MATH140 Exam 2 - Sample Test Solutions7 © LionTutors 2019 17. B. Take the antiderivative and then...

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MATH140 – Exam 2 - Sample Test 1 – Detailed Solutions

1. D. Create a first derivative number line

+ −

2. C. Create a second derivative number line. Remember that a number is only an inflection point if it changes concavity at that point.

+ − − +

-3 0 3

Since does not change signs at , it is not considered an inflection point.

3. B. For parts A & B, look at the first derivative number line. Parts C & D are false because the domain of the original function is all real numbers. Part E is false because the second derivative does not change signs around .

+ − +

-3/5 0

2 10 2 12 1

1 2

3

'( ) coscos

coscos /

f x xx

xx

x p

= -= -

==

=

4pæ öç ÷è ø 3

p2pæ öç ÷è ø

5 3

4 2

2 2

2

2 30 2010 90

0 10 90 10 3 30 3 3

'( )''( )

( )( )( )

, ,

f x x xf x x x

x xx x x

x x x

= - +

= -

= -

= - += = = -

''( )f x 0x =

0x =

10 6 010 6

3 5/

xx

x

+ == -

= -

133 00

xx

==


4. E. To find the absolute max and min, find the critical numbers that are within the interval. Take the critical numbers, and the endpoints of the interval and plug into the original function.

The critical numbers are however we only consider since it is the only number in the interval.

ß Minimum

ß Maximum

5. D. Simplify by multiplying by the conjugate

.

2

2

2 2

2

2

2

2 2 12

2 42

42

( )( ) ( )( )'( )( )

'( )( )

'( )( )

x x xf xx

x x xf xx

x xf xx

- -=

-

- -=

-

-=

-

0 2 4, ,x = 4x =

3 9( )f =4 8( )f =7 49 5( ) /f =

2 22 2

2 2

2 2

2 2 2 2

2 2

4 9 4 3lim 4 9 4 3 *4 9 4 3

4 9 (4 3 ) 12lim lim4 9 4 3 4 9 4 3

12 12 12lim lim 32 2 44 4

x

x x

x x

x x x xx x x xx x x x

x x x x xx x x x x x x x

x xx xx x

®¥

®¥ ®¥

®¥ ®¥

+ + -+ - -

+ + -

+ - -=

+ + - + + -

= = =++


6. E. Use the definition of the MVT.

(Note that you do not need to consider x = -4 because it is not in the interval)

7. D. This is an optimization problem. We want to minimize the distance between the graph and the point. First solve for y, then use the distance formula. Then take the derivative of the distance and set it equal to 0. Remember that you can square the distance equation before taking the derivative, to make it easier. This will not affect the answer.

2

2

2

2

16 116 1

1 1 11615

15116

15

1 116

164

( ) ( ) '( )f f f x

x

x

x

xx

-=

-

- -=

--

=

- -=

==

2 2

2

56 8

56 8

y x

y x

= -

= -

2 2 2

2 2

2 2 2

2 56 8 0

2 56 82 56 8

0 2 2 160 2 4 1614 42 7

( ) ( )

( )( )( )

/

d x x

d x xd x x

x xx xx

x

= - + - -

= - + -

= - + -= - -= - -

- == -


8. C. To find horizontal asymptotes look at the powers. Since the highest power is the same on top and bottom then divide the coefficients (don’t forget the signs). To find vertical asymptotes, SIMPLIFY, then set the denominator equal to 0.

9. B. Use long division to divide the numerator by the denominator.

10. B. To determine which graph this is look at intercepts, horizontal asymptotes, vertical asymptotes.

The graph has a H.A. at y = 0, so we know the power on bottom must be bigger. This eliminates answer choice D.

The graph has V.A. at x = 1 and x = -1. This eliminates are choices C & E because those functions only have a V.A. at x = 1 (remember to simplify the function and set the denominator = 0 to find V.A.)

The function has a y-intercept at x=0. Therefore when you plug in a 0 for x, you should get 0 for y. This eliminates answer choice A.

Therefore, by process of elimination this graph has to be answer choice B.

4 1 13 1

4 13

( )( )( )( )( )

( )( )

x xf xx x

xf xx

- +=

+ +

-=

+

3 03

xx+ == -

2

2

3 81 3 5

3 38

( )

xx x x

x xx

+- +

- -


11. D. Find the linearization, then plug in 2.1

12. E. To find calculate the differential. To find calculate the actual change.

13. A. This is an application of the Mean Value Theorem.

10 3 210 3 63 163 16

2 1 3 2 1 162 1 9 7

y xy xy xL x xLL

- = - -- = - += - +

= - += - +=

( )

( )( . ) ( . )( . ) .

dy yD

88 2 58

'( )( )( )( )(. )

dy f x dxdy x dxdydy

====

22

2 5 2

54 1 4 2 12

24 159

( . ) ( )

( )

y f f

y

yy

D = -é ùæ ö é ùê úD = - - -ç ÷ ë ûê úè øë û

D = -D =

10 810 8

10 3 72

10 3 1410 17

( ) ( ) '( )

( )

( )( )

f f f x

f

ff

-³

-

-³

- ³³


14. C. This is an application of the 2nd Derivative Test. Plug each of the critical numbers into the 2nd derivative. If you get a positive value then it is a local minimum. If you get a negative value then it is a local maximum.

, so there is a local maximum at x = 0

, so there is a local minimum at x = 3/2

, so there is a local minimum at x = -2

15. B. This is an optimization problem

16. C. First simplify, then take the antiderivative.

So the antiderivative is

0 3''( )f = -

273 24

''( / )f =

2 5''( )f - =

200200

xy

yx

=

=

1

2

2

2

2

2

2 42002 4

2 8002 8008000 2

8002

2 80040020

'

P x y

P xx

P x xP x

x

x

xxx

-

-

= +æ ö

= + ç ÷è ø

= +

= -

= -

=

=

==

20020

102 20 4 1080( ) ( )

y

yPP

=

== +=

2 22

2 2 2

1 1 1sin sin cscsin sin sinx x xx x x+

= + = +

cotx x C- +


17. B. Take the antiderivative and then use the point to solve for c.

The antiderivative of is

18.

This is an optimization problem. Find a formula for volume, then take the derivative and set it equal to 0.

However, x cannot equal 9. Since there is only 18 inches to work with, you cannot take away 9 inches from each side.

19. TRUE. This is the definition of Rolle’s Theorem. Also note that since 𝑓 is differentiable, then it must also be continuous according to the definition of differentiability.

20. FALSE. An absolute minimum can occur at the end point of the interval you are looking at.

21. FALSE. The second derivative also needs to change signs at that point. Look at the function 𝑓(𝑥) = 𝑥) as a counterexample.

2 sinx x- 2 cosx x C+ +2

2

2

5 0 05 1

4

42 2 2

42 4

( ) coscos( )

cos

f x x x CC

CC

f

f

p p p

p p

= + += + += +=

æ ö æ ö= + +ç ÷ ç ÷

è ø è ø

æ ö= +ç ÷

è ø

2

2

2

18 2 18 218 218 2 1 2 18 2 218 2 4 18 2

0 18 2 18 2 40 18 2 18 69 3

( )( )( )( )

' ( ) ( ) ( )( )( )( )' ( ) ( )( )[ ]( )( ),

V l w hV x x xV x xV x x xV x x x

x x xx x

x x

= ´ ´= - -

= -

= - + - -

= - - -= - - -= - -= =


22. It helps to create the 1st and 2nd derivative number lines before trying to graph.

a

0 3

0 2

23.

a) x - Intercept(s): (0,0) b) y – Intercept: (0,0) c) Vertical Asymptote(s): NONE d) Horizontal Asymptote(s): y =0 (since the power on bottom is bigger)

e) Increasing: (-1,1) − + − f) Decreasing: (-∞,1)U(1,∞) g) Relative maximum(s): x =1 -1 1 h) Relative minimum(s): x = -1

− + − + i) Concave Up: (-√𝟑, 0) U (√𝟑,∞) j) Concave Down: (- ∞,- √𝟑) U (0, √𝟑) -√𝟑 0 √𝟑 k) Inflection points: x = -√𝟑, 0, √𝟑


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MATH140 – Exam 2 - Sample Test 2 – Detailed Solutions

1. C. Remember that finding the linearization just means finding the equation of the tangent line. Find the point by plugging 0 into the original function and find the slope by taking the derivative and plugging in 0.

Point:

Slope:

Line:

2. A. Use differentials to approximate , the change going from 10 to 10.3 . Remember that in this case your dx = 0.3.

Add this to to get the answer.

105

( , )

1 25 /( ) ( )f x x -= +

3 2

3

1 52

1 1 102 5 2 5 5 5 10 5

/'( ) ( )

'( )( )

f x x

f

-= - +

- - -= = =

1 1 05 10 5

( )y x-- = -

15 10 5

15 10 5

15 10 5

( )

xy

xy

xL x

-- =

= -

= -

dy10( )f

108 32 4

'( )(. ).

dy f dxdydy

== -= -


3. C. Take the derivative, using the quotient rule. Then set the top and bottom equal to 0. Remember that 𝑥 = 0 cannot be a critical number since it is not in the domain of the original function.

4. D. Find the critical numbers and set up a first derivative number line.

− + − +

-3 0 1

5. E. Find the critical numbers in the interval. Take those numbers and the endpoints of the interval and plug into the original function.

10 3 1010 3 20 2 410 3 17 6

( . ) ( )( . ) ( . )( . ) .

f f dyff

» +» + -»

2 3

1 31 3 2 3

2 2 1 3 2 1 3 2

2

22 2 12 22 2 1 2 233 3 32 2

xf xx

x x xx x x x x xf x

x x x x x x

--

-=

é ù- - -ê ú- - - - + -

ë û= = = =- -

/

// /

/ /

( )( )

( ) ( )( ) ( ) *'( )

( ) ( )

1 12 0 2 63 3

2 0 2

x x x

x x

- = Þ = Þ =

- = Þ =

3 22 30 1 30 1 3

'( )( )( ), ,

f x x x xx x x

x x x

= + -= - += = = -

2 20 2 12 0 1

3 32 2 2

'( ) sin cos coscos (sin )

cos sin

,

f x x x xx x

x x

x xp p p

= += +

= = -

= =

0 0

32

3 12

2 0

( )

( )

f

f

f

f

p

p

p

=æ ö

=ç ÷è ø

æ ö= -ç ÷

è ø

=


6. E. Use the second derivative test to answer this question. Whenever the 2nd derivative is positive at a critical number, there is a local minimum. When the 2nd derivative is negative at a critical number, there is a local maximum. If the 2nd derivative is equal to zero, then you cannot tell.

ß local maximum

ß local minimum

7. C. To find inflection points look at the 2nd derivative. Remember that concavity must change at that point in order to be an inflection point.

- + + -

-1 0 1

has two inflection points (at x = 1 and x = -1)

8. A. Answer choice A is the false statement because x = -4 is not in the domain, so it cannot be considered a critical number.

9. B. Use the definition of the MVT to answer this question.

0 0''( )f =2 8 2 16''( ) ( )f = - = -6 24 6 144''( ) ( )f = =

3 5

2 4

2 2

2

5 20 1260 60

0 60 10 60 1 10 1 1

'( )''( )

( )( )( )

, ,

f x x xf x x x

x xx x x

x x x

= + -

= -

= -

= - += = = -

( )f x


10. E. To find VA set the denominator equal to zero. To find HA look at the highest power on top and bottom.

V.A.

H.A. The powers are the same on top and bottom so we take the coefficients and divide. The coefficient on top is -2 and the coefficient on bottom is 4 (because we have 2 squared). -2/4 = -2. So the H.A. is

11. C. To take the limit as x goes negative infinity, look at the most dominant terms on top and bottom.

12. A. To find V.A., set the denominator equal to 0.

13. D. Take the second derivative and set up a second derivative number line.

1 01 0

4 5 11 2

1 212

14

( ) ( ) '( )

( ) ( )

f f f x

x

x

x

x

-=

-

- - -=

=

=

=

22 2 0( )x + =2 2 02 2

1

xxx

+ == -= -

1 2/y = -

3 9

3

3

3

82

2 12

lim

lim

x

x

xx

xx

®-¥

®-¥=

4 04 04

xxx

- =- ==


− + −

14. B. First simplify the function, then take the antiderivative.

So the antiderivative is

Which simplifies to

15. D. Take the anti-derivative twice, solving for C each time.

3

03

2 2

'( ) sin''( ) coscos

,

f x xf x x

x

x p p

= - += -

- =

=

2p 3

2p

3

3 3

3

1 2

2

( )

( )

tf tt t

f t t-

= -

= -

2

22t t C-

- +-

2

1 22

t Ct-

- +


16. TRUE. First note that since the derivative exists, the function is continuous. If the derivative is non-zero, then it means the function is always increasing or always decreasing. Therefore it cannot have the same y-value twice.

17. TRUE. Every function has an absolute maximum and minimum on a closed interval.

18. FALSE. Functions do not necessarily have an absolute maximum on an open interval.

19. FALSE. If 𝑓 and 𝑔 are both increasing, it means that 𝑓0(𝑥) > 0 and 𝑔0(𝑥) > 0, however it does not necessarily mean that 𝑓 and 𝑔 are greater than 0. When you computer 𝑓𝑔, you have to use the product rule, so if the original function is less than 0, then the product may also be less than 0, and thus not necessarily increasing.

20. This is an optimization problem. Overall we want to minimize cost.

2

2

2

3 2

3 2

6 43 4

4 3 1 4 14 1

5

3 4 52 5

2 0 0 02

2 5 23 27 18 15 23 26

a t tv t t t C

CC

C

v t t ts t t t t C

CC

s t t t tss

= -

= - +

= - += - +=

= - +

= - + += - + +=

= - + += - + +=

( )( )

( ) ( )

( )( )

( )( )( )


21. Take the limits as x goes to infinity and negative infinity.

So the H.A. are y = 1, and y = -1.

22.

2

2

2

888

82

2

( )

Vx y

yx

y

y

=

=

=

=

=

2 2

2 2

39 4 9 3 13 1 3 3 3

39 4 9 3 13 1 3 3 3

lim lim lim lim

( )lim lim lim lim

x x x x

x x x x

xx x x xx x x x

xx x x xx x x x

®¥ ®¥ ®¥ ®¥

®-¥ ®-¥ ®-¥ ®-¥

+= = = =

+

+ -= = = = -

+

2 2

22

2

2 1

2

2

2

3

3

5 1 4 386 12

966

6 9612 96

960 12

9612

12 9682

( )( )

'

C x x xy

C x xx

C xx

C x xC x x

xx

xx

xxx

-

-

= + +æ ö

= + ç ÷è ø

= +

= +

= -

= -

=

=

==


a) Intercept(s): (0, 0) b) Vertical Asymptote(s): NONE c) Horizontal Asymptote(s): y = 0 d) Graph crosses the horizontal asymptote at x = 0 (set the equation to the asymptote to find this)

e) Increasing: (-1,1) _ + _ f) Decreasing: (-∞, -1) U (1, ∞) g) Relative maximum(s): (1, 1/2 ) -1 1 h) Relative minimum(s): (-1, -1/2)

i) Concave Up: (- ,0)U( ,∞) _ + _ +

j) Concave Down:(−∞,- )U(0, )

k) Inflection points: x = - , 0 , - 0

3 3

3 3

3 3 3 3


23. Looking at the graph of the derivative, create a first derivative number line by looking at where the graph is positive and negative. That is, where the graph is above or below the x – axis.

+ + − +

-1 1 4

To create the second derivative number line, look at where the graph is increasing and decreasing.

− + - +

-1 0 3

increasing: (-∞,-1)U(-1,1)U(4,∞) decreasing: (1,4)

concave up: (-1,0)U(3,∞) concave down: (∞, -1)U (0, 3)

local maximum: x = 1 local minimum: x =4 inflection point: x = -1, 0, 3

24. This is an optimization question. Overall you are trying to minimize cost.

2 2

2 2

22 2

2 1

2

2

2

3

3

2

10*2 8*220 16 20

20 2020 16

20 320320' 40

3200 40

320 40

320 4082

20 5(2)

C r rh V r hC r rh r h

C r r hr r

C r r

C rr

rr

rr

rr

r

h

p p p

p p p p

p p

p ppp

pp

p p

p p

-

= + =

= + =

= + =

= +

= -

= -

=

=

==

= =

MATH140 Exam 2 - Sample Test Solutions7 © LionTutors 2019 17. B. Take the antiderivative and then...

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