# math1013 calculus i midterm exam solution white-green ......

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Math1013 Calculus I

Midterm Exam Solution (White-Green Version)

Part I: Multiple Choice Questions

MC Answers: White-Green Version

Question 1 2 3 4 5 6 7 8 9 10 Total

Answer c d c e a b b d e e

MC Answers: Orange-Yellow Version

Question 1 2 3 4 5 6 7 8 9 10 Total

Answer d c d c d e b a c b

Solution (White-Green Version.)

1. Which of the following intervals is the domain of the function defined by the formula f(x) =x2 1

x3lnx+ 1

?

(a) (0,) (b) (e,) (c) (1e ,) (d) (1e ,) (e) (0, e)

Solution

The answer is (c), since lnx+ 1 > 0 x > e1.

2. Across its whole domain, which of the functions given by the following graphs has an inversefunction?

(a)

x

y

(b)

x

y

(c)

x

y

(d)

x

y

(e)

x

y

Solution

The answer is (d): the graph passes the horizontal line test, or the function is one-to-one on itsdomain.

1

3. Find the limit limx 1

2

4x2 + 2x4x2 8x+ 3

(a) 12

(b) 0 (c) 12

(d) 1 (e) Does not exist

Solution

The answer is (c): limx 1

2

4x2 + 2x4x2 8x+ 3 = limx 1

2

2x(2x 1)(2x 1)(x 3) = limx 1

2

2x2x 3 =

1

2

4. The temperature function of an object at t 0 hours is given by T (t) = 250t + 80t

25 + 5t, in C. As

t , approximately to what steady temperature will the object approach?

(a) 0 C (b) 10 C (c) 16 C (d) 40 C (e) 50 C

Solution

The answer is (e): limt

250t + 80t

25 + 5t= lim

t

250 + 80 1t

25t + 5

= 50

5. Using the graph of f given below, find limx1+

f(g(x)) where g(x) = 1 2x.

-4 4

3

-3

x

y

y = f(x)

(a) -2 (b) 0 (c) 1 (d) 3 (e) does not exist

Solution

The answer is (a): As x 1+, g(x) 1, hence f(g(x)) 2.

2

6. Find limx

(

x2 + x

x+

x

)

.

(a) (b) (c) -1 (d) 1 (e) 0

Solution

The answer is (b): limx

(

x2 + x

x+

x

)

= limx

(

1 +1

x+

x

)

= .

7. Find the equation of the tangent line to the graph of the function y = f(x) = 6x2 + 3x at thepoint (2, f(2)).

(a) y = 15x+ 12 (b) y = 21x+ 24 (c) y = 21x 60

(d) y = 12x 6 (e) y = 15x 48

Solution

The answer is (b): f (x) = 12x+ 3, f (2) = 21, f(2) = 18.Tangent line equation y = 21(x 2) 18 = 21x+ 24.

8. Compute g(1) for g(x) = esinx.

(a) e (b) (c) e (d) (e) e

Solution

The answer is (d): g(x) = esinx cos x, g(1) = esin cos = .

9. If f(x) = g(x2 + 7) and g(x) = x2/3, compute the derivative f (1).

(a) 23 (b) 12 (c) 12 (d) 13 (e) 23Solution

The answer is (e): f (x) = g(x2 + 7) 2x, g(x) = 23x1/3, hence f (1) = g(8) 2 = 2381/3 2 =23 .

3

10. The number of gallons of water in a tank t minutes after the tank started to drain is V (t) =25(30 2t)2. How fast is the water running out at the end of 8 minutes?

(a) 250 gal/min (b) 500 gal/min (c) 700 gal/min (d) 1000 gal/min (e) 1400 gal/min

Solution

The answer is (e): V (t) = 100(30 2t), V (8) = 1400 gal/min.

Part II: Answer each of the following questions.

11. [10 pts] The graphs of y = f(x), y = f (x), and y = f (x) are shown below.

x

y

Graph A

Graph B

Graph C

Graph C

Graph A

Graph B

(a) Match the following functions with their graphs: [4 pts]

The graph of y = f (x) is : Graph B

The graph of y = f (x) is : Graph A

(b) Give brief reason to support your answers above. [6 pts]

Solution

Derivative is the slope of the graph, or the rate of change of the function.

When x > 0, the positive slope of Graph C gradually increases and then decreases to 0, andturns into negative slope afterward. Graph B matches the changing slope of Graph C; i.e.C = B.

Similarly, Graph A matches the changing slope of Graph B; i.e., B = A.

4

12. [10 pts] A function f is defined and continuous on the interval 0 < x < 5. It is known that

f(x) =

2x+ 5 if 0 < x < 2

ax2 + bx+ c if 2 < x < 5

and the function has the following right-hand derivative and derivative at x = 2 and x = 3 respec-tively:

f +(2) = 0 and f(3) = 1.

(a) Find the function value f(2). Justify your answer for full credit. [3 pts]

Solution

Since f is continuous at x = 2,

f(2) = limx2

f(x) = limx2

(2x+ 5) = 2 2 + 5 = 9

(b) Express the right hand derivative f +(2) as the limit of an appropriate expression. [3 pts]

Solution

f +(2) = limh0+

[a(2 + h)2 + b(2 + h) + c] (4a+ 2b+ c)h

or= lim

h0+

[a(2 + h)2 + b(2 + h) + c] 9h

or= lim

h0+

4ah+ ah2 + bh

hor= lim

h0+(4a + b+ ah)

or= lim

x2+

(ax2 + bx+ c) (4a+ 2b+ c)x 2 = . . .

Or f +(2) = limx2+

(2ax+ b).

(c) Find the coefficients a, b, and c. [4 pts]

Solution

f(2) = 4a+ 2b+ c = 9 (1)

f +(2) = 4a+ b = 0 (2)

f (3) = 6a+ b = 1 (3)

(3) (2): 2a = 1, a = 12.

Hence by (2), b = 4a = 2, and by (1), c = 9 4a 2b = 9 2 + 4 = 11.

5

13. [15 pts] The functions f, g are differentiable everywhere. Using the given table of function values,

x -2 -1 0 1 2

f(x) 1 -4 -6 1 -2

f (x) 2 -5 1 3 5

g(x) -1 1 3 6 10

g(x) -5 -2 1 3 5

compute the following derivatives:

(a)d

dx

(

x2f(x)

g(x)

)

x=2[5 pts]

Solution

d

dx

(

x2f(x)

g(x)

)

x=2=

g(x)(x2f(x)) x2f(x)g(x)[g(x)]2

x=2

=g(2)[2(2)f(2) + (2)2f (2)] (2)2f(2)g(2)

[g(2)]2

=(1)[2(2)(1) + 4(2)] 4(5)

(1)2 = 16

(b)d

dxln([f(x)]2 + 2)

x=1

[5 pts]

Solution

d

dxln([f(x)]2 + 2)

x=1

=1

[f(x)]2 + 2

d

dx([f(x)]2 + 2)

x=1

=1

[f(1)]2 + 2 2f(1)f (1) = 2 1 3

12 + 2= 2

(c) (f g1) (1), assuming that g1 exists and is also differentiable. [5 pts]

Solution

(f g1) (1) = f (g1(1))(g1)(1) = f (g1(1)) 1g(g1(1))

= f (1) 1g(1) =

52 =

5

2

Note that g1(1) = 1 by the given table.

6

14. [15 pts] Answer the following true-false questions. Give brief reason to justify each answer.

(a) The equationx4 2x 2

x= |x| has at least two roots in the interval [-2,2]. [5 pts]

True False

Brief reason.

Solution

Let f(x) =x4 2x 2

x |x|. Then

f(2) =24 2 2 2

2 |2| = 3 > 0, f(1) = 1

4 2 1 21

|1| = 4 < 0.

Since f is continuous over the interval [1, 2], by the Intermediate Value Theorem, there is anx1 in the interval (1, 2) such that f(x1) = 0.

Similarly, f is continuous over the interval [2,12 ], and

f(2) = 16 + 4 22 2 = 11, f(1

2) =

116 + 1 2

12 1

2=

11

8> 0,

by the Intermediate Value Theorem, there must be another root x2 of f in (2,12 ) such thatf(x2) = 0.

Consequently, f has at least two roots in the interval [2, 2].(b) The function f(x) = |x| sin(2x) is not differentiable at x = 0. [5 pts]

True False

Brief reason.

Solution

The left-hand derivative and right-hand derivative are:

limh0+

f(h) f(0)h

= limh0+

h sin(2h)

h= lim

h0+

h sin(2h)

h= lim

h0+sin(2h) = 0

limh0

f(h) f(0)h

= limh0

|h| sin(2h)h

= limh0

h sin(2h)h

= limh0+

( sin(2h)) = 0

Thus f (0) = 0 exists, and f is differentiable at x = 0. Or,

f (0) = limh0

f(h) f(0)h

= limh0

|h| sin(2h)h

= limh0

2|h| limh0

sin(2h)

2h= 0 1 = 0

(c) If f (a) and g(a) do not exist, then the quotient functionf

gcan not be differentiable at x = a.

True False [5 pts]

Brief reason.

Solution

For example f(x) = |x|+1 and g(x) = |x|+1 are both not differentiable at x = 0, but f(x)g(x)

= 1

is clearly differentiable at x = 0.

7

15. [10 pts] Consider the curve defined by the equation x4 xy2 + y3 = 8.

(a) Find the equation of the tangent line to the curve at the point where the curve intersects they-axis. [6 pts]

Solution

Differentiating both sides of the equation, we have

4x3 y2 2xy dydx

+ 3y2dy

dx= 0

dy

dx=

y2 4x33y2 2xy

When x = 0, y = 2, and the slope of the tangent tothe curve at (0, 2) is

dy

dx

(0,2)

=22

3 22 =1

3

The equation of the tangent line to the curve at (0, 2)is

y =1

3x+ 2

3 3

3

3

x

y

(b) Is it true that the curve has at least four vertical tangent lines? Given reason for you answer.[4 pts]

Solution

True.

There is a vertical tangent line at (x, y) whendy

dx , or dx

dy= 0; in particular,

3y2 2xy = y(3y 2x) = 0, i.e., y = 0 or 3y 2x = 0..

Putting y = 0 into the equation of the given curve, we have x = 48. Hence there are two

vertical tangent lines at the points ( 48, 0) and ( 4

8, 0) respectively.

Putting y = 23x into the equation, we have

x4 49x3 +

8

27x3 = 8 x4 4

27x3 8 = 0

Consider the continuous function f(x) = x4 427x3 8. Note that f(3) = 81 + 4 8 > 0,f(0) = 8 < 0, and f(3) = 81 4 8 > 0. f(x) must have at least two roots in the interval(3, 3) by the Intermediate Value Theorem.Graphically speaking, the line y = 23x intersects the given curve at least twice.

Thus there are at least four p