 and t = 72hr. Then find the limit of N(t) as t → ∞.

Solution Using a calculator, the number of bacteria are

N(24) = 5000 +30000(24)2

100 + (24)2= 30, 562

N(48) = 5000 +30000(48)2

100 + (48)2= 33, 752

N(72) = 5000 +30000(72)2

100 + (72)2= 34, 432.

We evaluate the limit as follows:

limt→∞

�5000 +

30000t2

100 + t2·

1t2

1t2

�= lim

t→∞

�5000 +

30000100t2

+ 1

�

= 5000 +300000 + 1

= 35, 000

Thus, N(t) → 35, 000 bacteria as t → ∞. In Figure 7, we see a graph of the populationversus time.

✷


Try This 7

The population of a bacteria is approximated by N(t) = 10, 000

�1 +

4t2

100 + t2

�.

Find the number of bacteria after t = 24 hours. Also, find the limit of N(t) as t → ∞.

Infinite Limits at InfinityIf f(x) increases without bound as x → ∞, we say f(x) has an infinite limit at infinityand write lim

x→∞f(x) = ∞. Similarly, if f(x) decreases without bound as x → ∞, then

f(x) has a negative infinite limit at infinity and we write limx→∞

f(x) = −∞. The definition

below describes the notion of an infinite limit at infinity.

Definition 10 Infinite Limits at Infinity

Let y = f(x) be a function.

1. We writelim

x→∞f(x) = ∞

if for each positive number M > 0 there exists a positive number N > 0 such that

f(x) > M whenever x > N.

2. The statementlim

x→∞f(x) = −∞

means that for each negative number M < 0 there exists a positive number N > 0such that

f(x) < M whenever x > N.

We have similar statements for limx→−∞

f(x) = ∞ and limx→−∞

f(x) = −∞.

In evaluating infinite limits, the following observations are useful.If L > 0 is a positive number, lim

x→∞f(x) = L, and lim

x→∞g(x) = ∞, then

limx→∞

f(x)g(x) = ∞ (4)

and

limx→∞

g(x)f(x)

= ∞. (5)

The proofs of these observations are left as exercises at the end of the section.

Example 8 Evaluating Infinite Limits at Infinity

Evaluate the limits at infinity.

a) limx→∞

x3 b) limx→∞

�4x3 − 10x2 − 5

�

Solution

a) Since x3 > x when x > 1, it follows that x3 increases without bound as x → ∞. Then

limx→∞

x3 = ∞.


f �x�� 4x3�10x2�5

25 50 x

106

2�106

3�106

y

Figure 8As x increases withoutbound, f(x) increaseswithout bound.

b) Factor the monomial with the highest exponent:

4x3 − 10x2 − 5 = 4x3

�1− 5

2x− 5

4x3

�.

By part a), 4x3 → ∞ as x → ∞. Moreover, we find

limx→∞

�1− 5

2x− 5

4x3

�= 1− 0− 0 = 1.

Applying observation (4) in the previous page, we obtain

limx→∞

�4x3 − 10x2 − 5

�= lim

x→∞4x3 lim

x→∞

�1− 5

2x− 5

4x3

�

= ∞ · 1= ∞

Note, the few values of f(x) = 4x3 − 10x2 − 5 shown below

x 25 50 100f(x) 56, 245 474, 995 3, 899, 995

support the fact that f(x) increases without bound as x → ∞. The ‘right-end’behavior of the graph of f is seen in Figure 8.

✷

Try This 8Evaluate the limits at infinity.

a) limx→−∞

x3 b) limx→−∞

�x4 + x3�

�4 7 x

�10

15

y

Figure 9aThe graph of

y =x2 + 1x− 2

.

Example 9 Limits at Infinity of a Rational Function

Find the limits at infinity.

a) limx→∞

x2 + 1x− 2

b) limx→−∞

x5 + 12x3 + 1

Solution

a) The degree of the denominator inx2 + 1x− 2

is smaller than the degree of the numerator.

In such a case, multiply and divide by 1/xd where d is the degree of the denominator.

x2 + 1x− 2

· 1/x1/x

=x+ 1/x1− 2/x

.

Applying Theorem 3.10 and observation (5), we find

limx→∞

x2 + 1x− 2

= limx→∞

x+ 1/x1− 2/x

=∞+ 01− 0

=∞1

= ∞

In Figure 9a, we see the graph of y =x2 + 1x− 2

.


3�3 x

3

�3

y

Figure 9bThe graph of

y =x5 + 12x3 + 1

.

b) The degree of the denominator inx5 + 12x3 + 1

is smaller than the degree of the numerator.

We follows the strategy applied in part a). Then we multiply and divide by 1/x3.

x5 + 12x3 + 1

· 1/x3

1/x3=

x2 + 1/x3

2 + 1/x3

Then

limx→−∞

x5 + 12x3 + 1

= limx→−∞

x2 + 1/x3

2 + 1/x3

=∞+ 02 + 0

=∞2

= ∞

The graph of y =x5 + 12x3 + 1

is shown in Figure 9b.

✷

Try This 9Evaluate the limits at infinity.

a) limx→∞

x4

x2 + 1b) lim

x→−∞

x3

x2 + 1

3.5 Check-It Out

1. Evaluate the limits.

a) limx→∞

100xx2 + 1

b) limx→∞

3x4− 6x

c) limx→−∞

x2

x+ 1d) lim

x→−∞

√4x2 + 1x

2. Find the horizontal asymptote of f(x) =1− x2

x2 − 4.

3. Find a positive number N > 0 such that1x2

< 10−800 whenever x > N .

True or False. If false, revise the statement to make it true.

1. If limx→∞

f(x) = limx→∞

g(x) = ∞, then limx→∞

f(x)g(x)

= 1.

2. If limx→∞

f(x) = limx→∞


f(x) + limx→∞

g(x) = ∞.

3. If limx→∞

f(x) = 0.1 and limx→∞


(f(x)g(x)) = ∞.

4. limx→∞

sinxx

does not exist

5. limx→∞

x2 sin1x

= ∞ 6. limx→∞

x− cosxx

= 1

7. limx→−∞

(2− x) = ∞ 8. limx→∞

3x3 + 1x4 + x+ 3

= ∞

9. The graph of a function may have two horizontal asymptotes.

10. The graph of f(x) =x4 + 3x2x4 + 1

has no horizontal asymptote.



In Exercises 1-8, use the given graph to evaluate the limit.

a) limx→∞

f(x) b) limx→−∞

f(x)

1. f(x) =2x+ 11− x

2. f(x) =3x+ 1x+ 1

3. f(x) =2x

x2 + 14. f(x) =

10x2

x4 + 1

4�4 x2

�2

y

4�4 x

3

y

3�3 x

2

y

�5 x

5

y

For No. 1 For No. 2 For No. 3 For No. 4

5. f(x) =

√4x2 + 1x

6. f(x) =

√x4 + 1x2

7. f(x) =sinxx

8. f(x) =2 + cosx

x

4�4 x

�2

2

y

4�4 x12

y

�7 7 x

�1

2

y

10 x

2

y

For No. 5 For No. 6 For No. 7 For No. 8

In Exercises 9-34, evaluate the limit. If applicable, indicate if the limit is ∞ or −∞.

9. limx→−∞

�2 +

3x2

�10. lim

x→∞

�1− 2

x

�11. lim

x→∞

x2 − 6x2 − x− 2

12. limx→∞

x2 − 2x+ 7x2 − 4x+ 3

13. limx→−∞

3x3 − 12x3 + x2 + 3

14. limx→−∞

x3 + x2 + 3x+ 32x3 + 2x2 + 5x+ 5

15. limx→−∞

4− 5x3

x2 + 116. lim

x→∞

x− 2x2 + 4

17. limx→∞

x2 − 1x3 + 2x2

18. limx→−∞

x4

x2 + 119. lim

x→∞

√4x2 − 3x+ 1

x20. lim

x→∞

x√9x2 + 5

21. limx→−∞

4x√x2 + 3x+ 1

22. limx→−∞

5− 2x�(x− 4)2

23. limx→∞

3√8x3 + x2 − 53x− 1

24. limx→∞

√16x2 + 5x+ 23x+ 11

25. limx→∞

cos 2πxx

26. limx→−∞

sin 3xx

27. limx→−∞

12x+ sinx

28. limx→∞

3xx+ cos 4x

29. limx→∞

sin(πx)− 1x

30. limx→∞

x cos1x

31. limx→∞

x sin1x

32. limx→∞

x sin3x


33. limx→∞

cosxx

34. limx→∞

sinxx− π

Encounters with the indeterminate form ∞ − ∞.

In Exercises 35-40, evaluate the infinite limits.

35. limx→∞

��x2 + x− x

�36. lim

x→∞

��x2 + 6x− x

�37. lim

x→∞

�x−

�x2 + 4x

�

38. limx→∞

�√x+ 9−

√x+ 4

�39. lim

x→∞

�3�

x3 + 8x2 − x�

40. limx→−∞

�3�

x3 + 2x2 − x�

In Exercises 41-44, determine the horizontal asymptotes of the function.

41. f(x) =8− 27x3

9x3 + 242. h(t) =

√x2 + 1x

43. s(t) =4t

t2 + 344. R(x) = 2 +

sinxx

In Exercises 45-48, sketch the graph of a function that satisfies the given statements.

45. f(0) = 0, limx→∞

f(x) = 3, limx→−∞

f(x) = 3

46. f(0) = 1, limx→∞

f(x) = ∞, limx→−∞

f(x) = 0

47. f(0) = 0, limx→∞

f(x) = π/2, limx→−∞

f(x) = −π/2

48. f(0) = 2, f(1) = 4, f(−1) = 0, limx→∞

f(x) = 2, limx→−∞

f(x) = 2

Applications

49. During the spring season the proportion of adults with sinus problems due to pollen is P (t) =t2

4t2 + 1where

t is the number of weeks past March 21, the first day of spring. Determine the limit of P (t) as t → ∞. Whatis a practical interpretation of the limit?

50. The probability that a certain injury will heal in 12 hours is given by P (x) =

√x

2√x+ 1

where x is the number

milligrams of an ointment that should be administered right after the injury. Evaluate limx→∞

P (x).

51. A company has to spend $40,000 to set-up a manufacturing equipment. After the set-up, it will cost the company$5 to manufacture each item.

a) Find the total cost C(x) of manufacturing x items with the set-up cost included.

b) Find the limit at infinity of the average cost per item, i.e., the limit of C(x)x as x → ∞.

52. In a given day, the amount of nitrogen dioxide in the air is modeled by

A(t) =50t2 + 15(242)

2t2 + 242

where t is the number of hours past midnight, and A(t) is measured using the pollutant standard index (PSI ).

a) Find the amount of nitrogen dioxide in the air at 11 AM. b) Find the limit of A(t) as t → ∞.

Theory and Writing Proofs

53. Let p(x) and q(x) be nonzero polynomials of degree n and m, respectively. Suppose a and b are the leadingcoefficients of p(x) and q(x), respectively.

a) If n = m, prove limx→∞

p(x)q(x)

=ab= lim

x→−∞

p(x)q(x)

. b) If n < m, show limx→∞

p(x)q(x)

= 0 = limx→−∞

p(x)q(x)

.

54. Suppose h(x) ≤ f(x) ≤ g(x) for all x in (a,∞), and limx→∞

h(x) = limx→∞

g(x) = L. Prove limx→∞

f(x) = L.

55. Let L be a nonzero real number, limx→∞

f(x) = L, and limx→∞

g(x) = ∞. Prove limx→∞

f(x)g(x) = sign(L) ·∞ where

sign(L) denotes the sign of L.

3.6. SUMMARY OF CURVE SKETCHING 187

3.6 Summary of Curve Sketching

Curve Sketching TechniquesWe use the following features to sketch a detailed graph of a function. For a review of thefeatures, we list the pertinent sections in parentheses.

f �x�� x3

16� x2

�10 10 x

�10

10

y

Figure 1a

f �x�� x3

16� x2

�10 10 x

�25

25

y

Figure 1bViewing windowsfor the graph of

f(x) =x3

16− x2.

• Domain and range

• Symmetry

• x-intercepts and y-intercepts

• Continuity (Section 1.4)

• Vertical and horizontal asymptotes (Sections 1.5 and 3.5)

• Differentiability (Section 2.1)

• Relative extrema (Section 3.1)

• Concavity (Section 3.4)

• Points of inflection (Section 3.4)

• Limits at infinity (Section 3.5)

A graphing utility is helpful in obtaining different parts or sections of the graph of afunction. For example, the viewing windows in Figure 1a and 1b show certain parts of thegraph of

f(x) =x3

16− x2.

However, the second viewing window shows a holistic view of the graph.The guidelines listed below should help the student determine important features of

the graph. Consequently help her or him to choose a good viewing window. Not everyfeature is relevant to every function. For instance, a graph may have no asymptotes andthe range may not be determined before the relative extrema are identified.

Guidelines For Sketching the Graph of a Function

1. Determine the domain, and if possible the range.

2. Find the intercepts, symmetry, and horizontal and vertical asymptotes.

3. Find the x-values for which f �(x) and f ��(x) are either zero or undefined.

4. Determine the relative extrema and points of inflection from step 3.

�0,3�

�3,30��1,14�

1 3 x

30

�3

y


p(x) = x4 − 8x3 + 18x2 + 3.

Example 1 Sketching the Graph of a Polynomial Function

Sketch the graph of p(x) = x4 − 8x3 + 18x2 + 3.

Solution The first three features of the graph follow directly.

Domain: (−∞,∞)x-intercept: (0, 3)

Infinite limits: limx→∞

p(x) = ∞, limx→−∞

p(x) = ∞

Then we find

First derivative: p�(x) = 4x3 − 24x2 + 36x = 4x(x− 3)2

and by the product rule we obtain

Second derivative: p��(x) = 4 · (x− 3)2 + 4x · 2(x− 3)

= 4(x− 3)((x− 3) + 2x)

= 12(x− 3)(x− 1)


Then the critical numbers and the zeros of p��(x) determine the following test open inter-vals:

Critical numbers: x = 0, 3Test open intervals: (−∞, 0), (0, 1), (1, 3), (3,∞)

We list the signs of p�(x) and p��(x) on the test open intervals, determine whether thegraph of f is increasing or decreasing, and discuss concavity. Also, we list all relativeextrema and points of inflection.

p�(x) p��(x) Property of the Graph(−∞, 0) − + Decreasing, concave upward(0, 1) + + Increasing, concave upward(1, 3) + − Increasing, concave downward(3,∞) + + Increasing, concave upwardx = 0 0 + Relative Minimumx = 1 + 0 Point of Inflectionx = 3 0 0 Point of inflection

Applying the Second Derivative Test, the relative minimum value of p is p(0) = 3. Sincethe sign of p��(x) changes at x = 1 and x = 3, then (1, 14) and (3, 30) are inflection points.In Figure 2, the graph implies that the range is [3,∞).

✷

Try This 1Sketch the graph of F (w) = w5 − 5w4. Apply the sketching guidelines in page 187.

Example 2 The Graph of a Radical Function

Sketch the graph of f(x) =x√x− 1

.

�2,2� �4,4� 3 �2 4 8 x

2

4

y


f(x) =x√x− 1

.

Solution The first few features of the graph are

Domain: [1,∞)

Limit at infinity: limx→∞

f(x) = ∞

Vertical asymptote: x = 1

Applying the quotient rule, we find

First derivative: f �(x) =

√x− 1− x · 1

2√x−1

x− 1=

x− 2

2(x− 1)3/2

Similarly, we obtain

Second derivative: f ��(x) =2(x− 1)3/2 − (x− 2) · 3

√x− 1

4(x− 1)3

=2(x− 1)− 3(x− 2)

4(x− 1)5/2

=4− x

4(x− 1)5/2

Then the critical numbers and the test open intervals are as follows:

Critical number: x = 2Test open interval: (1, 2), (2, 4), (4,∞) Note f ��(4) = 0


We summarize the signs of f � and f �� on the test open intervals. From which, we findwhere f is increasing, decreasing, and concave upward or downward. Also, we identify therelative extrema by the Second Derivative Test and find the points of inflection.

f �(x) f ��(x) Property of the Graph(1, 2) − + Decreasing, concave upward(2, 4) + + Increasing, concave upward(4,∞) + − Increasing, concave downwardx = 2 0 + Relative Minimumx = 4 + 0 Point of inflection

The relative minimum value is f(2) = 2. Since the sign of f ��(x) changes at x = 4, thepoint (4, 4√

3) is an inflection point. From the graph in Figure 3, we conclude the range of

f is [2,∞).✷

Try This 2

Sketch the graph of R(x) =x+ 1√

x. Apply the guidelines in page 187.

Example 3 The Graph of a Rational Function

Sketch the graph of f(x) =x2 − 19− x2

.

Solution The following features can be computed directly:

Domain: {x : x �= ±3}Symmetry: With respect to the y-axisIntercepts: (±1, 0) ,

�0,− 1

9

�

Vertical asymptotes: x = ±3Limits at infinity: lim

x→∞f(x) = −1, lim

x→−∞f(x) = −1

Horizontal asymptote: y = −1

Applying the quotient rule, we obtain

1st derivative: f �(x) =(9− x2)(2x)− (x2 − 1)(−2x)

(9− x2)2=

16x(9− x2)2

Differentiating again, we find

2nd derivative: f ��(x) =(9− x2)2(16)− (16x)(−4x(9− x2))

(9− x2)4

=16(9− x2)((9− x2) + 4x2)

(9− x2)4

=48(x2 + 3)(9− x2)3

The following test open intervals are determined from the critical numbers and the domain.

Critical number: x = 0Test open intervals: (−∞,−3), (−3, 0), (0, 3), (3,∞)


�3 3�1 1x

�1

y

Figure 4Important features ofa graph are determinedby using calculus.

We determine the signs of f and f ��, and where f is increasing or decreasing. Also, wediscuss concavity and identify the relative extrema by using the Second Derivative Test.

f �(x) f ��(x) Property of the Graph(−∞,−3) − − Decreasing, concave downward(−3, 0) − + Decreasing, concave upward(0, 3) + + Increasing, concave upward(3,∞) + − Increasing, concave downwardx = 0 0 + Relative Minimum

The relative minimum value of f is f(0) = − 19 . Finally, the range is (−∞,−1)∪

�− 1

9 ,∞�

as seen from the graph of f in Figure 4.✷

Try This 3

Sketch the graph of g(t) =t2 − 4t2 − 1

. Use the guidelines in page 187.

��2, 3 43 ��0,0��1,6�

�5 x�10

50

y

Figure 5The graph off(x) = x2/3(x+ 5)

Example 4 Sketching the Graph of a Radical Function

Sketch the graph of f(x) = x2/3(x+ 5). Use the guidelines in page 187.

Solution The following features of the graph follow directly.

Domain: (−∞,∞)Intercepts: (0, 0), (−5, 0)

Limits at infinity: limx→∞

f(x) = ∞, limx→−∞

f(x) = −∞

Applying the product rule, we obtain

First derivative: f �(x) = x2/3 +2

3 3√x(x+ 5) =

5(x+ 2)

3 3√x

.

Differentiating again, by the quotient rule we find

Second derivative: f ��(x) =15 3

√x− 5(x+ 2) · 1

3√x2

9 3√x2

=10(x− 1)

9x4/3.

We list the test open intervals determined by the critical numbers and the the zero off ��(x), i.e., x = 1.

Critical number: x = −2, 0Test open intervals: (−∞,−2), (−2, 0), (0, 1), (1,∞)

We tabulate the signs of f � and f �� on the test open intervals. We determine where fis increasing or decreasing, and discuss concavity. Also, we identify the relative extremaand the points of inflection.

f �(x) f ��(x) Property of the Graph(−∞,−2) + − Increasing, concave downward(−2, 0) − − Decreasing, concave downward(0, 1) + − Increasing, concave downward(1,∞) + + Increasing, concave upwardx = −2 0 − Relative Maximumx = 0 undefined undefined Relative Minimumx = 1 + 0 Point of inflection


We have a local maximum value at x = −2 by the Second Derivative Test. A relativemaximum value is f(−2) = 3 3

√4. Note, the sign of f �(x) changes from negative to positive

at x = 0. Thus, by the First Derivative Test a relative minimum value is f(0) = 0.Since the sign of f ��(x) changes at x = 1 the point (1, 6) is an inflection point. In

Figure 5, we see the graph of f and the range is (−∞,∞).✷

Try This 4Sketch the graph of H(t) = 6 3

√t− 3t2/3. Apply the sketching guidelines in page 187.

Example 5 The Graph of a Trigonometric Functions

Sketch the graph of T (x) =sinx

2 + cosx.

�2 Π3,

33�

2Π

�4 Π3,�

33�

�Π x

1

y

Figure 6a

�2 Π3,

33��4 Π

3,

33�

�4 Π3,�

33��2 Π

3,�

33�

2Π�2Π x

1

y

Figure 6bApply the periodicityto extend the graph of

T (x) =sinx

2 + cosx.

Solution First, we sketch the graph for x in [0, 2π]. Since −1 ≤ cosx ≤ 1 and sin(kπ) = 0where k is an integer, we obtain

Domain: (−∞,∞)

x-intercepts: (0, 0), (π, 0), (2π, 0)

Applying the quotient rule and the identity cos2 x+ sin2 x = 1, we find

First derivative: T �(x) =(2 + cosx) cosx− (sinx)(− sinx)

(2 + cosx)2

=2 cosx+ 1(2 + cosx)2

Second derivative:

T ��(x) =(2 + cosx)2(−2 sinx)− (2 cosx+ 1)(−2 sinx(2 + cosx))

(2 + cosx)4

=2 sinx(2 + cosx)(−(2 + cosx) + (2 cosx+ 1))

(2 + cosx)4

=2 sinx(cosx− 1)

(2 + cosx)3

Note, x = 0,π, 2π satisfy T ��(x) = 0. Together with the critical numbers, we list the testopen intervals:

Critical numbers: x =2π3,4π3

Test open intervals:

�0,

2π3

�,

�2π3,π

�,

�π,

4π3

�,

�4π3, 2π

�

We tabulate the signs of T �(x) and T ��(x) on the test open intervals. Then determinewhere T is increasing or decreasing, and discuss concavity.

T �(x) T ��(x) Property of the Graph(0, 2π/3) + − Increasing, concave downward(2π/3,π) − − Decreasing, concave downward(π, 4π/3) − + Decreasing, concave upward

(4π/3, 2π) + + Increasing, concave upwardx = 2π/3 0 − Relative Maximum

x = π − 0 Point of Inflectionx = 4π/3 0 + Relative Minimum


By the Second Derivative Test, a local maximum value is T ( 2π3 ) =√3

3 , and a relative

minimum is T ( 4π3 ) = −√

33 . Since the sign of T ��(x) changes at x = π, (π, 0) is a point of

inflection. The graph of T in [0, 2π] is given in Figure 6a.Since T (x + 2π) = T (x), we extend the graph to [−2π, 2π], see Figure 6b. Using the

extrema, we obtain that the range is [−√3

3 ,√33 ].

✷

Try This 5

Sketch the graph of N(x) =cosx

2 + sinx. Apply the guidelines in page 187.

Example 6 Sketching the Graph of a Rational Function

Sketch the graph of f(x) =x3

1− 4x2.

� 32,�3 316�

�� 32,3 316�� 12

12

x

y


f(x) =x3

1− 4x2.

Solution The first few features of the graph can be obtained directly.

Domain: {x : x �= ± 12}

Vertical asymptotes: x = ± 12

Symmetry: with respect to the originIntercept: (0, 0)

Limits at infinity: limx→∞

f(x) = −∞, limx→−∞

f(x) = ∞

Applying the quotient rule, we find

1st derivative: f �(x) =(1− 4x2)(3x2)− x3(−8x)

(1− 4x2)2

=x2(3(1− 4x2) + 8x2)

(1− 4x2)2

f �(x) =x2(3− 4x2)(1− 4x2)2

=3x2 − 4x4

(1− 4x2)2

Critical numbers: x = 0,±√32

Differentiating again, we obtain

2nd derivative: f ��(x) =(1− 4x2)2(6x− 16x3)− (3x2 − 4x4)(−16x(1− 4x2))

(1− 4x2)4

=(1− 4x2)(6x− 16x3) + 16x(3x2 − 4x4)

(1− 4x2)3

=6x− 16x3 − 24x3 + 64x5 + 48x3 − 64x5

(1− 4x2)3

f ��(x) =8x3 + 6x(1− 4x2)3

=2x(4x2 + 3)(1− 4x2)3

We consider the test open intervals determined by the critical numbers, the zeros of f ��(x),and the domain:

Test open intervals:�−∞,−

√32

�,�−

√3

2 ,− 12

�,�− 1

2 , 0�,

�0, 1

2

�,�

12 ,

√32

�,�√

32 ,∞

�


Below, we list the signs of f � and f �� on the test open intervals. Then we identify anyrelative extrema by the Second Derivative Test. We describe concavity and the points ofinflection, too.

f �(x) f ��(x) Property of the Graph

(−∞−√3

2 ) − + Decreasing, concave upward

(−√3

2 ,− 12 ) + + Increasing, concave upward

(− 12 , 0) + − Increasing, concave downward

(0, 12 ) + + Increasing, concave upward

( 12 ,√3

2 ) + − Increasing, concave downward

(√32 ,∞) − − Decreasing, concave downward

x =√3

2 0 − Relative maximumx = 0 0 0 Point of inflection

x = −√

32 0 + Relative minimum

The relative maximum value of f is f�√

32

�= − 3

√3

16 and the relative minimum value is

f�−

√3

2

�= 3

√3

16 . Note, the sign of f �� changes at x = 0. Thus, (0, 0) is an inflection point.

Finally, the range of f is (−∞,∞) as seen in Figure 7.✷

Try This 6

Sketch the graph of h(t) =t2

t− 3using the guidelines in page 187.

y��14x

x

�1

1

y

Figure 8

The line y = −14x

is a slant asymptoteof the graph of

f(x) =x3

1− 4x2.

The graph in Example 6 has a slant asymptote or oblique asymptote. Precisely,a line y = mx+ b where m �= 0 is a slant asymptote of the graph of a function f if either

limx→∞

[f(x)− (mx+ b)] = 0

orlim

x→−∞[f(x)− (mx+ b)] = 0.

If R(x) = p(x)/q(x) is a rational function and the degree of p(x) is one more than thedegree of q(x), then the graph of y = R(x) has a slant asymptote. To find the slantasymptote, use long division to express R(x) as a sum of linear function and anotherrational function. For example, using long division we find

f(x) =x3

1− 4x2= −1

4x+

x/41− 4x2

.

Then we obtain

limx→∞

�f(x)−

�−14x

��= lim

x→∞

x/41− 4x2

= 0

Thus, y = −x4 is a slant asymptote of the graph of f , see Figure 8.

3.6 Check-It Out

1. Find the domain, asymptotes, and the symmetry of the graph of R(x) =x

x2 − 1.

2. Find the domain, range, and open intervals where the graph of f(x) =x2

x+ 1is increasing, decreasing, concave upward, or concave downward.

3. Find the critical numbers and open intervals in [0, 2π] where g(t) = sin t + cos t isincreasing or decreasing.



In Exercises 1-6, use the graph below to answer the problems.

�2 1 2 3 4 x

2

4

�2

y

Figure for nos 1-6: The graph of f(x) =x2

x− 1

1. f ��(x) > 0 on (0,∞) 2. f is increasing on (−∞, 0)

3. limx→1

f(x) = ∞ 4. The range of f is (−∞,∞).

5. The graph has no slant asymptote. 6. (0, 0) is a local maximum point.

7. The domain of y =sinx

1 + cosxis {x : x �= kπ where k is an integer }.

8. The graph of y = x2 sinx is symmetric about the y-axis.

9. The graph of y = tanx is symmetric about the origin.

10. If f �(x0) = 0, then f(x0) is a relative extreme value of f .


In Exercises 1-4, an approximate sketch of the graph of a function is given. Find therelative extrema, open intervals where the function is increasing or decreasing, limits atinfinity, and range.

1. f(x) = 6x2 − x4 + 2 2. g(x) =x3

x2 − 1

�1�2 1 2 x

6

12

y

�3 �2 32 x

4

�4

y

For No. 1 For No. 2


3. p(t) = t5/3 − 5t2/3 4. C(t) = sin t+ cos t, 0 ≤ t ≤ 2π

�2 1 2 3 t1

y

Π 2Π t

�1

1

y

For No. 3 For No. 4

In Exercises 5-14, sketch the graph of the function. Find the intercepts, relative extrema,points of inflection, asymptotes, and the range.

5. g(t) =t2

t2 − 16. h(t) =

8t4− t2

7. f(x) =2x3

x2 − 98. R(x) =

x4

x2 − 1

9. r(w) =w√w − 2

10. C(w) =w√

w4 + 1

11. p(x) = 8x1/3 − x4/3 12. F (x) =19

�36x1/3 − x4/3

�

13. y =x4 − 24x2

2414. y =

x5

20− 2x3

3

In Exercises 15-18, sketch the graph of the function over the indicated interval. Find thecoordinates of the local extrema and points of inflection.

15. T (x) =4 sinx

2− cosx, [0, 2π] 16. S(θ) =

5 cos θ2− sin θ

, [0, 2π]

17. R(x) = 4x− tanx,�−π2,π2

�18. f(θ) = 3 tan 2θ − 8θ,

�−π4,π4

�

In Exercises 19-22, find a function with the given asymptotes. There are several possibleanswers.

19. Vertical asymptote x = 3, horizontal asymptote y = 1

20. Vertical asymptotes x = ±1, no horizontal asymptote

21. Vertical asymptote x = 0, slant asymptote y = x+ 1

22. Vertical asymptote x = −1, slant asymptotes y = x, y = −x

In Exercises 23-26, the graphs of y = f �(x) and y = f ��(x) are given. Sketch the graph off if the graph passes through the indicated point P .

23. P (0, 0) 24. P (0, 0)

y� f '�x�

y� f ''�x�1 x

1

3

y

y� f '�x�

y� f ''�x�1 2 x

1

2

y



25. P (0, 1) 26. P (1, 0)

y� f ''�x�

y� f '�x�1 x

1

2

y

y� f '�x�

y� f ''�x�

3 x�1

�3

y


Applications

v � s'�t�3 6.25 t

v

Figure for 27

27. Free-falling Object

A ball is thrown from a certain height s0 above the ground. The graph shown isthat of the velocity function v(t) in ft/sec over [0 sec, 6.25 sec], which is the timeinterval when the ball is above the ground. If v(3) = 0, find s0 and the maximumheight reached by the ball.

28. FM Radio

Sketch the graph and find all relative extrema of F (θ) = cos (θ + sin 2θ) over theinterval [0,π]. A function such as F is used in frequency modulation FM synthesis.

29. Fourier sine series

Find all the relative extrema of S(θ) =12sin θ +

16sin 3θ over the interval [0,π],

and sketch the graph of S. The function S is called a Fourier sine series and hasapplications in signal processing.

30. Harmonic oscillator

Find the critical numbers of the function y =34cos(4t) + cos(3t) over the interval

[0,π]. Moreover, show that the function satisfies the differential equation y��+16y =39 cos 3t. Such a differential equation is said to model the motion of a weight thatis attached to a spring.

Theory and Writing Proofs

31. Let y = ax2+bx+c be a quadratic function where a �= 0. Use calculus to determinethe coordinates of the vertex of the parabola. When is the graph concave upward?Concave downward?

32. Determine the local extrema and points of inflection of the cubic polynomial c(x) =x3 + kx2 where k is a constant.

33. The signs of the first two derivatives of y = G(t) on certain open intervals are listedbelow. If the graph of G passes through the points A(−1, 3), B(1, 1), and (3,−1),then sketch a possible graph of G.

Open interval G�(t) G��(t)(−∞,−1) + −(−1, 1) − −(1, 3) − +(3,∞) + +

34. The derivative of a polynomial y = p(x) is p�(x) = (x− 1)2(2−x). Find the relativeextrema and points of inflection of the graph of p.

35. Let p be a polynomial with an even degree n ≥ 2. Prove that p has at least onerelative extreme value.

3.7. OPTIMIZATION PROBLEMS 197

3.7 Optimization Problems

Solving Optimization ProblemsThe applications of derivatives in finding minimum and maximum values are useful insolving optimization problems. For instance, we may want to know the maximum area,shortest distance, maximum profit, and least cost among others. In the process, we haveto convert word problems into equations as the following examples illustrate.

Wall

y

x x

Figure 1aA rectangular enclosurewhere the dotted side repre-sents the side that uses nofencing material.

�125, 31250�

x

A

Figure 1bA sketch ofthe graph ofA = 500x− 2x2 in thefeasible domain [0, 250].

Example 1 Maximizing the Area of a Rectangle

A homeowner plans to build a rectangular enclosure along the side of her house using 500feet of fencing material, see Figure 1a. Find the dimensions of the enclosure that has thelargest area assuming no fencing material is needed along the side of the house.

Solution The area A of the rectangle is

A = xy. Primary equation

Since there are 500 feet of fencing material and only three sides of the rectangle requirefencing, we have

2x+ y = 500. Secondary equation

Substitutingy = 500− 2x. (6)

into the primary equation, we obtain

A = xy = x(500− 2x) = 500x− 2x2.

Since x and y must be nonnegative, using (6) we find that x ≤ 250. Then the feasibledomain of A as a function of x is [0, 250]. The graph of A as a function of x is seen inFigure 1b.

Now, we find the critical numbers of A in (0, 250).

dAdx

= 500− 4x = 0 Set the derivative to zero.

x = 125 Critical number

Next, evaluate A at the critical number and at the endpoints of [0, 125].

x 0 125 150A 0 31, 250 0

Thus, the maximum value of A occurs when x = 125 ft. Correspondingly, we get y = 250ft by using (6). Therefore, the dimensions of the rectangular enclosure with the largestpossible area are

125 ft by 250 ft.

✷

In Example 1, with the given 500 feet of fencing we must have realized that therewere infinitely many ways to construct a rectangular enclosure. For instance, we list a fewpossible dimensions of such rectangular enclosures with their corresponding areas.

Dimensions Area50 ft by 50 ft by 400 ft 20, 000 ft2

80 ft by 80 ft by 340 ft 27, 200 ft2

150 ft by 150 ft by 200 ft 30, 000 ft2


Area � 20,000 ft 2

400

50 50

Area � 27,200 ft2

340

80 80

Area � 30,000 ft2

200

150 150

Figure 2Rectangular enclosures with the same perimeters but different areas.

Try This 1Find the dimensions of the rectangle with the maximum possible area that has a perimeterof 100 feet.

�1,0�

�x,y�f �x�� x

x

y

Figure 3aFind the point (x, y) onthe graph of f(x) =

√x

that is nearest to (1, 0).

The following guidelines are useful in solving optimization problems, as illustrated inExample 1.

Guidelines for Solving Optimization Problems

1. Identify all quantities and draw a picture of the problem. A quantity could be a variableor constant such as area, perimeter, etc.

2. Write a primary equation that involves the variable to be optimized, i.e., maximizedor minimized.

3. Express the quantity to be optimized as a function of one variable. A substitution anda secondary equation may be needed to do this.

4. Find the feasible domain, i.e., the set of numbers for which the optimization problemmakes sense.

5. Find the maximum or minimum value of the quantity to be optimized using the calculustechniques in Sections 3.1 through 3.6.

Example 2 Minimizing the Distance from a Point to a Graph

Find the point on the graph of f(x) =√x that is closest to the point (1, 0). What is the

minimum distance between the graph of f and (1, 0)?

Solution Let d be the distance between (1, 0) and a point (x, y) on the graph of f , seeFigure 3a.

d =�

(x− 1)2 + (y − 0)2

=�

(x2 − 2x+ 1) + y2 Primary equation

Since (x, y) lies on the graph of f(x) =√x, we can substitute

y =√x Secondary equation

into the primary equation. Then we write

d(x) =�

(x2 − 2x+ 1) + (√x)2 =

�x2 − x+ 1.


�12,

32�

d�x�� x2 � x � 1

x

y

Figure 3bThe shortest distancebetween (1, 0) and thegraph of f is

d(1/2) =√3/2.

Using the domain of f(x) =√x, we find that the feasible domain of d(x) is [0,∞).

Applying the Chain Rule, we obtain

d�(x) =ddx

��x2 − x+ 1

�=

12(x2 − x+ 1)−1/2(2x− 1) =

2x− 1

2√x2 − x+ 1

.

Now, if d�(x) = 0 then

2x− 1 = 0

x =12

Critical number

Using the First Derivative Test, we find that the minimum value of d occurs when x = 1/2as seen in Figure 3b. Hence, the minimum distance between the graph of f and (1, 0) is

d(1/2) =

��12

�2

−�12

�+ 1 =

√32

.

Finally, the point on the graph of f(x) =√x that is closest to (1, 0) is

�12, f(1/2)

�=

�12,

�12

�, or

�12,

√22

�.

✷

Try This 2Find the point on the graph of f(x) =

√x that is closest to the point (4, 0).

P�x,y�y� 1� x2

�1 1 x

y

Figure 4aInscribing a rectangle in asemicircle with radius 1.

Example 3 Maximizing the Area of an Inscribed Rectangle

A rectangle is inscribed in the region bounded by the semicircle y =√1− x2 and the

x-axis, see Figure 4a. Find the length and width of the rectangle with the maximumpossible area.

Solution Let P (x, y) be a vertex of a rectangle such that P lies on the semicircle in thefirst quadrant, see Figure 4a. Since 2x and y are the length and width of the rectangle,respectively, the area of the rectangle is

A = 2xy. Primary equation

Sincey =

�1− x2, Secondary equation

we obtainA = 2x

�1− x2.

A feasible domain of A is [0, 1]. Using the Product Rule, we obtain

dAdx

= 2x · −2x

2√1− x2

+ 2�

1− x2

=−2x2

√1− x2

+ 2�

1− x2

=2√

1− x2

�−x2 + (1− x2)

�Factor

2√1− x2

dAdx

=2(1− 2x2)√

1− x2Simplify


� 2 �2,1�A�2x 1� x2

1 x

y

Figure 4bThe maximum possiblevalue of A is 1 and thisoccurs when x =

√2/2.

If dA/dx = 0, then

1− 2x2 = 0 or x =

√22

Critical number

The values of A at the critical number and the endpoints of [0, 1] are

A(0) = 0, A(√2/2) = 1, and A(1) = 0.

Thus, the maximum value of A occurs when x =√2/2, see Figure 4b. Hence, the length

and width of the rectangle with the maximum area are

length = 2x = 2

�√22

�=

√2

and

width = y =�

1− x2 =

�

1−�√

22

�2

=

�1− 1

2=

√22

.

✷

Try This 3A rectangle is inscribed in a region bounded by the semicircle y =

√16− x2 and the x-axis.

Find the length and width of the rectangle that has the maximum possible area.

Example 4 Minimizing the Total Length

A 3-feet high post and a 5-feet high post are 10 feet apart, see Figure 5a. The posts are tobe supported by two wires staked at a common point C in the horizontal ground betweenthe posts. Where should the wires be staked to minimize the sum of the lengths of thewires?

5 ft3 ftA

B C D

E

BD�10 ft

Figure 5aA 3-ft post and a 5-ft postare supported by wires thatare staked at point C.

Solution Let x = BC and y = CD be the lengths of the line segments shown in Figure5a. By the Pythagorean Theorem, we obtain

AC =�

9 + x2 and CE =�

25 + y2.

Then the sum L of the lengths of the wires is

L = AC + CE =�

9 + x2 +�

25 + y2. Primary equation

Since the posts are 10 feet apart, we have

y = 10− x. Secondary equation

Substituting into the primary equation, we find

L =�

9 + x2 +�

25 + (10− x)2

=�

9 + x2 +�

x2 − 20x+ 125.

See Figure 5b. Since x and 10 − x are nonnegative, let [0, 10] be the feasible domain ofL. Using the Chain Rule, we find

dLdx

=x√

9 + x2+

x− 10√x2 − 20x+ 125

.


L� 9� x2 � 25� �10� x�2

3.75 10 x

5

10

15

y

Figure 5bThe minimum total length Loccurs when x = 3.75 ft.

If dL/dx = 0, then

�x√

9 + x2

�2

=

�− x− 10√

x2 − 20x+ 125

�2

x2

9 + x2=

x2 − 20x+ 100x2 − 20x+ 125

Then cross-multiply and set one side of the equation to zero.

16x2 + 180x− 900 = 0

4(4x− 15)(x+ 15) = 0

x = 15/4 Critical number in [0, 10]

The values of L at the endpoints of [0, 10] and the critical number are

L(0) ≈ 14.2, L(3.75) ≈ 12.8, and L(10) ≈ 15.4.

Hence, the wires must be staked at a point in the ground that is x = 3.75 ft from the 3-ftpost.

✷

Try This 4If the posts in Example 4 are 16 feet apart, then where should the wires be staked tominimize the sum of the lengths of the wires?

Figure 6aA right circular cylinder witha top and bottom. Let r bethe radius and h the height.

A�2Πr2�2000�r

500Π

3r

300

600

A

Figure 6bThe graph of the surfacearea A as a functionof its radius r.

Example 5 Using the Least Material

Find the radius and height of a 1-liter can with the least surface area. Assume the can isa right circular cylinder that has a cover at the top and bottom.

Solution Let r and h be the radius and height of the can, respectively. Since the canhas a top and bottom, its surface area A is

A = 2πr2 + 2πrh. Primary equation

Since the volume is 1 liter or equivalently 1000 cm3, we find

πr2h = 1000. Secondary equation

Substituting

h =1000πr2

into the primary equation, we obtain

A = 2πr2 + 2πr

�1000πr2

�= 2πr2 +

2000r

.

Since r must be positive, the feasible domain of A is (0,∞) as in Figure 6b. Then

dAdr

= 4πr − 2000r2

=4πr3 − 2000

r2.

If dA/dr = 0, then

4πr3 − 2000 = 0

r = 3

�500π

=5 3√4π2

πCritical number


Applying the First Derivative Test, we find that the minimum surface area A occurs atthe critical number. Hence, the radius and height of the can with the least possible surfacearea are

radius = r =5 3√4π2

π≈ 5.4 cm, and

height = h =1000πr2

=1000

π

�5 3√4π2

π

�2 =10 3

√4π2

π≈ 10.8 cm.

✷

Try This 5Find the radius and height of a 1-liter can with the least surface area. Assume the can isa right circular cylinder that has a cover at the bottom but is open at the top.

3.7 Check-It Out

1. Gary wants to build a fence on two adjacent lots, as shown in Figure 1. No fencingis needed along the boundary represented by the dotted lines. If 1200 feet of fencingmaterial is available, then find x and y as indicated in the figure if the area enclosedis the maximum.

x

y

Figure for No. 1. The length from the left-most side to the

right-most side is y, and the width is x.

2. Find the minimum value of p if the primary equation is p = 2x+ y,secondary equation is xy = 32, and the feasible domain is x > 0.


1. The distance between (a, b) and a point (x, y) on the parabola y = x2 is

�(x+ a)2 + (x2 + b)2.

2. If a right circular cylinder has a cover at the top and bottom, then its surface areais S = 2πrh+ 2πr2 where r and h are the radius and height, respectively.

3. If the perimeter of a rectangle is 4 feet, then the area of such a rectangle is 1 ft2.

4. If the sum of two numbers is 10, then the maximum possible product of two suchnumbers is 100.

5. If the area of a rectangle is 12 square inches, then the dimensions of such a rectangleare 4 inches by 3 inches.


6. The two positive numbers that have a product of 15 and the minimum possible sum are√15 and

√15.

7. If two negative numbers differ by 1, then the minimum product of two such numbers does not exist.

8. A box with a square base and open top has a surface area of 1200 square inches. Then the shape of the boxthat has the maximum volume is a cube, i.e., the height of the box is the same as side of the base.

9. A rectangle is inscribed in a semicircle with radius 1, see figure below. If the indicated radial line makes anangle θ with the positive x-axis where 0 < θ < π/2, then the area of the rectangle is sin 2θ.

Θ�1 1 x

y

Figure for No. 9 and 10

10. In the above figure, the minimum perimeter of the inscribed rectangle is 2.


In Exercises 1-6, optimize the indicated variable given the primary andsecondary equations, and feasible domain.

1. Minimize D, primary equation D = (y−1.5)2 +x2, secondary equation y = x2, feasible domain −∞ < x < ∞.

2. Minimize D, primary equation D =�

(x− 1)2 + y2, secondary equation y = x+ 1,feasible domain −∞ < x < ∞.

3. Maximize A, primary equation A = l · w, secondary equation 2l + 2w = 20, feasible domain 0 ≤ w ≤ 10.

4. Maximize A, primary equation A =12bh, secondary equation b2 + h2 = 27, feasible domain 0 ≤ b ≤

√27.

5. Minimize S, primary equation S = 2πrh+ πr2, secondary equation r2h = π, feasible domain 0 < r < ∞.

6. Maximize V , primary equation V = πr2h, secondary equation 2πrh+ 2πr2 = 24π, feasible domain 0 < r < ∞.

Optimization Problems

7. Find the point on the graph of f(x) =√x that is nearest the given point.

a) (9, 0) b) ( 12 , 0)

8. Find the point on the graph of f(x) = 2√x that is closest to the indicated point.

a) (4, 0) b) (1, 0)

9. Find the number which exceeds its square by the greatest value.

10. Find the positive number that exceeds its cube by the greatest amount.


11. A homeowner plans to fence a rectangular region that must have an area of 30,000 square feet. In addition,

the homeowner plans to subdivide the region into three equal rectangles by inserting two smaller fences,

as seen below. Find the dimensions of the rectangular region that will use the least amount of fence.

y

x 2r

h

Figure for No. 11. Figure for No. 12.

12. A window consists of a rectangle and a semicircle that is mounted on top of the rectangle, see figure above.

Let h be the height from the base of the rectangle to the base of the semicircle. If the perimeter of the

window is 20 feet, find the radius of the semicircle so that the window has the maximum area.

13. From an aluminum square sheet, 12 inches on each side, a box with an open top is to be made by cutting

off small squares from each corner and bending up the sides. How large a square should be cut from each

corner so that the box has the maximum possible volume?

Figure for No. 13

14. The sum of the perimeters of a square and an equilateral triangle is 4 feet. Find the length of the side

of the square that minimizes the sum of the areas of the square and triangle.

15. A right triangle in the first quadrant is bounded by coordinate axes, and a line through point (4, 2), as seen

below. Find the coordinates of the vertices if the triangle has the minimum area. Does the triangle have

the minimum perimeter?

�4,2�x

y


16. Find the area of the largest isosceles triangle that can be inscribed in a circle with radius 2 units.

17. Find the dimensions of the right circular cylinder with a volume of 30 cubic inches and that has the

minimum surface area. Assume the cylinder has a top and bottom.


Figure for No. 18

18. A track and field has the shape of a rectangle and two semicircles attached to the sides of the rectangle. Theperimeter of the track and field is 400 meters. Find the length of the rectangle and the radius of the semicircleif the rectangular region has the maximum possible area.

19. Two cities A and B are at a distance of√116 miles from each other, see figure below. Let x = PC denote

the distance from P to C. The cities plan to build a recreation center C along a certain highway. Suppose theperpendicular distances from A and B to the highway are 3 miles and 7 miles, respectively. Find the value ofx that minimizes the sum AC + CB of the distances. The figure is not drawn to scale.

!

"

#

$%

&&'

(

A

B CP

3 km


20. A forest ranger must travel from point A to point C through some point P , see figure above. There is a roadfrom P to C that runs from west to east. The point B on the road nearest A is 3 km away, and the distancefrom B to C is 10 km. The forest ranger can travel diagonally from A to P at the speed of 15 kph, and drivefaster from P to C at 30 kph. How far from B should P be located so that the forest ranger travels the leastamount of time from A to P to B?

21. Mark plans to make a rectangular garden using x feet of fencing material. He figures that only three sides ofthe garden will be fenced and the fourth side will be left open without a fence. What is the maximum area ofthe garden that that he can enclose?

22. If the base b and perimeter p of a triangle are fixed, then determine the remaining two sides of the triangle withthe maximum area.

23. A crew is needed to service a manufacturing plant. It has been found that the number of hours needed to dothe job is 15 + 64/x where x is the size of the crew. If each worker is paid $16 per hour and the use of anequipment is $60 per hour, find the most economical size of the crew.

24. A field trip will cost each student $60 if 200 students join the trip. If more than 200 students join, then the costof the trip per student will be reduced by 25 cents times the number of students in excess of 200. How manystudents should join the trip to realize the largest gross income?

25. A contractor wishes to find the dimensions of a rectangular piece of land where she can build a house that willhave an area of 22,000 square feet. Moreover, the land must have 50 feet of space from the exterior of the houseto the front and back property lines, and 27.5 feet of space from the exterior to the left and right propertyboundaries. Find the dimensions of the land with the least area on which the house can be built.

26. What are the dimensions of the rectangle with the greatest area that can be inscribed in a triangle with base40 ft and height 15 ft, if one side of the rectangle lies on the base of the triangle? See figure below.

xy

Figure for No. 26

27. Find the minimum distance between a point (p, q) and a line y = mx+ b.


28. A right circular cone is made by cutting a sector from a unit circle, see figure below. Let θ be the central

angle of the sector. Find θ if the cone has the maximum volume.

!


29. A rain gutter is made by folding up the edges of an 18-inch metal sheet, see figure above. Suppose each of the

left and right edges of the gutter are 6 inches long. Find the angle that the edges must be folded so that the

gutter holds the maximum volume.

30. Find the volume of the largest circular cone that can be inscribed in a sphere of radius r.

31. A 4-by-6 rectangular card is folded in such a way that vertex A is placed at some point B on the opposite

longer side, see figure below. Let c be the length of the crease that is formed. Find the minimum value of c.

x

x

c

A

B

R1 x

1

y


32. A rectangle region R in the first quadrant has a vertex at the origin, the opposite vertex is a point in the

parabola y = (x− 1)2, and two sides that lie in the coordinate axes, see figure above. Find the maximum

area of such a rectangle.

33. A rope of length L is formed into a sector of a circle. Find the central angle of the sector that gives the

maximum area of the sector.

Odd Ball Problems

34. Determine the constant C if G(x) = x2 +Cx

has a relative minimum at x = 3.

35. Determine the constant K if the function H(t) = t3 +Kt

has points of inflection at t = ±1

36. Find the maximum and minimum values of f(x) = A cos(2x) +B cosx assuming B ≥ 4|A| > 0.

37. Find the maximum and minimum values of f(x) = A cos(2x) +B cosx assuming 4|A| > B > 0.

38. A circle of radius 2 is centered at (0, 2). A second circle of radius 3 lies in the first quadrant, and is tangent

to the first circle and x-axis. In the figure above, we see a third circle that is tangent to the first and second

circles. If the center of the third circle lies in the y-axis, find its radius. What is the limit of the common point

between the first and third circles as the radius of the third circle increases to ∞?

3.8. NEWTON’S METHOD 207

3.8 Newton’s Method

• Aproximating a Real Zero of a Function • A Sufficient Condition for Con-vergence

Aproximating a Real Zero of a Function

x2 x1c x

y

Figure 1a

�x1, f �x1��x2, f �x2��

x2 x1c x

y

Figure 1bNewton’s Method findssequential approximationsx1, x2, x3, ... to anx-intercept of the graphof a function.

x2

f �x��x4�2x1 x

y

Figure 2The tangent line at thepoint where x1 = 1 hasx-intercept x2 = 1.25.

The solutions of f(x) = 0 are not easy to solve. In fact not all solutions are necessarilyreal numbers. If c is a real zero of f , we can approximate c to any degree of precisionunder certain conditions. We will discuss and apply Newton’s method to approximatec.

The first step is to choose an arbitrary number x1 that is approximately c, as in Figure1a. The graph of f will help you choose x1. The tangent line at (x1, f(x1)) is given by

y − f(x1) = f �(x1)(x− x1)

y = f �(x1)(x− x1) + f(x1).

If (x2, 0) is the x-intercept of the tangent line, then

0 = f �(x1)(x2 − x1) + f(x1)

x2 = x1 −f(x1)f �(x1)

if f �(x1) �= 0

Now, we have two approximations for c, namely, x1 and x2. We continue the process.Similarly, we can find the x-intercept (x3, 0) of the tangent line at (x2, f(x2)). In general,the nth approximation to c is

xn+1 = xn − f(xn)f �(xn)

if f �(xn) �= 0

The repeated applications of the above procedure is called Newton’s Method.

A Summary of Newton’s Method

Let f be a differentiable function such that f(c) = 0.

Step 1. Find an initial estimate x1 of c.

Step 2. The (n+ 1)st approximation is


.

Step 3. If |xn+1 − xn| is less than the desired accuracy, let xn+1 be the final estimate ofc. Otherwise, return to Step 2 and find a new approximation.

Under certain conditions, the approximations become more precise as n increases.

Example 1 Approximating 4√2

Apply Newton’s Method to approximate the real root of x4 − 2 = 0 as follows: use x1 = 1and estimate x4.

Solution Let f(x) = x4 − 2. Since f �(x) = 4x3, the iterative process is given by theformula


= xn − x4n − 24x3

n.


Since x1 = 1, we find

x2 = x1 −x41 − 24x3

1

= 1− 14 − 24 · 13

= 1.25.

Then the third estimate is

x3 = x2 −x42 − 24x3

2

= 1.25− (1.25)4 − 24(1.25)3

≈ 1.1935

Continuing the process, we obtain

x4 = x3 −x43 − 24x3

3

≈ 1.18923.

The first iteration of Newton’s Method is shown in Figure 2. Using a calculator, we find4√2 ≈ 1.189207 which agrees with x4 up to four decimal places

✷

Try This 1

Repeat Example 1 but approximate the real root of x3 − 2 = 0.

Example 2 Finding a Zero of a Function

The functionf(x) = x5 − x4 + x− 3

has a zero in [1, 2], see Figure 3. Apply Newton’s Method with x1 = 1.5 to find xn+1 if|xn+1 − xn| < 0.001. Then find the value of f(xn+1).

�1.5, f �1.5��

1 2 x

3

�3

y

Figure 3The tangent line to thegraph of f at the pointwith x1 = 1.5 hasx-intercept x2 = 1.41951.

Figure 4

The graph of

f(x) = x3 − 2x+ 2.

Solution Since f �(x) = 5x4 − 4x3 + 1, the iterative process is given by the formula


= xn − x5n − x4

n + xn − 35x4

n − 4x3n + 1

.

Using a calculator with x1 = 1.5, we obtain the following estimates

x2 ≈ 1.41951

x3 ≈ 1.40705 |x3 − x2| ≈ 0.01246

x4 ≈ 1.40678 |x4 − x3| ≈ 0.00027

We stop iterating since |x4 − x3| < 0.001. Then the zero of f(x) in [1, 2] is approximatelyx4 ≈ 1.40678. Finally, x4 satisfies

f(x4) ≈ 1.406785 − 1.406784 + 1.40678− 3 ≈ 0.000001.

✷

Try This 2

The function g(x) = cosx− x has a zero in [0, 1]. Apply Newton’s Method to find xn+1 ifx1 = 1 and |xn+1 − xn| < 0.0001.


Newton’s Method does not always guarantee that the values of


will approximate a zero of f(x) to any accuracy desired. For instance, let f(x) = x3−2x+2whose graph is given in Figure 4. The tangent line at (1, 1) has x-intercept (0, 0), andthe tangent line at (0, 2) has x-intercept (1, 0). Applying Newton’s Method with initialestimate x1 = 1 will generate the following values: x1 = x3 = x5 = · · · 1 and x2 = x4 =x6 = · · · 0.

A Sufficient Condition for ConvergenceThe next theorem will give a condition when Newton’s Method is guaranteed to work.That is, the values of xn’s generated by the method will approximate a given zero c of afunction to any desired precision. In such a case, we write

limn→∞

xn = c.

Also, we say xn converges to c as n → ∞. The proof of the theorem relies on the MeanValue Theorem. Before stating the theorem, if a < b, we observe that the interval [a, b] isthe middle third of [2a− b, 2b− a].

Theorem 3.11Let f(c) = 0 and suppose ��

f(x)f ��(x)(f �(x))2

�� ≤ α < 1 (7)

for all x in (c − β, c + β) for some positive constants α,β. If x1 lies in [c − β, c + β] and

xn+1 = xn − f(xn)f �(xn) , then lim

n→∞xn = c.

Proof Let N(x) = x− f(x)f �(x)

. Observe N(c) = c and

N �(x) = 1− (f �(x))2 − f(x)f ��(x)(f �(x))2

=f(x)f ��(x)(f �(x))2

.

In particular, |N �(x)| ≤ α for all x in (c− β, c+ β) . Applying the Mean Value Theorem,we find

|x2 − c| = |N(x1)−N(c)| ≤ α |x1 − c| .Since 0 < α < 1 and |x1−c| ≤ β, it follows that |x2−c| < β. Thus, x2 lies in (c−β, c+β).Similarly, by the Mean Value Theorem we obtain

|x3 − c| = |N(x2)−N(c)| ≤ α |x2 − c| ≤ α2|x1 − c|.

In general, for any integer n ≥ 1 we have

|xn+1 − c| ≤ αn|x1 − c|

and xn+1 lies in (c− β, c+ β).Let ε > 0 be a given degree of accuracy. Then |xn+1 − c| < ε whenever n is sufficiently

large . Hence, limn→∞

xn = c.

✷


Example 3 Convergent Approximations Using Newton’s Method

Let f(x) = x3 − 3 and x1 = 43 . Prove the values of xn generated by Newton’s Method

satisfylim

n→∞xn = 3

√3

Solution The graph of f is seen in Figure 4a, and 3√3 is the real zero of f(x). The first

two derivatives aref �(x) = 3x2 and f ��(x) = 6x.

Then inequality (7) in Theorem 3.11 reduces to��f(x)f ��(x)[f �(x)]2

�� =��(x3 − 3)(6x)

9x4

�� =23

��1−3x3

��

The graph of g(x) = 23

��1− 3x3

�� is shown in Figure 4b. We find g(2) = 512 , g(

43 ) < g(2),

and g( 3

�65 ) = 1.

1 43

53

x

1

y

43

53 26

53

x

512

1

y

Figure 4a: f(x) = x3 − 3. Figure 4b: g(x) =23

��1−3x3

��.

From Figure 4b, we see that g(x) < 512 for all x in [ 43 , 2]. Let β = 3

√3− 4

3 > 0. Then

g(x) < 512 for all x in ( 3

√3− β, 3

√3 + β). Thus, we can apply Theorem 3.11 with x1 = 4

3 .

Hence, the values of xn generated by Newton’s Method satisfy limn→∞

xn = 3√3.

✷

Try This 3

Let f(x) = x2 − 60 and x1 = 8. Prove the values of xn generated by Newton’s Methodsatisfy lim

n→∞xn =

√60.

Example 4 Convergent Approximations Using Newton’s Method

Let f(x) = cos(x) − x and x1 = 1. If f(c) = 0, prove the values of xn generated byNewton’s Method satisfy lim

n→∞xn = c. Then find x2 to the nearest thousandth.

Solution The graph of f is seen in Figure 5, and its zero c is near 34 . The first two

derivatives aref �(x) = − sin(x)− 1 and f ��(x) = − cosx.

Then inequality (7) in Theorem 3.11 reduces to

g(x) =

��f(x)f ��(x)[f �(x)]2

��

=

��(cos(x)− x) cosx

(1 + sinx)2

��


In Figure 5, we find g(x) ≤ 12 for all x in [ 12 , 1].

y� f �x� y�g�x�12

34 1

x

12

�12

y

Figure 5: f(x) = cos(x)− x and g(x) =

��f(x)f ��(x)(f �(x))2

��.

Then Theorem 3.11 applies with x1 = 1 or for any x1 in [ 12 , 1] . Thus, the values of xn generated by Newton’s Methodwill satisfy lim

n→∞xn = c. Finally, since x1 = 1, we find

x2 = x1 −f(x1)f �(x1)

= 1− cos(1)− 1− sin(1)− 1

≈ 0.739

✷

Try This 4

As in Example 4, but f(x) = sinx+ x− 2. As done earlier, let x1 = 1.

An equation f(x) = 0 is said to be solvable by radicals if its solutions can be determined in a finite number ofsteps using addition, subtraction, multiplication, division, and the radicals n

√x. For example, ax2 + bx+ c = 0 with

a �= 0 is solvable by radicals for the solutions are

x =−b±

√b2 − 4ac2a

.

Moreover, 3rd and 4th degree polynomials are solvable by radicals as proved by Geronimo Cardano (1501-1576). Forinstance, the real solution to x3 + px+ q = 0 where p > 0 is

x =3

��q2

4+

p3

27− q

2− 3

��q2

4+

p3

27+

q2

The verification of this cubic solution is an exercise at the end of this section. However, the zeros of 5th and higherdegree polynomials are not solvable by radicals. The latter fact was established by Evariste Galois (1811-1832).

Newton’s Method is a robust numerical algorithm for solving equations whether solvable by radicals or not.

3.8 Check-It Out

Use Newton’s Method to find xn given n, x1, and f(x) = 0.

1. n = 2, x1 = 1, x3 + x− 1 = 0 2. n = 3, x1 = 1, x2 − 2 = 0


1. In Newton’s Method, the (n+ 1)st estimate is xn+1 = xn − f(xn)f �(xn)

2. If f(x) = mx + b, m �= 0, and x1 is any real number, then the second value generated by Newton’s Methodsatisfies f(x2) = 0.

3. If p(x) is a 3rd degree polynomial and x1 = 0, then the values xn generated by Newton’s Method converge to azero of p(x).

4. Let x1 be a first estimate of the zero of f(x) = 1x −a. Then the second estimate of the zero of f(x) by Newton’s

Method is x2 = x1(2− ax1).


5. Let x1 = 1 be a first estimate of the zero of f(x) = x2 − a. By Newton’s Method the second estimate of thezero is x2 = 1−a

2 .

6. Let f(c) = 0 and suppose f ��(x) is continuous. If�� f(c)f

��(c)(f �(c))2

�� < 1, then there exists x1 satisfying limn→∞

xn = c.


In Exercises 1-8, apply Newton’s method using the given x1 to find the next two estimates x2 and x3 that solves theindicated equation f(x) = 0. Round the answers to the nearest thousandth.

1. x1 = 2, x2 − x− 5 = 0 2. x1 = −1, x2 − 2x− 10 = 0 3. x1 = 2, x3 − 2 = 0

4. x1 = 1, x5 − 2 = 0 5. x1 = 2, cos(πx)− x = 0 6. x1 = 3, 2 sin(x)− x+ 2 = 0

7. x1 = 3, sin(x) + cos(x) + 1 = 0 8. x1 = 14 , cos(2πx)− πx = 0

In Exercises 9-12, explain why Newton’s Method fails to find a zero of f(x) given the initial estimate x1.

9. x1 = −1, f(x) = x3 − 5x 10. x1 = 1, f(x) = 3√x 11. x1 = 1, f(x) =

� √x if x ≥ 0

−√−x if x < 0

12. x1 = −3, f(x) = 2x3 + 9x2 − 1 13. x1 = 5, f(x) = 1x − 1 14. x1 = 2.5, f(x) = 1

x2 − 1

In Exercises 15-18, you are given an equation f(x) = 0 and an initial estimate x1 of a zero of f(x). Apply Newton’sMethod to find xn+1 that satisfies |xn+1 − xn| < 0.001. Round xn+1 to four decimal places.

15. x1 = 2, x3 − 5 = 0 16. x1 = 4, x5 + 15x− 1000 = 0

17. x1 = 3, sin(x6 )−12 = 0 18. x1 = 3, tan(x3 )−

√3 = 0

In Exercises 19-22, given x1 show that the values xn generated by Newton’s Method converges to a zero of the givenfunction f . Then approximate x4 to the nearest thousandth. Apply Theorem 3.11 and see Examples 3 and 4.

19. x1 = 3, f(x) = x2 − 10 20. x1 = −1, f(x) = x3 + 5x+ 10

21. x1 = 0.35, f(x) = cos(πx2) 22. x1 = 7, f(x) = x2 sin(x)− 2x

Theory and Proofs

23. Let f(x) = x3 + x+ 1 and let x1 be any real number. Prove limn→∞

f(xn) = 0 where xn+1 = xn − f(xn)f �(xn) .

24. Cardano’s Formulas Let x3 + px+ q = 0 where p > 0.

a) Verify the identity: (a− b)3 + 3ab(a− b)− (a3 − b3) = 0

b) If a =3

��q2

4 + p3

27 − q2 and b =

3

��q2

4 + p3

27 + q2 , prove 3ab = p and a3 − b3 = −q

c) Using the notation in part b), show x = a− b is the only real solution of x3 + px+ q = 0.

25. Let f(x) = ax2 + bx+ c, a > 0, and b2 − 4ac ≥ 0. If x1 �= − b2a and xn+1 = xn − f(xn)

f �(xn) , prove limn→∞

f(xn) = 0.

26. Let f(c) = 0 where a < c < b. Suppose for some α we have��f(x)f ��(x)(f �(x))2

�� ≤ α < 1

for all x in (2a− b, 2b− a). If x1 lies in [a, b] and xn+1 = xn − f(xn)f �(xn) , prove lim

n→∞xn = c.

3.9. LINEAR APPROXIMATIONS AND DIFFERENTIALS 213

3.9 Linear Approximations and Differentials

• Linear Approximations • Differentials • Approximating a Functional Value• Propagated Error

Linear AproximationsIf f is differentiable at c, the tangent line to the graph of f at (c, f(c)) is

L(x) = f �(c)(x− c) + f(c). (8)

The linear function in (8) is called the linear approximation, linearization, or tangentline approximation of f at c. Clearly, L(x) → f(c) as x → c. Also, f(x) → f(c) asx → c by continuity. Combining, we obtain

f(x) ≈ f �(c)(x− c) + f(c). (9)

whenever x is near c.

Example 1 Finding a Linear Approximation

Find the tangent line approximation of f(x) = sinx at (0, 0).Then approximate the values of sin(0.1) and sin(0.05).

f �x�� sin �x�y � x

0.5 x

0.5

y

Figure 1The tangent line y = xapproximates f(x) = sinxnear x = 0.

Solution Note, f �(x) = cosx and f �(0) = cos 0 = 1. Using (8) with c = 0, the tangentline to f(x) = sinx at (0, 0) is given by

L(x) = f �(0)(x− 0) + f(0)

= 1(x− 0) + 0

L(x) = x.

Thus, the tangent line approximation to f(x) = sinx at (0, 0) is

sinx ≈ x provided x is near 0.

Hence, sin(0.1) ≈ 0.1 and sin(0.05) ≈ 0.05. These are good approximations for sin(0.1) ≈0.09983 and sin(0.05) ≈ 0.04998 by using a calculator.

✷

Try This 1

Apply (8) to find the linearization of f(x) = tanx at (0, 0).Then apply (9) to approximate the value of tan( π

24 ) using c = 0 and x = π24 .

Differentials

Definition 11 The Differential of a Function

Let y = f(x) be differentiable on an open interval containing c.

a) The differential dx of x is an independent variable.

b) The differential dy of y is a function of x and dx, and defined by

dy = f �(x)dx.


Also, the independent variable dx is denoted by �x and is called the change in x. Inthe definition of f �(x) in Section 2.1 we encountered the expression

�y = f(x+�x)− f(x)

We call �y the change in y. If �x = x− c, then the linearization in (9) may be rewrittenas follows:

f(x) ≈ f �(c)(x− c) + f(c) for x near c

f(c+�x) ≈ f �(c)�x+ f(c) for �x ≈ 0

f(c+�x)− f(c) ≈ f �(c)�x

Since c is an arbitrary number, we may replace c by x:

f(x+�x)− f(x) ≈ f �(x)�x for �x ≈ 0 (10)

�y ≈ dy

In Figure 2, we see a (dotted) tangent line at x to the graph of f .

y� f �x�

Tangent line at x

�y dy

x x��xx

y

Figure 2: �y ≈ dy if �x ≈ 0.

The quantity �y represents a change in the functional values of f . While dy representsa change in the functional values of the tangent line.

f �x�� 1� x

Tangent line at 3

�1 3 x

2

y

Figure 3dy is the change in the valuesof the tangent line, and �yis the change in the values ofy = f(x).

Example 2 Comparing �y With dy

If y =√1 + x, evaluate dy and �y if x = 3 and �x = 0.1.

Solution Let f(x) = (1 + x)1/2. By the Chain Rule we find

f �(x) =1

2√1 + x

.

Then we obtain

dy = f �(x)dx =dx

2√1 + x

and

�y = f(x+�x)− f(x)

=�

1 + x+�x−√1 + x.

Substitute x = 3 and dx = �x = 0.1. Then we find

dy =0.1

2√4= 0.025

and�y =

√4.1−

√4 ≈ 0.0248.

Finally, note dy ≈ �y since dx = 0.1 is approximately zero. In Figure 3, we seethe tangent line at x = 3.

✷


Try This 2

If y =1√x, evaluate dy and �y given x = 1 and �x = 0.1.

Approximating a Functional Value

Differentials may be used to approximate the values of a function. We rewrite (10) asfollows:

f(x+�x) ≈ f(x) + f �(x)�x (11)

where �x ≈ 0. The above approximation shows how to estimate f(x +�x) if we knowthe values of f(x), f �(x), and �x.

Example 3 Approximating a Square Root

Approximate√10 by applying (11).

Solution In (11), let f(x) =√x, x = 9, and �x = 1. Then

√10 = f(10) = f(9) + f �(9)�x

= 3 +16(1) for f �(x) =

1

2√x

Thus, √10 ≈ 19

8= 2.375

The latter decimal is obtained by using a calculator. ✷

Try This 3

Estimate√15 by applying (11) with x = 16 and �x = −1.

f �x��tan �x�Π4

x

1

y

Figure 4The linear approximationat π

4 underestimatesf(x) = tanx since thetangent line lies belowthe graph of f locally.

Example 4 Approximating a Functional Value

Approximate tan 47◦ by applying (11).

Solution We recall

tan(45◦) = tan�π4

�= 1 and 2◦ = 2

� π180

�=

π90

.

Let x = π4 , dx = π

90 , and f(x) = tanx. Then f �(x) = sec2 x .Applying (11) we find

f�π4+

π90

�≈ f

�π4

�+ f �

�π4

� π90

= tan�π4

�+ sec2

�π4

� π90

= 1 + 2� π90

�

f�π4+

π90

�= 1 +

π45

Since tan(47◦) = f�π4 + π

90

�, we obtain

tan(47◦) ≈ 1 +π45

≈ 1.0698

The latter decimal is obtained by using a calculator. ✷


Try This 4

Estimate the value of sin 29◦ by applying (11) with x = π6 and �x = − π

180 .

Propagated ErrorMany scientific experiments involve approximations of position, rate, etc. Suppose x is anapproximation, and x+�x is the true value of a quantity. We say �x is the error in theapproximation.

Let y = f(x) be a function of x. The propagated error is defined as |�f | =

|f(x+�x)− f(x)|, the relative error in f(x) is given by�� ff(x)

��, and the percentage

relative error in f(x) is�� ff(x)

�� · 100%.

Example 5 Approximating a Propagated Error

The approximate radius of a sphere is 6 in. with an error of at most ±0.05 in. of the trueradius. Use differentials to approximate the propagated error and the percentage relativeerror in calculating the surface area of the sphere.

Figure 5A sphere with atrue radius of 6 +�x.

Solution The surface area of a sphere of radius x is

f(x) = 4πx2.

If the true radius is 6 +�x, then the propagated error is

|�f | = |f(6 +�x)− f(6)|

where −0.05 ≤ �x ≤ 0.05.Note f �(x) = 8πx. Applying the differential in (10), we obtain

|f(6 +�x)− f(6)| ≈��f �(6)�x

��

= |48π�x|≈ 48π(0.05)

≈ 7.5 sq. in.

Thus, the propagated error is |�f | ≈ 7.5 sq. in. Finally, the percentagerelative error is ��

�ff(6)

�� 100 ≈ 48π(0.05)144π

· 100% ≈ 1.7%.

✷

Try This 5

A rectangular box has a square base with an exact area of 16 sq. in. The height of thebox is 6 in. with an error of at most ±0.1 in. Find the propagated error and percentagerelative error in the volume of the box.


3.9 Check-It Out

1. Find the linear approximation of f(x) =√x at (4, 2). Then use it to estimate

√4.1.

2. Let y = tanx. Evaluate dy if x = π4 and dx = 0.005.

3. The side of a cube is measured to be 12 inches with an error of at most ± 116 inch. Use differentials to find the

propagated error in the volume of the cube.


1. If L(x) is the linear approximation of y = f(x) at x = c,then L(c) = f(c) and L�(c) = f �(c).

2. f(c+�x) ≈ f(c) + f �(c)�x whenever �x ≈ 0

3. �f = f(x+�x) + f(x)

4. If y =�

f(x), then dy =�

f �(x)dx.

5. If x > 0 and �x ≈ 0, then√x+�x ≈ √

x+ �x2√x.

6. For any x, (x+�x)5 − x5 ≈ (�x)5 if �x ≈ 0

7. If x is the measurement of the side of a cube with a possible error of �x, then the propagated error in thevolume of the cube is

��(x+�x)3 − x3��.

8. If L(x) is the linear approximation of y = f(x) at x = c, then f(x) ≈ L(x) for values of x near c.

9. The linear approximation of f(x) = mx+ b at x = 0 is L(x) = mx.

10. The tangent line approximation of f(x) = cos(x+ π2 ) at x = 0 is L(x) = −x.


In Exercises 1-10, find the linear approximation of f(x) at the given point. Then use it to estimate the indicatednumber. See Example 1.

1. f(x) = x3, (2, 8); estimate (2.1)3 2. f(x) =1x2

, (2, 14 ); estimate 1

(1.9)2

3. f(x) =√x, (289, 17); estimate

√285 4. f(x) = 3

√x, (64, 4); estimate 3

√70

5. f(x) = tanx, (π4 , 1); estimate tan(π4 + 0.1) 6. f(x) = secx, (π6 ,2√3); estimate sec(π6 + 0.2)

7. f(x) = 3√8− x, (0, 2); estimate 3

√7 8. f(x) =

2√x+ 16

, (0, 12 ); estimate 2√

17

9. f(x) = cosx, (π3 ,12 ); estimate cos(61◦) 10. f(x) = sinx, (π4 ,

√2

2 ); estimate sin(44◦)

In Exercises 11-16, evaluate �y and dy for the given function and constants x and �x = dx. See Example 2.

11. y = x2, x = 5, �x = 0.1 12. y =1√x, x = 4, dx = 1

10

13. y = cotx, x = π4 , dx = π

180 14. y = cscx, x = π6 , �x = π

90

15. y = |x|3, x = −2, �x = 0.01 16. y = 1x − [[ 1x ]], x = 3

4 , �x = 0.01

In Exercises 17-22, use differentials to approximate the following. See Examples 3 and 4.

17.√26 18. 3

√7 19. sin 121◦

20. sec 59.5◦ 21. (2.01)5 22. (27.5)2/3


23. The side of a square is measured to be 12 inches with a possible error of at most 0.125 inches. Use differentialsto approximate the propagated error in the area of the square. Approximate to the nearest square inch.

24. The radius of a circle is measured to be 2 in. with a possible error of at most 0.0625 in. Use differentials toapproximate the propagated error in the area of the circle. Approximate to the nearest thousandth of a sq. in.

25. A sphere has a radius that is measured to be 20 in. with an error of at most 0.01 in. Use differentials toapproximate the propagated error in the volume of the cube. Round to the nearest cubic inch.

26. A cube has six faces. If an edge is measured to be 5 inches with an error of at most 0.05 inch, use differentials toapproximate the propagated error in the surface area of the cube. Round to the nearest hundredth of a squareinch.

27. A projectile is fired at an angle of elevation of 60◦ with an error of at most ±1◦. Let v0 = 800 ft/sec be the initialvelocity of the firing. Use differentials to approximate the propagated change in the range of the projectile.

Assume the range is R =v20

32 sin(2θ) where θ is the angle of elevation. Round to the nearest foot.

28. A circular cylinder has a height of 6 inches. The radius is 2 inches with a possible error of �x. If the propagatederror in the volume is 1 cubic inch, use differentials to approximate �x. Round to the nearest thousandth ofan inch.

29. A sphere has a radius of 5 inches with a possible error of �r. If the propagated error in the surface area of thesphere is 10 square inches, use differentials to approximate �r. Round to the nearest hundredth of an inch.

30. A homeowner is standing 10 ft from a tree. She knows that the true angle of elevation θ of the top of the treeis between 45◦ and 60◦. She likes to approximate the height of the tree to the nearest foot. Use differentialsto estimate the maximum error �θ for measuring the true angle of elevation? Round to the nearest tenth of adegree.

Chapter 3 Multiple Choice Test

Choose the best answer.

1. If f(c) = 10 and f �(c) does not exist, then

A. c is a critical number of f B. x = c is an asymptote of the graph of f

C. f(c) is a relative extremum of f D. The point (c, f(c)) is an inflection point of the graph of f

2. The graph y =2x

x+ 2has an asymptote, namely

A. x = −2 B. y = 1 C. x = 0 D. None of the above

3. The function f(x) = x2 increases on the interval

A. (−∞,∞) B. (−∞, 0) C. (0,∞) D. Nowhere

4. The value of the second derivative of f(x) = x3 at x = −1 is

A. −3 B. 3 C. −6 D. 6

5. limx→∞

2x2

x2 + 3equals

A. ∞ B. 2 C. −2 D. 3

6. limx→−∞

3x3 − 24x3 + 1

equals

A. ∞ B. −∞ C.34

D. −2

7. The horizontal tangent line to the graph of y = x3 − 1 passes through the point

A. (0, 0) B. (0,−1) C. (0, 1) D. (1, 1)

8. In (−∞, 0) ∪ (0,∞), the function y =1x

is

A. Increasing B. Decreasing C. Constant D. Increasing and decreasing

9. The graph of y = 3√x has a point of inflection at x equals

A. −1 B. 1 C. 0 D. ∞


10. If f(x) =x− 22− x

, then f can be made continuous everywhere by defining f(2) equal to

A. 0 B. 1 C. −1 D. 2

11. The graph of y = x(x2 − 1) is concave downward on the interval

A. (−∞, 0) B. (0, 1) C. (0,∞) D. (−1, 1)

12. If y = secx(1 + sinx) thendydx

at x =π4

equals

A. 0 B. 1 +√2 C. 2 D. 2 +

√2

13. If xy + y2 = 1 defines y as an implicit function of x, thendydx

equals

A.−1

x+ 2yB.

−2yx

C.−y

x+ 2yD. None of the above

14. If y = u1/3 and u = tanx, thendydx

at x =π4

equals

A. 1 B.√2 C.

23

D. ∞

15. If f(x) = x|x|, then f �(x) equals

A. 2x B. −2x C. 2|x| D. −2|x|

16. If f(x) = x7, then limx→1

f(x)− f(1)x− 1

equals

A. 1 B. 7 C. −1 D. Does not exist

17. The slope of the curve xy2 − x− 3 = 0 at the point (1, 2) is

A. 0 B. −1 C. −34

D.34

18. limh→0

cos�π2 + h

�

hequals

A. −1 B. 1 C. 0 D. Does not exist

19. The graphs of 2x+ y = 1 and x− 2y = 1 are

A. Parallel B. Perpendicular C. Coincident D. None of the above

20. The shortest distance between the graph of y = 12− x2 and the origin is

A.√47 B.

√472

C.√23 D.

√232

Investigation Projects

Inventory: Delivery in One Batch per Order

We store goods in an inventory in order to meet a demand. If a company has a large inventory, the cost of holding thegoods is high. However, with a low inventory the demand may not be satisfied, customers can go to the competition,and there is a loss in revenue.

Leonardo, Inc., is a bookstore with an annual demand of D units for a certain children’s book. The bookstore submitsseveral orders for the book in a year to a third party publisher. The size of an order is Q units. Given an order thepublisher delivers the books to the bookstore in one batch.

Then the ratio D/Q represents the number of orders. Let Co be the cost to the bookstore for preparing an order.The annual ordering cost is

O1(Q) = (Number of orders per year)× (Ordering cost)

=DQ

× Co


Since the inventory level fluctuates due to the demand, we assume the average annual inventory is one-half the highestinventory. That is, the average annual inventory is Q/2 units. Let Ch denote the annual holding cost per unit. Thenthe annual holding cost is

H1(Q) = (Average inventory )× (Holding cost) =Q2

× Ch

To minimize the annual inventory cost we do not consider the cost of the books since the demand is constant. Theannual inventory cost is

I1(Q) = O1(Q) +H1(Q)

Let Q∗1 be the order quantity that minimizes I1(Q). Note, Q∗

1 is a critical number of I(Q). In Management Science,Q∗

1 is called the Economic Order Quantity.

Exercises

1. Prove O1(Q∗1) = H1(Q

∗1), i.e., annual ordering cost equals holding cost. 2. Show that Q∗

1 =

�2DCo

Ch

3. For the bookstore Leonardo, Inc., suppose the annual demand is 10, 000 units, ordering cost is $50 per order,and holding cost is $0.25 per book/year. Determine the economic order quantity Q∗

1.

The Production Run ModelA production run will be a period when goods are produced and sold simultaneously. There is a setup cost Cs

associated to a production run. The setup cost replaces the ordering cost in the above inventory model.

Let Q be the number of units produced during a production run. Denote the production rate by p units/day, andthe demand rate by d units/day. If t is the length of a production run, then Q = pt. We declare the variables:

(Annual demand) = D units

(Number of units per production run) = Q units

(Setup cost) = Cs per production run

(Carrying cost) = Ch per unit per year

(Daily production rate) = p units/day

(Daily demand rate) = d units/day

(Length of a production run) = t days

Then the annual setup cost is

Sp(Q) = (Number of production runs)× Cs

=DQ

× Cs

Similarly, the annual holding cost depends on one-half of the maximum inventory. The annual holding is calculatedas follows:

Hp(Q) =(Maximum inventory)

2× Ch =

pt− dt2

× Ch

=Q2

�1− d

p

�× Ch since t =

Qp

The annual inventory cost isIp(Q) = Sp(Q) +Hp(Q).

Let Q∗p be the optimal value of Q which minimizes Ip(Q).

Exercises

4. Prove Sp(Q∗) = Hp(Q

∗) 5. Show that Q∗p =

��2DCs

Ch

�1− d

p

� .

chapter 3 · 2018. 9. 18. · chapter 3 applications of derivatives 3.1 extrema of functions on...

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