chapter 4 derivatives and their applications€¦ · derivatives and their applications in this...

Chapter 4

Derivatives and TheirApplications

In this chapter, we will study one of two central concepts in Calculus- thederivative of a function. We will also show some applications of the derivative.

4.1 Definition of the Derivative

Definition 4.1.1. (a) Let f be a function, and let x be in the domain of f .Let ∆x be any change (increase or decrease) in the value of x, so thatx + ∆x is to the right of x on the real line if ∆x > 0, and x + ∆x is tothe left of x when ∆x < 0. The derivative of f with respect to x isdefined to be

f ′(x) = lim∆x→0

f(x + ∆x)− f(x)

∆x

if this limit exists.

(b) If f ′(x) exists, we say that f is differentiable at x. The process ofcomputing the derivative of f is called differentiation.

(c) If f is defined by an equation of the form y = f(x), then the derivative of

f may also be denoted by any of the following notations: y′,dy

dx, Dxy.

Example 4.1.1.

91

92 CHAPTER 4. DERIVATIVES AND THEIR APPLICATIONS

(a) Let f be defined by f(x) = x2. We have

f ′(x) = lim∆x→0

f(x + ∆x)− f(x)

∆x= lim

∆x→0

(x + ∆x)2 − x2

∆x

= lim∆x→0

x2 + 2x∆x + (∆x)2 − x2

∆x= lim

∆x→0

∆x(2x + ∆x)

∆x= lim

∆x→02x + ∆x = 2x

This shows that f ′(x) = 2x for any real number x, and f is differentiableat every real number. Let f(x) =

√x. Using the definition, we have

f ′(x) = lim∆x→0

√x + ∆x−√x

∆x

= lim∆x→0

(√

x + ∆x−√x)(√

x + ∆x +√

x)

∆x(√

x + ∆x +√

x)

= lim∆x→0

∆x

∆x(√

x + ∆x +√

x)=

1

2√

x

Thus, f ′(x) =1

2√

xand f is differentiable for all x > 0.

Theorem 4.1.1. If f is differentiable at x, then f is continuous at x.

4.2 Rules for Differentiation

Using the definition of the derivative, the following rules for finding the deriv-ative of a function can be obtained. We will only show the proof for thederivatives of the power function and the sum of two differentiable functions.The proofs for the other functions are similar.

1. Derivative of a constant functionIf f(x) = k, k a constant, then f ′(x) = 0Example:If f(x) = −3, then f ′(x) = 0

2. Derivative of the power functionIf f(x) = xn where n is an integer, then f ′(x) = nxn−1.

4.2. RULES FOR DIFFERENTIATION 93

Proof. By definition, we have

f ′(x) = lim∆x→0

f(x + ∆x)− f(x)

∆x= lim

∆x→0

(x + ∆x)n − xn

∆x

= lim∆x→0

xn + nxn−1∆x +

(

n2

)

xn−2(∆x)2 + · · ·+ (∆x)n − xn

∆x

= lim∆x→0

nxn−1∆x +

(

n2

)

xn−2(∆x)2 + · · ·+ (∆x)n

∆x

= lim∆x→0

nxn−1 +

(

n2

)

xn−2∆x + · · ·+ (∆x)n−1

Each of the terms containing the factor ∆x will will have a limit of 0 as∆x approaches 0, so only the first trm remains, and we get

f ′(x) =d(xn)

dx= nxn−1

Example: If f(x) = x3, then n = 3 and f ′(x) = 3(x2) = 3x2

3. Derivative of a constant multiple of a functionIf f(x) = k g(x) where k is constant and g is a differentiable function,then

f ′(x) =d(k g(x)

dx= k · g′(x)

Example: If f(x) =5

x4= 5x−4, then f ′(x) = (5)(−4x−5) = −20

x5

4. Derivative of a sum/difference of two or more functions

d(f(x)± g(x)

dx= f ′(x)± g′(x)

Proof. Let y = f(x) + g(x). Then we have

dy

dx= lim

∆x→0

(f + g)(x + ∆x)− (f + g)(x)

∆x

= lim∆x→0

f(x + ∆x) + g(x + ∆x)− (f(x) + g(x))

∆x

= lim∆x→0

f(x + ∆x)− f(x)

∆x+ lim

∆x→0

g(x + ∆x)− g(x)

∆x= f ′(x) + g′(x)


Example: If y = x2 − 5x +2

x, then

dy

dx= 2x− 5− 2

x2.

5. Derivative of the product of two or more functions

d(f(x)g(x)

dx= f(x)g′(x) + g(x)f ′(x)

Example: If y = (x2 + 3)(5− x + x3), then

dy

dx= (x2 + 3)(−1 + 3x2) + (5− x + x3)(2x) = 5x4 + 6x2 + 10x− 3

6. Derivative of the quotient of two functions

d

dx

(

f(x)

g(x)

)

=g(x)f ′(x)− f(x)g′(x)

(g(x))2

Example: If y =3x− 2

4− x, then

dy

dx=

(4− x)(3)− (3x− 2)(−1)

(4− x)2=

10

(4− x)2

7. Generalized Power Rule If a function is defined by an equation ofthe form y = [f(x)]n, where n is an integer, then the derivative of thegeneralized power function is given by

dy

dx= n [f(x)]n−1

Example: If y = (x2−3x+2)5, thendy

dx= 5(x2−3x+2)4(2x−3). In this

example, 5(x2 − 3x + 2)4 is the derivative of the outer power function,while (2x− 3) is the derivative of the inner function (x2 − 3x + 2).

Remark 4.2.1.

(a) The rule for the derivative of a power function may be extended to non-integer exponents. For example, if f(x) =

√x = x1/2 so that n = 1/2,

then the power rule gives us the derivative

f ′(x) = nxn−1 = (1/2)x1/2−1 = (1/2)x−1/2 =1

2x1/2=

1

2√

x

which is the same result we got earlier.

4.3. TANGENT AND NORMAL LINES 95

(b) The quotient rule may be recalled more easily using the mnemonic device

dy

dx=

lodehi - hidelo

(lo)2

where lodehi means the lower function (i.e. g(x)) multiplied by the deriv-ative of the upper function (i.e. f ′(x)) and hidelo means the upperfunction (f(x)) multiplied by the derivative of the lower function (g′(x)).

Exercises. Find the derivative of each of the following functions:

1. f(x) = 5− 3x2 + 4x3

2. f(x) =1

2x4 − 2

3x3 − 8

3. f(x) =4

x3+

6

x4

4. f(x) = 6x√

x− 3x2 3√

x2

5. f(x) =2√x− 4x

√x

6. f(x) =√

x(4x3 − 3x2 + 5)

7. f(x) =3x2 − 4x

2x + 5

8. f(x) =3x− x3

x2 − 4

9. f(x) = (x2 + 2x)(x3 − 1)

10. f(x) = (3x− 2)4

11. f(x) =

√

2

x− x2

3

12. f(x) = 3

√

2x− 1

x + 1

13. f(x) =x + 1

2x− 3· (5− 4x)

4.3 Tangent and Normal Lines

In this section, we give a geometric interpretation for the derivative. Let f becontinuous on an interval containing x1. Also let P = (x1, f(x1)) be a pointon the curve. If ∆x is an arbitrary small number, then Q(x1 +∆x, f(x1 +∆x)is another point on the curve. The difference quotient

f(x1 + ∆x)− f(x1)

∆x


is the slope of the secant line that passes through the points P and Q.

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tangent line at Psecant line through P and Q

P

Q

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x1 x1 + ∆x

When ∆x gets smaller, the point Q slides along the curve towards P , so thatthe position of the secant line through these points gets closer and closer tothe tangent line to the curve at P . Thus, the limiting value of the slope of thesecant line as ∆x gets smaller and smaller is the slope of the tangent line atP . This slope is then equal to

mtan = lim∆x→0

f(x1 + ∆x)− f(x1)

∆x

if this limit exists. Observe But this limit is just the same as f ′(x1), whichmeans that the derivative may be interpreted geometrically as the slope ofthe tangent line to the curve at a point on the curve, if this tangentline is nonvertical.

Example 4.3.1.

(a) To find the equation of the tangent line to the curve y = (x − 1)2 at thepoint P (0, 1), we first compute the derivative and evaluate it at x = 0.Thus:

dy

dx= 2(x− 1)

is the slope of the tangent line to the curve at any point. When x = 0,the slope is 2(0− 1) = −2. Using the point-slope form of the equation ofa line, we get

y − 1 = −2(x− 0) or, equivalently, 2x + y − 1 = 0

4.3. TANGENT AND NORMAL LINES 97

(b) To find the equation of the tangent line to the curve y =√

x which isparallel to the line x− 4y + 5 = 0, we use the fact that parallel lines havethe same slope. The given equation can be expressed in the slope- interceptform as y = 1/4x+5/4, so that the slope of the given line and the tangent

line is 1/4. From Example 4.1.1(2), we have mtan = f ′(x) =1

2√

x=

1

4.

This gives us x = 4, y =√

4 = 2, so the point of tangency is at P (4, 2).The equation of the tangent line is obtained by the point-slope form, thus:

y − 2 = 1/4(x− 4) or x− 4y + 4 = 0

Definition 4.3.1. Let P be a point on a curve y = f(x) where the tangentline exists. The normal line at P is the line perpendicular to the the normal

line at the point of tangency. Its slope is given by mnorm = − 1

mtan

.

Example 4.3.2. To find the equation of the normal line(s) to the curve y =x3 − 3x2 + 1 which are vertical, we first determine the points on the curvewhere the tangent line is horizontal, so that the slope is equal to zero. Sincethe derivative represents the slope of the tangent line, we have

dy

dx= 3x2 − 6x = 3x(x− 2) = 0 ⇒ x = 0, x = 2

This means that the tangent lines are horizontal when x is either 0 or 2, andthe equations of the vertical normal lines at these points have the equationsx = 0, which is the y-axis, and x = 2.

Exercises.

1. Find an equation of each of the following lines:

(a) the tangent line to the curve y = 3− x3 at the point (-1,4).

(b) the tangent line to the curve y = ds2x+13x−2

at the point (1,3).

(c) the tangent line to the curve y = 4x2 − 5x + 1 which is parallel to theline y = 11x− 3.

(d) the normal line to the curve y = 12x− x4 which is parallel to the linex + 2oy − 20 = 0.

(e) each of the lines tangent to the curve y = dx +1

xand parallel to the

line 2x + 4y − 1 = 0.


(f) The line with slope -6 and normal to the curve y =√

x.

2. Answer the following:

(a) Determine the points on the graph of the function defined y = x2+6x+2where the tangent line is horizontal.

(b) Determine the points on the graph of the function defined by y =x3 − 12x where the normal lines are vertical.

(c) For the function defined by y = 3x2 − 6x, determine the points where

(i) the tangent line has slope is -6

(ii) the tangent line is parallel to the line 12x + y − 3 = 0.

(iii) the tangent line is perpendicular line x + 6y − 2 = 0.

4.4 Implicit Differentiation

Definition 4.4.1. When a function is defined by an equation of the formy = f(x), where f(x) is an expression entirely in terms of x, we say that f isexplicitly defined as a function of x. An equation of the form g(x, y) = 0for which there exist at least one function y = f(x) satisfying the equation issaid to define y implicitly as a function of x.

Example 4.4.1. Consider the equation x2+y2−1 = 0, which may be expressedin the form x2 + y2 = 1. This shows that the graph of the equation is a circleand, clearly, the equation does not define a function. However, if we solve fory in terms of x, we get

y = ±√

1− x2

which shows that there are two functions, namely:

y = f1(x) =√

1− x2 and y = f2(x) = −√

1− x2

that satisfy the given equation. Thus we say that the equation x2 + y2−1 = 0implicitly defines y as a function of x.

If we are given an equation of the form g(x, y) = 0, we assume that thereexists an unknown function defined by y = f(x) even if we can not actuallysolve for f(x) like in the above example. To find f ′(x), we use a process knownas implicit differentiation, which is described below:

4.4. IMPLICIT DIFFERENTIATION 99

• Differentiate each term of the defining equation with respect to x,treating y as a function of x, and any function of y as a compositefunction.

• Solve for y′ in the equation obtained in the preceding step.

Example 4.4.2.

1. Finddy

dx= y′ if the function is defined by the equation x2 + y2 = 4 using

implicit differentiation.

We have

(Differentiate each term with respect to x)2x + 2yy′ = 0

(Solve for y′ =dy

dxis the resulting equation)

2yy′ = −2x, so thatdy

dx= y′ =

−2x

2y= −x

y

2. Use implicit differentiation to finddy

dxif x2y − xy2 = y3.

We have

2xy + x2y′ − (y2 + x(2yy′)) = 3y2y′

(Differentiate term by term)

y′(x2 − 2xy − 3y2) = y2 − 2xy

(Combine terms containing y′)

y′ =y2 − 2xy

x2 − 2xy − 3y2

(Solve for y′)

Observe that the terms x2y = (x2)(y) and xy2 = (x)(y2) are treated asproducts with y representing some unknown function.

Exercises.

1. Use implicit differentiation to finddy

dxfor each of the functions defined

below:


(a) (x + y)2 = y

(b) xy +x

y= y2

(c)√

x +√

y = 1

2. Find the equations of the tangent and normal lines to the curve x2+y2 = 25at the point (-4,3).

4.5 Higher Order Derivatives

If f is a function defined by y = f(x), then f ′(x) = y′ =dy

dxis again a function,

which may have its own derivative, denoted by f ′′(x) = y′′ =d2y

dx2, which is

called the second derivative of f . The second derivative is again a function,so if its derivative exists, then this derivative is called the third derivativeof f . This process may be repeated any number of times for as long as therequired derivative exists.

Definition 4.5.1.

(a) If n is a positive integer, then the n-th derivative of a function f

defined by y = f(x) is given by f (n)(x) = y(n) =dny

dxn. It is obtained

by differentiating the original function and successively differentiating thederivatives until n differentiations have been performed.

(b) The order of a derivative is the number of times differentiation is per-formed on the original function and its successive derivatives in orderto obtain this derivative. If k is a positive integer, then the order of

f (k)(x) =dky

dxkis k.

Example 4.5.1.

1. If f(x) = x4, then the first derivative is f ′(x) = 4x3, the second derivativeis f ′′(x) = 4(3x2) = 12x2, the third derivative is f (3)(x) = 12(2x) = 24x,the fourth derivative is f (4)(x) = 24, and the kth-derivative is f (k)(x) = 0when k > 4, since the fourth-derivative is already a constant function.

2. If x2 + y2 = 4, then as shown in the example of the preceding section,

we havedy

dx= y′ = −x

y=−x

y. The second derivative can be obtained

4.6. APPLICATIONS OF THE DERIVATIVE 101

by the quotient rule and implicit differentiation as follows:

y′′ =d2y

dx2=

y(−1)− (−x)y′

y2

=−y + x

(

−xy

)

y2=−(x2 + y2)

y3= − 4

y3

since x2 + y2 = 4.

Exercises.

1. Findd2y

dx2for each of the following functions:

(a) y = x3 − 3x2 + 6

(b) y = (x2 + 1)(3x3 − 2)

(c) y =2x− 1

5− 3x

(d) y = (2x− 3)4

(e) y = 3√

4x− x3

2. If y = (2x− 10)6, findd4y

dx4.

3. Findd3y

dx3when y =

2

3x− 5.

4. If x2 − y2 = 4, show thatd2y

dx2= − 4

y3.

5. If x2 − 3y2 = 4, findd2y

dx2.

4.6 Applications of the Derivative

In this section, we discuss some applications of the derivative. We have alreadygiven a geometric interpretation of the derivative as the slope of the tangentline to the graph of the function at any point on the graph. It is time toinvestigate what information about the function can we get from the derivative.

4.6.1 The Derivative as Rate of Change

(a) The quotientf(x + ∆x)− f(x)

∆xrepresents the average rate of change of

f(x) corresponding to changes in the values of the independent variable.


If x = x0, the derivative

f ′(x0) = lim∆x→0

f(x0 + ∆x)− f(x0)

∆x

represents the instantaneous rate of change of f(x) with respect to x. Thisrepresents the amount by which f(x) will change if the value of x increasesfrom x0 to x0 + 1, assuming it maintains its rate of change at x0.

(b) Ifdy

dx> 0, then y increases as x increases, and we say that f is an in-

creasing function. On the other hand ,dy

dx< 0 implies that the value of

y decreases as x increases, and f is a decreasing function. The magni-tude of the derivative, on the other hand, tells us how fast the derivativeis changing with respect to the independent variable.

Example 4.6.1.

(a) A stone is dropped onto a still pond, causing circular ripples to spread out.The area of the disturbed region in the water is given by A = πr2, wherer is the radius of the outermost circle. Here, our independent variable is rrepresenting the radius, and our dependent variable is A which representsthe area of the disturbed region. The rate at which the area is changing

when the radius is 12 cm is computed as follows: We first computedA

dr=

2πr. When r = 12, we getdA

dr= 2π(12) = 24πcm2 > 0, which means that

the area increases by this amount if the radius were to increase by 1 cm.

(b) Sand is being dropped onto a conical pile such that the height of the pileis always twice the radius of the base of the pile. The formula for the

volume of sand at any time is given by V =1

3πr2h, where r is the base

radius and h is the height of the pile. If we want to know how fast thevolume of sand is changing with respect to the height when the height ofthe pile is h = 5 feet, we first rewrite the formula for the volume in termsof h alone. Since h = 2r, we get

V =1

3πr2h =

1

3π(h/2)2h =

1

12πh3

This gives usdV

dh=

1

4πh2


When h = 5 feet, we getdV

dh=

1

4π(25) =

25π

4ft3 when the height increase

by one foot.

Exercises.

1. If y = x4 − 3x2, find the instantaneous rate of change of y with respect tox when x = 2.

2. Find the rate of change of the area of a circle with respect to the diameterwhen the radius is 6 inches.

3. Find the rate of change of the lateral surface area of a cylinder with respectto its radius when the radius is 8 cm and the height is always twice theradius.

4. Find rate of change of the volume of a sphere with respect to its radius atthe instant when the radius is 10 inches.

5. What is the length e of each edge of a cube when the volume of the cubeis increasing with respect to e at 27 in3 per inch of change in e?

6. The height of a right circular cone is three times its radius. Find the rateat which the volume of the cone is changing with respect to the diameterwhen the diameter is 10 cm.

4.6.2 Related Rates

In the preceding section, we showed that the derivative represents the instan-taneous rate of change of a function with respect to the independent variable.In some cases, the quantities represented by both the independent and depen-dent variables are changing with time. We may then differentiate the givenequation implicitly with respect to time to obtain an equation that relatesthe rates of change with respect to time of the dependent and independentvariables. When one of these rates is unknown, we can use the equation ofrelated rates to solve for this unknown value.

Example 4.6.2.

(a) A balloon is rising vertically above a straight horizontal road at a constantrate of 1 ft/sec. When the balloon is 65 feet above the ground, a bicycle


moving at a constant rate of 17 ft/sec passes under it. How fast is distancebetween the bicycle and the balloon increasing 4 seconds later?

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x

y

y(t)

x(t)

s(t)

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Let y(t) be the vertical distance of the balloon from the ground and letx(t) represent the distance of the bicycle from the point on the grounddirectly above the balloon. Also let s(t) represent the distance betweenthe balloon and the bicycle t seconds after the bicycle passes under theballoon. Then we have the Pythagorean relation

x2 + y2 = t2

Differentiating implicitly with respect to t gives us

2xdx

dt+ 2y

dy

dt= 2s

ds

dt⇒ x

dx

dt+ y

dy

dt= s

ds

dt

The last equation is called the equation of related rates. Note that wehave

dx

dt= 17 ft/sec,

dy

dt= 1 ft/sec

When t = 3, we have

y = 65 + t = 68, x = 17t = 51, s =√

x2 + y2 = 85

We get

ds

dt=

1

s

(

xdx

dt+ y

dy

dt

)

=1

85[(15)(17) + (68)(1)] =

19

5= 3.8ft/sec


(b) A man 6 feet tall walks along a horizontal road at the rate of 5 ft/sectoward a street light which is 16 feet above the ground.At what rate isthe length of his shadow changing when he is 10 feet from the base of thelight?

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x y

Let x be the distance of the man from the base of the light, and let y bethe length of his shadow. Since the man is walking towards the light, x is

decreasing, so thatdx

dt= −5 ft/sec. We have the proportion

x + y

16=

y

6⇒ 6x = 10y

This gives us

dy

dt=

(

3

5

)

dx

dt=

(

3

5

)

(−5) = −3ft/sec

This shows that the length of the shadow decreases as the man gets closerto the light.

(c) Sand is being poured onto a conical pile at the rate of 5 cubic meters perminute. If the height of the pile is always twice the base radius, how fastis the base radius increasing when the radius is 4 meters?

The formula for the volume is given by

V =1

3πr2h

where h is the height of the pile and r is the base radius. Since we are

being asked fordr

dt, we can rewrite the formula for the volume in terms of


the base radius only, thus:

V =1

3πr2(2r) =

2πr3

3⇒ dV

dt= (2πr2)

dr

dt

Solving fordr

dt, we get

dr

dt=

1

2πr2

dV

dt=

5

2(16)π=

5

32πm3/min

Exercises. Solve the following problems:

1. A woman on the ground is watching an airplane through a telescope as theplane approaches at the rate of 15 km per minute while it maintains analtitude of 10 kms. At what rate (in radians per minute) is the angle θ ofthe telescope changing when the horizontal distance between the womanand the plane is 30 kms?

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airplane

telescope

2. A kite flying at a height of 300 feet is being blown by a horizontal wind. Ifwe assume that the string remains taut and is being paid out at the rate of20 feet per minute, how fast is the kite moving horizontally when 500 feetof string is out?

3. A spherical snowball is melting at the rate of 15cm3/min. Assuming thatit keeps its spherical shape as it melts, how fast is the diameter decreasingwhen the diameter is 18 inches?

4. A water tank is in the form of an inverted cone which is 6 feet across thetop and 8 feet high. If water pours into the cone at the rate of 4 cubicfeet per minute, at what rate is the depth of the water increasing when thewater in the tank is 4 feet deep?

5. A man 6 feet tall is walking away from a lamp hung from a post 20 feethigh at the rate of 5 feet per minute.

(a) At what rate is the length of his shadow changing when he is 10 feetfrom the base of the lamp?


(b) At what rate is the tip of his shadow moving when he is 15 feet awayfrom the base of the lamp?

6. Two cars start out simultaneously from the same point, one moving north at80 kms per hours and the other moving east at 60 kms per hour. Assumingthat the cars maintain their speeds, at what rate are the cars separatingtwo hours after they started out?

7. Air from a spherical balloon is being let out at the rate of 1 cubic inch persecond. Assuming that it maintains its shape, at what rate is the radiuschanging when the radius is 4 inches?

8. A ladder 17 feet long is leaning against a vertical wall, and its base restson horizontal ground. If the base of the ladder is being pulled horizontallyaway from the wall at the rate of 1 foot per minute, how fast is the top ofthe ladder sliding down the wall when the foot of the ladder is 8 feet awayfrom the wall?

4.6.3 Maximum and Minimum Values of a Function

Definition 4.6.1. Let f be a function defined on some open interval contain-ing the number c. Then

(a) f has a relative maximum at x = c if there exists an open interval (a, b)that contains c such that f(c) ≥ f(x) for all x ∈ (a, b). The number f(c)is a relative maximum for f on the interval (a, b).

(b) f has a relative minimum at x = c if there exists an open interval (a, b)that contains c such that f(c) ≤ f(x) for all x ∈ (a, b). The number f(c)is a relative minimum value for f on the open interval (a, b).

In the figure below, we illustrate two functions which have relative maximumvalues at a number c. We see that the graphs have peaks at the points wherex = c. For relative minimum values, the graphs will have ”valleys“ at thenumbers where the relative mimimum exist.

In the above figure, tangent line at x = c is horizontal for the graph on theleft, and vertical for the graph on the right. Thus, the slope of the tangentline is 0 or is undefined. This illustrates the fact that at the points where afunction f attains a relative maximum or minimum value, then tangent line iseither horizontal or vertical, and hence the derivative, which gives the slope ofthe tangent line, is either 0 or fails to exist. We give a name to the numberswhere these occur.


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Figure 4.1: Relative maximum at x = c

Definition 4.6.2. If f is defined at a number c and either

• f ′(c) = 0 or

• f ′(c) does not exist

then c is called a critical number of f .

Example 4.6.3. To find the critical numbers of the function defined by y =x3 − 3x2 + 5, we first find the derivative, which is

dy

dx= 3x2 − 6x

Since the derivative is defined by a quadratic polynomial, it exists everywhere,so that the only critical numbers are those for which the derivative is 0. Wehave

f ′(x) = 3x2 − 6x = 3x(x− 2) = 0 ⇒ x = 0 or x = 2

This shows that the given function has two critical numbers, namely 0 and 2.

Remark 4.6.1.

• If we wish to determine where a given function f assumes relative max-imum or minimum values, we only have to determine the elements inthe domain of f where either f ′(x) = 0 or f ′(x) fails to exist. Theseare precisely the critical numbers of f . If f has relative maximum orminimum values, they can only occur at the critical numbers.


• If c is a critical number of a function f , we use the following test todetermine if a relative maximum or minimum is attained at c.

First Derivative Test for Relative Extrema

• Let c be a critical number of f .

• If f ′(x) < 0 when x < c and f ′(x) > 0 when x > c, then f isdecreasing and the graph is falling as x approaches c from theleft. After passing through c, the function is increasing and thegraph is rising to the right of c. Thus, the function would haveattained a relative minimum value of f(c) at x = c.

• If f ′(x) > 0 when x < c and f ′(x) < 0 when x > c, then the graphis rising as x approaches c then falls as x moves to the right of c.This shows that f attains a relative maximum value f(c) atc.

Example 4.6.4. To determine the relative maximum and minimum values ofthe function defined by y = x3 − 3x2 + 5, we use the first derivative test onthe critical numbers 0 and 2 obtained in the preceding example. We need todetermine the signs of the derivative at points to the left and to the right ofeach of these critical numbers. Hence, we subdivide the domain of f , which isR, into three intervals, consisting of the following:

(∞, 0) = { x ∈ R : x < 0 } = the points to the left of 0

(0, 2) = { x ∈ R : 0 < x < 2 } = the points between 0 and 2

(2, +∞) = { x ∈ R : x > 2 } = the points to the right of 2

The following table summarizes our results:

x f(x) f ′(x) Conclusionx < 0 + f is increasingx = 0 5 0 Relative maximum

0 < x < 2 - f is decreasingx = 2 1 0 Relative minimumx > 2 + f is increasing


Figure 4.2: Graph of y = x3 − 3x2 + 5

The graph of f shown below confirms that a relative maximum of 5 occurs atx = 0 and a relative minimum of 1 is attained at x = 2.

In Figure 4.6.4, the relative maximum value of 5 is not the maximum valuethat the function can attain, since we see that the graph of the function risesindefinitely as x increases. Similarly, the relative minimum value of 1 is notthe smallest value that f can attain, as the graph of f extends indefinitelydownward to the left of 0. These relative extrema (or extreme values) are onlythe largest or smallest within a small interval (a, b) containing c, comparedwith the values of f(x) near the number c. If we want the maximum orminimum value that f(x) can have over a larger interval or over the entiredomain of f , then we get the absolute maximum/minimum values of f(x).

Definition 4.6.3.

• A function f is said to have an absolute maximum on an interval [a, b]at the number c, where a ≤ c ≤ b , if f(c) is the largest value attainedby f(x) on this interval.

• A function f is said to have an absolute minimum on an interval [a, b]at the number d, where a ≤ d ≤ b , if f(d) is the smallest value attainedby f(x) on this interval.

The following theorem guarantees the existence of both the absolute max-imum and absolute minimum values of a continuous function on a closed in-terval.


Theorem 4.6.1. (Extreme Value Theorem)If a function f is continuous and differentiable on a closed interval [a, b], thenf will have both an absolute maximum value and an absolute minimum valueon the interval. To obtain these values, do the following:

(a) Find all the critical numbers of f inside the interval.

(b) Evaluate f(x) at each of the critical numbers obtained in (a) and at theendpoints a and b.

(c) Compare the function values obtained in (b). The largest is the absolutemaximum value of f(x) on the interval, while the smallest is the absoluteminimum value of f(x) on the same interval.

Remark 4.6.2. Observe that a continuous function can have an absolutemaximum or minimum value over an interval either at an interior point orat an endpoint of the given interval. This explains the above procedure fordetermining the absolute maximum and minimum values of f in the interval.

Example 4.6.5. Let f be the function defined by 2y = 12− x2, and supposethat we want to determine the absolute maximum/minimum values of f onthe interval I = [−3, 4]. We have

dy

dx= −x = 0 ⇒ x = 0

The critical number 0 lies inside the interval I. We compare the functionvalues at 0 and at the endpoints of the interval:

f(−3) = 6− (−3)2

2= 6− 9

2=

3

2

f(0) = 6− 02

2= 6− 0 = 6

f(4) = 6− 32

2= 6− 16

2= −2

This shows that f has an absolute maximum of f(0) = 6 at 0 and an absolute


minimum of -2 at 4. The graph is shown below:

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• absolute minimum

absolute maximum

When solving word problems that involve absolute maximum or minimumvalues of a function, we often have to set up the function that gives the valuesof the quantity whose absolute maximum/minimum value is to be determined,and the interval where the computation is to be carried out. We illustrate thisin the following examples:

Example 4.6.6.

(a) The distance d in feet travelled in t seconds by an object which is allowedto fall freely is given by the equation d = 16t2. The instantaneous speedof the object at any time t is given by

d′(t) = 32t

For example, at the instant three seconds after the object started to fall(t = 3), the speed of the object is d′(3) = 32(3) = 96 feet per second. Onthe other hand, since the distance travelled by the object after 3 secondsis d = 16t2 = 16(9) = 144 feet, the average speed of the object during thefirst three seconds is

∆d

∆t=

144

3= 48ft/sec


(b) The volume of a balloon in the shape of a sphere is given by v = 4/3πr3,where r is the radius. If, while being inflated, the balloon retains itsspherical shape, then the rate of change in the volume with respect to theradius is given

v′(r) =dv

dr= 4/3π(3r2) = 4πr2

When the radius is 6 inches, the volume is increasing at the rate of v′(r) =4π(62) = 144πin2/in.

(c) To find two nonnegative numbers whose sum is 12 and whose productis a maximum, we note that the numbers can be represented by x and12 − x, respectively. Since both numbers must be nonnegative, we canonly choose x in the interval [0, 12]. The product is given by f(x) =x(12 − x) = 12x − x2, and we wish to find its maximum. The criticalnumbers are obtained from

f ′(x) = 12− 2x = 2(6− x) = 0

which shows that 6 is the only critical number, and it is inside the interval.We have

f(0) = f(12) = 0, f(6) = 12(6)− 62 = 36

so the maximum product of 36 is obtained when both numbers are equalto 6.

Exercises.

1. Use the first derivative test to determine the relative maximum/minimumvalues of each of the given functions:

(a) f(x) = x3 − 3x2 + 3x

(b) f(x) = x4 − 4x + 3

(c) f(x) = 2x3 + 3x2 − 6x

(d) f(x) = x4 − 2x2 + 3

2. Find the absolute maximum and minimum values of each of the given func-tions in the indicated intervals:

(a) f(x) =√

25− x2, [0, 4]

(b) f(x) =2x + 1

x− 3, [−3, 2]

(c) f(x) =4√

x− 2, [3, 6]

(d) f(x) =x

x2 − 1, [2, 5]

3. Solve the following problems:


(a) A man has 100 feet of wire to build a fence about a rectangular garden.If one side of the garden is along a vertical wall, find the dimensions ofthe largest garden that can be enclosed by the fence.

(b) Find two nonnegative numbers whose sum is 12 such that the sum oftheir squares is a maximum.

(c) Find a number in interval [0.5, 2.5] such that the sum of the numberand its reciprocal is a minimum.

(d) An open square box is to be formed by cutting off equally-sized squaresfrom the four corners and turning up the flaps. Find the volume of thelargest possible box.

(e) Find the dimensions of the largest rectangle that can be inscribed in acircle of radius 8 inches.

(f) Find the dimensions of a right circular cylindrical can with an open topand a volume of 100 cubic cm. having the least surface area.

4.7 The Mean Value Theorem

One of the most important theorems in Calculus is the Mean Value Theo-rem. Its proof uses a theorem which actually represents a special case of theMean Value Theorem. This is Rolle’s Theorem, which is stated and provenbelow:

Theorem 4.7.1. (Rolle’s Theorem)Let f be a function satisfying the following conditions:

• f is continuous on a closed interval [a, b];

• f is differentiable on the open interval (a, b); and

• f(a) = f(b) = 0

Then there exists a number c ∈ (a, b) such that f ′(c) = 0

Proof. We consider two cases, namely:

Case 1: f(x) = 0 for all x ∈ [a, b]In this case, f is a constant function, so that f ′(x) = 0 for all x ∈ [a, b], andwe can choose c to be any element of [a, b].

4.7. THE MEAN VALUE THEOREM 115

Case 2: f is not identically zero in [a, b]In this case, either f has a positive absolute maximum or a negative absoluteminimum at some number c inside the interval. This is guaranteed by theExtreme Value Theorem. This absolute extremum must then be a relativeextremum as well. Since f ′ exists in (a, b), we must have f ′(c) = 0.

Remark 4.7.1. Rolle’s theorem has a geometric interpretation. If f is afunction satisfying the hypothesis of Rolle’s theorem, then there is at least onepoint in the graph of the function over the interval [a, b] where the tangentline is horizontal.

Example 4.7.1. Consider the function defined by y = 8 + 2x − x2, and letI = [a, b] = [−2, 4]. Clearly, f is continuous everywhere, and hence continuousin the interval I. Moreover, f ′(x) = 2−2x, so that f is differentiable in (-2, 4).Also,

f(−2) = f(4) = 0

Setting the derivative to zero, we get

0 = f ′(x) = 2− 2c ⇒ c = 1 ∈ [−2, 4]

The graph of f and the horizontal tangent line at c = 1 are shown below:

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We use Rolle’s Theorem to prove the Mean Value Theorem which is statedbelow:

Theorem 4.7.2. (Mean Value Theorem)Let f be a function satisfying the following conditions:

• f is continuous in the closed interval [a, b];

• f is differentiable in the open interval (a, b); and


Then there exists a number c ∈ (a, b) such that

f ′(c) =f(b)− f(a)

b− a

Proof. Note that the secant line through the points (a, f(a)) and (b, f(b)) hasthe equation

y − f(a) =f(b)− f(a)

b− a(x− a) or y =

f(b)− f(a)

b− a(x− a)− f(a)

Define the function

F (x) = f(x)− f(b)− f(a)

b− a(x− a)− f(a)

Geometrically, this function represents the vertical distance between a point(x, f(x)) on the graph of f and the corresponding point on the secant line.Since f is continuous on [a, b] and differentiable on (a, b), so is F . Moreover,it can be verified that

F (a) = F (b) = 0

This shows that F satisfies the hypothesis of Rolle’s theorem. Thus, from thistheorem, we know that there exists a number c ∈ (a, b) such that

F ′(c) = f ′(c)− f(b)− f(a)

b− a= 0 or f ′(c) =

f(b)− f(a)

b− a

Remark 4.7.2. Geometrically, the quantity f ′(c) represents the slope of thetangent line to the graph of f at the point (c, f(c)). Hence, the mean value the-orem says that there is a point on the graph of f over the interval [a, b] wherethe tangent line is parallel to the secant line joining (a, f(a)) and (b, f(b)).

Example 4.7.2. Consider the function defined by y = x3 + x2 − x and theinterval [-2,1]. We have f ′(x) = 3x2 + 2x − 1, so that f is differentiable over(-2,1). Since f is a polynomial function, it is continuous everywhere and hencecontinuous on the closed interval [-2,1]. Thus, the hypothesis of the mean valuetheorem is satisfied, and there exists a number c ∈ (−2, 1) such that

f ′(c) = 3c2 + 2c− 1 =f(1)− f(−2)

1− (−2)=

1− (−2)

1− (−2)= 1

4.7. THE MEAN VALUE THEOREM 117

By the quadratic formula, we have

c =−2±

√

4− 4(3)(−2)

6=−2±

√28

6=−1±

√7

3

Since both values are inside the interval [-2,1], we have two values for c satis-fying the conclusion of the mean value theorem.

Exercises.

1. In each of the following, show that the hypothesis of Rolle’s theorem issatisfied. Then find a number c inside the given interval satisfying theconclusion of the theorem.

(a) f(x) = x2 − 4x + 3, [1, 3]

(b) f(x) = x4 − 4x2, [−2, 2]

(c) f(x) = x3 − 2x2 − x + 2, [1, 2]

(d) f(x) = x3 − 5x2 − 6x, [0, 6]

2. In each of the following, verify the hypothesis of the mean value theoremand find a number c in the indicated interval which satisfies the conclusionof the theorem.

(a) f(x) = x3 − 3x2 + 4, [−1, 4]

(b) f(x) =√

x2 − 4x− 5, [−3, − 2]

(c) f(x) =2x− 4

x2 − 5x + 6, [−1, 1]

(d) f(x) = x2/3, [0, 1]

3. If f(x) = x4 − 2x3 + 2x2 − x, then f ′(x) = 4x3 − 6x2 + 4x− 1. Use Rolle’stheorem to show that the equation

4x3 − 6x2 + 4x− 1 = 0

has at least one real root inside the interval (0, 1).

4. Use the mean value theorem to show that if a < b, then the arithmetic meana + b

2of a and b lies in the open interval (a, b). (Hint: Use f(x) = x2)


4.8 Curve Sketching

As a final application of derivatives to be discussed in this chapter, we turnto sketching the graph of a continuous function. We already learned that thefirst derivative tells us if a function is increasing or decreasing, and hence if itsgraph is rising or falling. In order to sketch the graph, we also need to knowif a portion of the graph opens upward or downward.

Definition 4.8.1. Let f be a function with a continuous derivative on aninterval I. We say that the graph of f is

(a) concave downward on I if the tangent line to the curve lies above thecurve on the interval.

(b) concave upward on I if the tangent line to the curve lies below thecurve on the interval.

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a b

Figure 4.3: Graph is concave downward on the interval [a, b]

Note that when the graph is concave downward on an interval, the slopesof the tangent lines changes from positive to zero to negative. This means thatthe the function f ′ is decreasing on this interval, and hence the derivative of f ′,which is f ′′, must be negative. Similarly, if the graph of a twice-differentiablefunction f is concave upward on an interval I, then the slopes of the tangentlines to the curve on this interval change from negative to zero to positive,so that f ′ is an increasing function on the interval. The second derivative f ′′

must then be positive on the interval I. We have

4.8. CURVE SKETCHING 119

Second Derivative Test for Concavity

Let f be a function whose second derivative exists on an interval I.

• If f ′′(x) > 0 for all x ∈ I, then the graph of f is concave upwardon I,

• If f ′′(x) < 0 for every x ∈ I, then the graph of f is concavedownward on the interval I.

Example 4.8.1. Let f be the function defined by y = −x3 + 9x2 − 15x + 5.We have

f ′(x) = −3x2 +18x−15, f ′′(x) = −6x+18 = −6(x−3) =

{

> 0 if x < 3< 0 if x > 3

This shows that the graph of f is concave upward on the interval (−∞, 3) andconcave downward on the interval (3, +∞). The graph of f is shown below:

Figure 4.4: Graph of y = −x3 + 9x2 − 15x + 5

Definition 4.8.2. The concavity of a graph may be upward on some intervaland downward on another interval. A point (c, f(c)) on the graph of a functionf is called a point of inflection of the graph if the concavity changes at thispoint.

Example 4.8.2. In the following figure, we see that the graph is concavedownward when x < 0 and concave upward when x > 0. This shows that theorigin (0,0) is a point of inflection.


Remark 4.8.1. If (c, f(c)) is a point of inflection of the graph of a functionf , then either of the following is true:

• The graph is concave upward when x < c and concave downward whenx > c; or

• The graph is concave downward when x < c and concave upward whenx > c.

In terms of f ′, this translates to

• f ′ > 0 when x < c and f ′ < 0 when x > c; or

• f ′ < 0 when x < c and f ′ > 0 when x > c.

This shows that f ′ must have either a relative minimum or a relative maximumat x = c, i.e. c is a critical number of f ′. Thus, either f ′′(c) = 0 or f ′′(c) failsto exist. This shows that the only possible points where f can have a point ofinflection are those points for which either f ′′(c) = 0 or f ′′(c) fails to exist.

Example 4.8.3. Consider the function defined by y = x3. We have f ′(x) =3x2 and f ′′(x) = 6x. Since the second derivative exists everywhere, the onlypossible point of inflection occurs where f ′′(x) = 6x = 0. This happens atx = 0. Note that f ′′(x) > 0 when x > 0, and f ′′(x) < 0 when x < 0. Thisshows that the graph is concave upward when x > 0 and concave downwardwhen x < 0, and hence a point of inflection exists at (0,0). This can be verifiedfrom the graph of y = x3 which is shown in Example 4.8.2


Aside from helping us determine the concavity of the graph of a function ona given interval, the second derivative also gives us another method to test ifthe function has a relative maximum or minimum value at a critical number.This is given in the following:

Second Derivative Test or Relative Extrema

Let f be twice differentiable on an interval I, and let c ∈ I be a criticalnumber of f for which f ′(c) = 0. Then

• If f ′′(c) > 0, then f has a relative minimum at c.

• If f ′′(c) < 0, then f has a relative maximum at c.

• If f ′′(c) = 0, the test fails. In this case, apply the first derivativetest.

Example 4.8.4. Consider the function defined by y = x3 − 3x2 + 5, whichwas discussed in Example 4.6.4. We have

dy

dx= 3x2 − 6x = 3x(x− 2) = 0 if either x = 0 or x = 2

Thus, 0 and 2 are critical numbers of this function. To apply the secondderivative test, we get f ′′(x) = 6x− 6. Then

(a) f ′′(0) = 6(0) − 6 = −6 < 0, which shows that f has a relative maximumwhen x = 0.

(b) f ′′(2) = 6(2)− 6 = 6 > 0, and so f has a relative minimum when x = 2

These results agree with our findings in Example 4.6.4.

Exercises.

1. For each of the following functions, determine all the critical numbers andapply the second derivative test for relative extrema:

(a) f(x) = x4 + 2x3

(b) f(x) = x3 − 6x2 + 4(c) f(x) =

x2 + 3

x2 − 4


2. For the functions in the preceding number, use the first derivative test todetermine the relative maximum and minimum values of f .

3. Use the second derivative to identify possible points of inflection of thegraph of the given function.

(a) f(x) = 3x4 + x3

(b) f(x) = x4 − 4x + 3

(c) f(x) = 2x3 + 3x2

Curve Sketching

We now use the first and second derivatives of a function to help us sketch thegraph of the function. Recall that

(a) f ′(x) tells us whether the function is increasing or decreasing, and hencewhether the graph is rising or falling, on an interval I.

(b) f ′′(x) tells us if the graph of f is concave upward or downward on aninterval I.

Using these facts, we now have the following procedure for sketching the graphof a continuous function f .

Steps for Sketching the Graph of a Continuous Function

• Find f ′(x) and use this to determine the critical numbers of f .

• Find f ′′(x) and use it to determine the possible points of inflection ofthe graph of f .

• Using the numbers obtained from the first two steps as endpoints, par-tition the domain of f into intervals.

• In each interval, determine if f is increasing/decreasing and if the graphof f is concave upward/downward.

• Evaluate f at the numbers obtained in the first two steps, and locatethese points in the coordinate plane.

• Using the information obtained from the fourth step, join the pointsplotted in step five to obtain a sketch of the graph of f .


Example 4.8.5. We use the above procedure to sketch the graph of the func-tion defined by y = x4 − 4x2. We have

f ′(x) = 4x3 − 8x = 4x(x2 − 2) = 0 if either x = 0 or x = ±√

2

This shows that f has three critical numbers. Next

f ′′(x) = 12x2 − 8 = 4(3x2 − 2) = 0 if x = ±√

2/3

Next we partition the the domain, which is R, into the following subintervals:

(−∞,−√

2), (−√

2,−√

2/3), (−√

2/3, 0), (0,√

2/3), (√

2/3,√

2), (√

2, +∞)

In each of these intervals, we use the first and the second derivatives to deter-mine if the function is increasing/decreasing and the concavity of the graph.Our results are summarized in the following table:

x f(x) f ′(x) f ′′(x) Conclusion

(−∞,−√

2) - + dec., con. upward

x = −√

2 -4 0 + relative min.

(−√

2,−√

2/3) + + inc., con. upward

x = −√

2/3 -20/9 + 0 point of inf.

−√

2/3, 0) + - inc., con. downwardx = 0 0 0 - relative max.

(0,√

2/3) - - dec., con. downward

x =√

2/3 -20/9 - 0 point of inf.

(√

2/3,√

2) - + dec., con. upward

x =√

2 -4 0 + relative min.

(√

2, +∞) + + inc., con. upward

Exercises. For each of the following functions, do the following:

• Determine all the critical numbers.

• Determine the intervals where f is increasing/decreasing.

• Determine the intervals where the graph is concave upward/downward.

• Determine the relative maximum/minimum values of f .

• Determine the points of inflection of the graph of f .


Figure 4.5: Graph of y = x4 − 4x2

• Sketch the graph of the function.

(a) f(x) = x4 − 4x3 + 6x2

(b) f(x) = 2x3 + 9x2 − 3

(c) f(x) = x3 + 3x2 − 4

(d) f(x) = x4 + 4x3 + 6x2 + 1


REVIEW EXERCISES

Directions: Choose the letter corresponding to the correct answer.

4.2 Tangent and Normal Lines

1. The slope of the tangent line to the curve y = 3x2 + 4 at the point(1, 7) is equal to

A. 3 B. 6 C. 1 D. 7

2. The equation of the tangent line to the curve y =8√x

at the point

(4,−4) is

A. x + 2y + 4 = 0

B. x + 2y + 12 = 0

C. x− 2y − 12 = 0

D. x− 2y + 4 = 0

3. The slope of the normal line to the curve y = x3/2 at the point(1, 1) is

A. 1 B. -1 C. 3/2 D. -2/3

4. The slope function for the curve y =x2

4is

A. y′ =x2

2B. y′ =

x

4C. y′ =

x

2 D. y′ =x2

4

5. The tangent line to the graph of a function f at a point P is 4x +y − 5 = 0. What is the slope of the normal line to the graph of fat the same point P?

A. 4 B. -4 C. 1/4 D. -1/4

6. The equation of the normal line to the curve y = 4x − x2 at thepoint (0, 0) is

A. y = 4x B. 4x + y = 0 C. x = 4y D. 4y + x = 0


7. Which of the following is the equation of the tangent line to the

curve y =3

xat the point (−1, − 3)?

A. 3x + y + 6 = 0

B. 3x− y − 6 = 0

C. 3x + y = 0

D. 3x− y = 0

8. If the slope of a line tangent to the curve y = 3x2 − 6x + 1 is 6,then the equation of this tangent line is

A. x− 6y + 4 = 0

B. x + 6y − 8 = 0

C. 6x + y − 13 = 0

D. 6x− y − 11 = 0

9. Which of the following represents a line normal to the curve y =−√x at the point (4, − 2)?

A. 4x + y + 14 = 0

B. x− 4y − 12 = 0

C. x + 4y + 4 = 0

D. 4x− y − 18 = 0

10. Which of the following is an equation of a line parallel to the linex + 4y − 2 = 0 and normal to the curve y = x3 − 2x2 + 3?

A. 4x− y − 5 = 0

B. 4x + y − 11 = 0

C. x− 4y + 10 = 0

D. x + 4y − 14 = 0

4.3 Rules for Differentiation

11. If f(x) =√

x, then f ′(x) equals

A.1

2√

x

B. 2√

x

C.1√x

D. does not exist

12. Given that f(x) =√

2x + 3, the value of f ′(1/2) is

A. ±2 B. 2 C.1

2D.

1

4


13. If f(x) =

√

4x− 5

7− 2x, then f ′(x) is equal to

A.18(4x− 5)1/2

(7− 2x)5/2

B.9

(4x− 5)1/2(7− 2x)3/2

C.9

(4x− 5)1/2(7− 2x)5/2

D.18(4x− 5)1/2

(7− 2x)3/2

14. If f(x) = 3x√

x and g(x) =3

x4, then (f + g)′(x) is equal to

A. 9√

x +12

x5

B. 9√

x− 12

x5

C.9√

x

2− 12

x5

D.9√

x

2+

12

x5

15. If f(x) = (x2 − 9)2/3, then f ′(x) fails to exist at the number

A. 9 B. 3 C. 1 D. 0

4.4/4.5 Implicit Differentiation and Higher Order Derivatives

16. If f(x) = 3x4 − 5x3 + 4x− 3, then f (3)(0) is equal to

A. -3 B. 4 C. 0 D. -30

17. If x2 + y2 = 1, thend2y

dx2is equal to

A.y2 − x2

y3B.

x2 − y2

y3C.

1

y3D. − 1

y3

18. If y = 3x2, which of the following is true about y′′?

A. The graph of y′′ intersects the x-axis.

B. The graph of y′′ is a vertical line.

C. The graph of y′′ is a horizontal line.

D. The graph of y′′ is a diagonal line.

19. If the equation xy − 3y2 = 2x + y defines a function of x, then the

value ofdy

dxat the point (-4, 1) is


A. −7

4B. − 1

11C.

1

11D.

7

4

20. An equation of the tangent line to the curve x2+4y2−4x−8y+3 = 0at the point (0, 1/2) is

A. 2x− 2y − 1 = 0

B. 2x + 2y − 1 = 0

C. 2x− 2y + 1 = 0

D. 2x + 2y + 1 = 0

21. If x3 + y3 = 1, thend2y

dx2is equal to

A. −2x

y3B.

2x

y3C.

2x

y5D. −2x

y5

4.6 Applications of the Derivative

22. Which of the following is a critical number of f(x) = x +16

x

A. 0 B. 1 C. 16 D. -4

23. A stone thrown into a still pond creates circular ripples that growoutward. Assuming that the ripples retain their circular shape, atwhat rate is the area A of the disturbed region of the pond changingwith respect to the radius r of this region at the instant when theradius is r = 4 feet?

A. 4πsq.ft per foot

B. 8πsq.ft per foot

C. 2πsq.ft per foot

D. 16πsq.ft per foot

24. Given that the equation of motion of a body along a horizontal lineis s = f(t) where s is the distance and t is time. The expression

f(t1 + ∆t)− f(t1)

∆trepresents

A. the average velocity of the body during the time interval [t1, t1+∆t]

B. the instantaneous velocity of the body at time t = t1C. the average distance travelled by the body during the time

interval [t1, t1 + ∆t]

D. the acceleration of the body

25. The function f(x) = 1 + 6x− x2 has a maximum value at x equalto


A. -2 B. 0 C. -3 D. 3

26. The area of the largest rectangle having a perimeter of 40 cm. is

A. 100cm2 B. 400cm2 C. 200cm2 D. 80cm2

27. If V (x) cubic centimeters is the volume of a cube having edges of xcentimeters, what is the instantaneous rate of change of V (x) withrespect to x when x = 2?

A. 8 B. 4 C. 12 D. 6

28. The minimum value of the function f(x) = |x| on the interval[−π/4, π/3] occurs at

A. x = −π/4 B. x = π/3 C. x = 0 D. x = 1

29. For the function described in the preceding item, the absolute max-imum value of f on the given interval is

A. 0 B. π/3 C. π/4 D. 1

30. Which of the following is true about the function defined by f(x) =x− 3

x + 7?

A. f is defined at all real numbers

B. f has no critical number

C. -7 is a critical number of f

D. f is continuous at all real numbers

31. A ladder 13 feet long is leaning against a vertical wall. If the baseof the ladder is being pulled horizontally away from the wall at2ft/sec, then when the top of the ladder is 5 feet from the ground,the top of the ladder is sliding down the wall at the rate of

A. 5/6ft/sec

B. -5/6ft/sec

C. 13/6 ft/sec

D. -13/6ft/sec

32. If the edge of a cube is increasing at the rate of 3 cm/sec, how fastis the volume increasing when the edge is 5 cm?


A. 125 cm3/sec

B. 375 cm3/sec

C. 225 cm3/sec

D. 75 cm3/sec

33. For the function defined by y = x3−3x2 +1, which of the followingis true?

A. The function is concave upward at x = 1.

B. The function has a relative maximum at x = 2.

C. The function is increasing when x is negative.

D. The function is decreasing when x is positive.

34. For the function defined in No. 33, in which interval is the functionincreasing and the graph is concave upward?

A. (−∞, 0)

B. (2, +∞)

C. (1, 2)

D. (0, 1)

35. Which of the following functions has a local minimum?

A. y = 4− x2

B. y = 2− |x + 1|C. y = 3x− 2

D. y = x2 − 2x

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