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

x3√lnx+ 1

?

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

Solution

The answer is (c), since lnx+ 1 > 0 ⇐⇒ x > e−1.

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 + 2x

4x2 − 8x+ 3

(a) − 1√2

(b) 0 (c) 1√2

(d) 1 (e) Does not exist

Solution

The answer is (c): limx→ 1

2

√

−4x2 + 2x

4x2 − 8x+ 3= lim

x→ 1

2

√

−2x(2x− 1)

(2x− 1)(x− 3)= lim

x→ 1

2

√

−2x

2x− 3=

√

1

2

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

√t

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 + 80√t

25 + 5t= lim

t→∞

250 + 80 1√t

25t + 5

= 50

5. Using the graph of f given below, find limx→1+

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) = esinπx.

(a) πe (b) π (c) −πe (d) −π (e) e

Solution

The answer is (d): g′(x) = esinπx · π cos πx, g′(1) = esinππ cos π = −π.

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

(a) −23 (b) −1

2 (c) 12 (d) 1

3 (e) 23

Solution

The answer is (e): f ′(x) = g′(x2 + 7) · 2x, g′(x) = 23x

−1/3, hence f ′(1) = g′(8) · 2 = 23·81/3

· 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) = limx→2−

f(x) = limx→2−

(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) = lim

h→0+

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

h

or= lim

h→0+

[a(2 + h)2 + b(2 + h) + c]− 9

h

or= lim

h→0+

4ah+ ah2 + bh

hor= lim

h→0+(4a + b+ ah)

or= lim

x→2+

(ax2 + bx+ c)− (4a+ 2b+ c)

x− 2= . . .

Or f ′+(2) = lim

x→2+(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 =1

2.

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 · 312 + 2

= 2

(c) (f ◦ g−1)′ (1), assuming that g−1 exists and is also differentiable. [5 pts]

Solution

(f ◦ g−1)′ (1) = f ′(g−1(1))(g−1)′(1) = f ′(g−1(1))1

g′(g−1(1))

= f ′(−1) · 1

g′(−1)=

−5

−2=

5

2

Note that g−1(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) =

14 − 2 · 1− 2

1− |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− 2

−2− 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 that

f(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:

limh→0+

f(h)− f(0)

h= lim

h→0+

h sin(2h)

h= lim

h→0+

h sin(2h)

h= lim

h→0+sin(2h) = 0

limh→0−

f(h)− f(0)

h= lim

h→0−

|h| sin(2h)h

= limh→0−

−h sin(2h)

h= lim

h→0+(− sin(2h)) = 0

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

f ′(0) = limh→0

f(h)− f(0)

h= lim

h→0

|h| sin(2h)h

= limh→0

2|h| · limh→0

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, butf(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 − 2xydy

dx+ 3y2

dy

dx= 0

dy

dx=

y2 − 4x3

3y2 − 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 = ± 4√8. Hence there are two

vertical tangent lines at the points (− 4√8, 0) and ( 4

√8, 0) respectively.

Putting y = 23x into the equation, we have

x4 − 4

9x3 +

8

27x3 = 8 ⇐⇒ x4 − 4

27x3 − 8 = 0

Consider the continuous function f(x) = x4 − 427x

3 − 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 points on the curve with vertical tangent line.

Exercise Consider the interval of increase/decrease of f(x) = x4 − 427x

3 − 8 and show thatthe equation x4 − 4

27x3 − 8 = 0 has exactly two real roots.