Math 151 Sample Final Exam

1. (8 points) Find the following limits.

(a)

limx→9

x2 − 81√x− 3

= limx→9

(x + 9)(x− 9)√x− 3

·√

x + 3√x + 3

= limx→9

(x + 9)(x− 9)(√

x + 3)

x− 9

= limx→9

(x + 9)(√

x + 3)

= 108

(b)

limx→−∞

3x3 − 27x

2− 4x2= lim

x→−∞

3x3 − 27x

2− 4x2· 1/x2

1/x2

= limx→−∞

3x− 27x

2x2 − 4

= limx→−∞

3x

−4= ∞

2. (6 points) Using the definition of the derivative, find the derivative of g(x) = 13x−4

.

g′(x) = limt→x

g(t)− g(x)

t− x

= limt→x

13t−4− 1

3x−4

t− x

= limt→x

(1

3t− 4− 1

3x− 4

)· 1

t− x

= limt→x

(3x− 4)− (3t− 4)

(3t− 4)(3x− 4)· 1

t− x

= limt→x

(3x− 3t)

(3t− 4)(3x− 4)(t− x)

= limt→x

−3(t− x)

(3t− 4)(3x− 4)(t− x)

= limt→x

−3

(3t− 4)(3x− 4)

=−3

(3x− 4)2

OR

g′(x) = limh→0

g(x + h)− g(x)

h

= limt→x

13(x+h)−4

− 13x−4

h

= limt→x

(1

3x + 3h− 4− 1

3x− 4

)· 1

h

= limt→x

(3x− 4)− (3x + 3h− 4)

(3x + 3h− 4)(3x− 4)· 1

h

= limt→x

(3x− 4− 3x− 3h + 4)

(3t− 4)(3x− 4)h

= limt→x

−3h

(3x + 3h− 4)(3x− 4)h

= limt→x

−3

(3x + 3h− 4)(3x− 4)

=−3

(3x− 4)2

3. (12 points) Find the following derivatives.

(a) f(x) = x3 − 2x2 + 4√

x

f ′(x) = 3x2 − 4x + 2x−1/2.

(b) g(t) = sin(t3).

g′(t) = cos(t3) · 3t2 = 3t2 cos(t3)

g′′(t) = 3t2(− sin(t3) · 3t2) + cos(t3) · 6t = −9t4 sin(t3) + 6t cos(t3).

(c) y =x2 − 3

x2 + 1dy

dx=

(x2 + 1)(2x)− (x2 − 3)(2x)

(x2 + 1)2.

4. (8 points) Find the tangent line to the curve x2 + 2xy − y2 = 1 at the point (1, 2).

d

dx(x2 + 2xy − y2) =

d

dx(1)

2x + 2xdy

dx+ 2y − 2y

dy

dx= 0

2 + 2dy

dx+ 4− 4

dy

dx= 0

−2dy

dx= −6

dy

dx= 3

5. (8 points)

(a) State the Mean Value Theorem.

If f(x) is continuous on the interval [a, b] and differentiable on the interval (a, b), thenthere is a number c with a < c < b such that

f ′(c) =f(b)− f(a)

b− a,

(or equivalently, f ′(c)(b− a) = f(b)− f(c)).

(b) For f(x) = 1x

on the interval [1, 9], give the number c guaranteed by the Mean ValueTheorem.

f(b)− f(a)

b− a=

1/9− 1/1

9− 1=(

1

9− 1

1

)· 1

9− 1=−8

9· 1

8=−1

9

f ′(c) =−1

c2=−1

9

c2 = 9, so c = ±3. The only value of c between 1 and 9 is 3.

6. (16 points) Let f(x) = 2− 6x

+ 6x2 .

(a) Find all the asymptotes of the graph y = f(x).

Vertical asymptote: x = 0

Horizontal asymptote: y = 2

(b) Find the intervals on which f is increasing and on which f is decreasing.

f ′(x) = 6x2 − 12

x3 = 6x−12x3 = 2(x−2)

x3 .

f(x) is increasing on (−∞, 0) and on (2,∞) and decreasing on (0, 2).

(c) Find the intervals on which f is concave up and on which f is concave down.

f ′′(x) = − 12x3 + 36

x4 = −12x+36x4 = −12(x−3)

x4 .

f(x) is concave up on (−∞, 0) and on (0, 3) and concave down on (3,∞).

(d) Sketch the graph y = f(x). Be sure to label all asymptotes, local maxima, localminima, and inflection points.

The graph has a local minimum at x = 2 and an inflection point at x = 3.

7. (10 points) A rectangle with its base along the x-axis has its top vertices on the circlex2 + y2 = 9. What is the largest possible area of the rectangle?

Base of the rectangle = 2x.

Height of the rectangle = y.

We are maximizing the area of the rectangle, A = 2xy, and x2 + y2 = 9, so y =√

9− x2.The possible values of x are 0 ≤ x ≤ 3.

A(x) = 2x√

9− x2 = 2x(9− x2)1/2.

A′(x) = 2x · 12(9− x2)−1/2(−2x) + 2(9− x2)1/2

= 2

(−x2

√9− x2

+√

9− x2

)

= 2

(−x2

√9− x2

+√

9− x2

)

= 2

(−x2 + 9− x2

√9− x2

)

= 2

(9− 2x2

√9− x2

)

The critical numbers in [0, 3] are 0 (because A′(0) is undefined) and 3/√

2 (becauseA′(3/

√2) = 0).

A(0) = 0

A(3/√

2) = 3

A(3) = 0

So, by Fermat’s method, the maximum possible area is 3.

8. (6 points) For the function f(x) = x2 − x + 1 on the interval [0, 8], compute the Riemannsum with 4 pieces using midpoints.

Riemann sum = f(1) · 2 + f(3) · 2 + f(5) · 2 + f(7) · 2 = 1 · 2 + 7 · 2 + 21 · 2 + 43 · 2 = 144.

9. (16 points) Compute the following.

(a)∫

cos 4t dt =1

4sin 4t + C

(b)∫ 2

1

1

x3dx =

[−1

2x2

]21

=−1

8− −1

2=

3

8

(c) ddx

∫ x

3sec3 t dt = sec3 x

(d) The average value of√

x over the interval [1, 9] =1

9− 1

∫ 9

1

√x dx =

1

8

∫ 9

1x1/2 dx =

1

8

[2

3x3/2

]91

=1

8

[2

3· 27− 2

3· 1]

=13

6.

10. (10 points) Find the area between the curve y = x4 − 2x3 and the x-axis.

y = x4 − 2x3 intersects the x-axis at x = 0 and x = 2, and y = x4 − 2x3 is below thex-axis.

Area = −∫ 2

0(x4 − 2x3) dx = −

[1

5x5 − 1

2x4]20

= −(

32

5− 8− 0

)=

8

5.