Solutions to Sample Questions Final Exam

8/3/2019 Solutions to Sample Questions Final Exam

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www.TutoringZone.comMAC2311FinalExamSolutionstoSampleQuestions

1)B.( )( ) f g x =

21 1 1 5 6

( ( )) 15 5 55 5

x x f g x f

x x xx x

+ + = = + = + = + + ++ +

Domainoftheinside1

5x + x> 5. BecauseALSOinthedenominator.

Domainoverall:xcannotequal 5.

2)E.( )( ) f g x =

1( ( ( ))

4 f g x

x=

Domainoftheinside: 0x

Domainoverall: 0x becauseofthesquareroot. ALSOthedenominatorcannot

equalzero,soxcannotequal16either.

Domainoverall: 0x andnotequalto16.

3)D.2 2( )( ) ( ( )) (1 ) 1 f g x f g x f x x= = =

Domainoftheinside: 21 x ,norestriction

Domainoverall 21 x ,

21 0

(1 )(1 ) 0

x

x x

+

needtosetupanumberline!


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4)E.2( )( ) ( ( )) ( ) 1 ( ) 1g f x g f x g x x x= = = =

Domainoftheinside: 0x

Domainoftheoverall:norestriction.

5)C.Tofindthedomainofasquareroot,settheinsidegreaterthanorequaltozero

andsolve.Makesureifyoudivideormultiplybyanegative,toswitchthe

directionoftheinequalitysign.

6) E.

To find domain, set the inside greater than or equal to zero, and now you have aninequality where you need to set up a number line.

7)B.3 4

14 2

x

x

getonesidetobezero:

3 4

1 04 2

x

x

now

find

a

common

denominator

3 4 4 20

4 2 4 2

x x

x x

3 4 (4 2 )0

4 2

x x

x

3 4 4 20

4 2

x x

x

+

1 20

4 2

x

x

Nowsetthetopandbottom=0tosetupanumberline.

+ + Choosenumbersbeforeandaftereachpoint.

______ _______ ______

1

2 2

Wewanteverythingpositive,writetheinterval.


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8) BStarteverylimitbypluggingin. Inthisquestionyouwillobtain

0

0,meaning

youwillneedtosimplify. Todothis,findacommondenominator,whichis 24x

2

2 2

2

1 1 4

4 4lim

2x

x

x x

x

=

2

22

2 2 2 22 2 2 2

44 1 ( 2)( 2) 1 2 2 2 4 14lim lim lim lim

2 4 2 4 2 4 4(2) 16 4 x x x x

x

x x x xx

x x x x x x

+ + +

= = = = = =

i i

9) AStarteverylimitbypluggingin. Inthisquestionyouwillobtain

0

0,meaning

youwillneedtosimplify. Todothis,LOOKINSIDEtheabsolutevalueto

determinewhetheritispositiveornegative. Again,youhavetoLOOKINSIDE.

3meansnumberslessthan3. Pluginanynumberlessthan3INSIDEof|x3|

anditwillbenegativeinside,inthiscase. Thusweneedtomultiplybya

negativesotheexpressionstayspositive:

2 23 3 3 3

| 3 | ( 3) ( 3) 1 1lim lim lim lim

9 9 ( 3)( 3) 3 6 x x x x

x x x

x x x x x

= = = =

+ +

10)D.2

4

7 3lim

4x

x

x

+

plug

in

to

get

0

0 ,

so

simplify

by

using

the

conjugate:

2 2 2 2

2 2 24 4 4

7 3 7 3 7 9 16lim lim lim

4 7 3 ( 4)( 7 3) ( 4)( 7 3) x x x

x x x x

x x x x x x

+ = =

+ + + + + +

2 24 4

( 4)( 4) ( 4) 8 4lim lim

6 3( 4)( 7 3) ( 7 3)x x

x x x

x x x

+ = = =

+ + +


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11) D.

3

1 1

3lim3x

x

x

plugintoobtain 0

0,sosimplifybyfindingacommondenominator.

3 3 3 3

3 1 1 33 1 1 13 3 3lim lim lim lim

3 3 3 3 3 9 x x x x

x x

xx x x

x x x x x

= = = =

12)C.Soyouarereallyfinding

9

3lim

9x

x

x

=

0

0 whichmeanssimplifywiththe

conjugate.

9 9 9

3 3 9 1 1lim lim lim

9 63 ( 9)( 3) 3 x x x

x x x

x x x x x

+ = = =

+ + +

13)C.Thisisa2sidedlimit,soyouapproachwherex>0andfromwherex


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14)B.3

lim ( )x

r f x+

= Starteverylimitbypluggingin,andinthiscaseyou

obtain0

0

,sosimplifybyfactoring,oneyoudosoplug 3backin.

2lim ( )

xs f x

=

Again,starteverylimitbypluggingin,inthiscase

youwillobtaina#

0,soyoumustchooseaclosenumberlessthan 2,try

pluggingin 2.1forxtodetermineiftheanswerisplugorminusinfinity.

15)D.Thewaytoevaluatethislimitistofoiloutthenumeratorandthencomparethe

termswiththehighestdegreesinthenumeratoranddenominator. Youendup

comparing2x2(numerator)and x(denominator),whichbecomes 2x. Now,

thinkaboutplugginginlargenegativevaluesforx.Whathappens? 2x

becomesinfinitelylargerpositively,soyougetpositiveinfinity.

16)CThisproblemisrelatingtheideaofcontinuity.Noticeall3pieceswillmeetatx=

1. Thereforepluginx=1intoall3parts,givingyou:

p+3

1

2+q

Nowwhenyoupluginx,youobtainayvalueasaresult. Inthiscaseall3y

valueshavetobeequalforthefunctiontobecontinuousatx=1. Thereforeset

themallequalto 1,solveforpandq.


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17)D.Factorf(x)toobtain:

3( 1)( 1)( )

( 1)( 2)

x xf x

x x

+=

+ +

Sox= 1isahole(RD),x= 2isaVerticalAsymptote(NRD).

18) A.Forthisproblemyoushouldfirstfactorf(x). Youllget (3 1)( 1)( 1)( 1)

x x

x x

+

+

Youllnoticethatthefactor(x+1)willcancelout. Thisleavesyouwith(3 1)

( 1)

x

x

.

Usethissimplifiedfunctiontoevaluateallthelimitsbutdontforgetthatsince(x

+1)cancelledoutx= 1isaremovablediscontinuity.A.NOTTRUE,becauseifthenumeratoranddenominatorhavethesamedegree,whichinthiscaseis1,youjusttake

theratioofthecoefficientsofxandyouget3/1whichis3.

B. TRUE,becausethefactor(x+1)cancelledoutfromthenumerator

and

denominator.

To

find

the

actual

xvalue

justsetx+1=0andsolveforx,whichisx= 1.

C. TRUE.Useyoursimplifiedfunction(3x 1)/(x1)andjustplugin1forx.

D.TRUE. Sinceyoucanpluginx=3intheoriginalequationandgetanumberthefunctioniscontinuousatx=3.

Remember,afunction

is

only

discontinuous

for

whatever

xvaluesmakethedenominatorzero.


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19)D.Forthisproblemjustrememberhowtofindleftandrighthandlimits.

Rememberthatafunctioniscontinuousatacertainxvalueifitisdefinedfor

thatxvalue

and

the

left

and

right

hand

limits

are

equal

to

each

other

there.

Letstryit.A.TRUE. Findthelefthandlimitasx=>2 soyouusethemiddlepartofthefunction,x,andplugin2. Youshouldget2. Findtheright

handlimitasx=>2soyouuse2x2andplugin2andyouget2.

Fromthisweknowthatthelimasx=>2exists. Nowevaluatef(2).

The

only

place

you

can

plug

x

=

2

into

is

2x

2

so

you

get

f

(2)

=

2.

Sincethelimasx=>2=f(2)=2thenwecansaythatf(x)is

continuousatx=2.

B. TRUE. Thiswasmeanttobealittletrickybutdoesnthavetobe.Justrememberthatx=1isintheinterval00whichis0. Sincetheleftandrighthandlimitsarentequalto

eachother

f(x)

cannot

be

continuous

at

x=0.

E. TRUE. RefertohowwefoundtheleftandrighthandlimitsinpartA.


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20)B.P)False,simplifyandyouwillseeyouhaveoneholeandoneverticalasymptote.

Q)True. x=4isaholeandtodefineityouplugx=4intothesimplifiedform.

21)C.( )( ) f g x =

1(ln )

lnf x

x=

Insidelnx>0

Overalllnx>0 alsolnxcannot=0So

xcannot

equal

1.

22) E.

Solution:To find the derivative of f(x), we will need to use the product rule since the equation is 2 separate

functions, e

2x

and cot(2x), being multiplied by each other. Furthermore, you must alsosee that we will be using the chain rule as we work through the product rule:

f (x) = (first)(derivative of the second) + (second)(derivative of the first)f (x) = (e2x) (-csc2(2x) (2) ) + (cot(2x))( e2x(2) )

After you take a derivative, you need to factor out what is in common:

f (x) = 2e2x[-csc2(2x) + cot(2x)]

So the answer is E since the inside of the brackets above were just switched.


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23) B.

Solution:In this question you have to notice tan is the outside function for sin(3x), it is not being

multiplied, but rather sin(3x) is inside of tan, making this the chain rule. So we will takethe derivative of the outside function, tan, then multiply by the derivative inside function,sin (3x).

f (x) = sec2( sin (3x) ) ( cos (3x) ) (3)

You may notice the 3 on the last part is due to the derivative of sin (3x) being cos (3x)times 3, another chain rule!

So the answer is B.

24) C

Solution:

Here we will use the quotient rule:2

'( )LdH HdL

f xL

=

3 2

6

(4 3) (1) 3(4 3) 4'( )

(4 3)

x x xf x

x

+ +=

+

Next we need to factor:2

6

(4 3) [(4 3) 3(4)]'( )

(4 3)

x x xf x

x

+ + =

+

Now reduce the 4x+3 in the numerator and denominator, and clean up inside thebrackets:

4

3 8'( )

(4 3)

xf x

x

=

+So the answer is C.


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25) E.Solution:

Here you have to recall 12

1(tan )

1

dx

dx x

=+

So using the chain rule with this derivative, we get:

1

2

1 1(tan )

2 21

2

d x

dx x

=

+

i

Now we need to use our algebra skills to clean this up, I will start with the denominator:

2 2 2

1 1 1 1 1 1

2 2 24 41

4 4 4 4

x x x

= =

+ + +

i i i

Now 1 divided by anything is equivalent to multiplying by the reciprocal, meaning:

2 2

1 1 4 1

2 4 24

4

xx

= ++

i i and now reduce the fraction to get:2 2

4 1 2

4 2 4x x

= + +

i

So the answer is E.

26) D.

Solution:Here I would use the quotient rule:

2'( )

LdH HdLf x

L

=

Also I would change the square root symbol to the power.

1

22 1d x

dx x

=

1 1

2 2

2

1(2( ) ) (2 1)(1)

2 x x x

x

Now we need to clean up the numerator:

1 1

2 2

2

1(2( ) ) (2 1)(1)

2 x x x

x

=

1 1

2 2

2

( ) (2 1)(1) x x x

x

=

1 1

2 2

2

2 1x x

x

+=

1

2

2

1x

x

+

Lastly, notice the power can be rewritten as a square root, so the answer is D.


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27) C.

Solution:

Here, you have to be careful with your rules, to split up the problem (since everything is

being added), I will look at each term and take the derivative:

( )xd

xdx

use the product rule, so the derivative = ( )( ln ) ( )(1)x xx +

( )d

xdx

use the power rule, so the derivative = 1x

( )d

dx

this is a constant, so the derivative = 0.

Add all these together and the answer is C.

28) B.

Solution:

Here we will take the derivative and plug in 0. You have to recall

1

(log ) lna

d

xdx x a=

Now we take the derivative as follows:

2 2

9 2

1(log ( 1)) (2)

( 1) ln 9

x x

x

de e

dx e+ =

+i

In this case we also needed to use the chain rule for 2 1xe + , where 2 2( 1) (2)x xd

e edx

+ =

Now we do not need to simplify, but rather just plug in x = 0 into the derivative, so2(0)

2(0)

1 1 1'(0) (2) (2)

( 1) ln 9 (2) ln 9 ln 9f e

e= = =

+i i , so the answer is B.


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29) B.

Solution:

Here we will use the quotient rule:2

'( )LdH HdL

f x

L

=

1 423 3

2

46 4( ) (3 4 )(1)

3'( )

x x x x x

f xx

=

At this point, all we are doing for this question is trying to find f (1), so plug in x = 1into the derivative, and then simplify.

1 423 3

2

41 6(1) 4( )(1) (3(1) 4(1) )(1)3

'(1)(1)

f

= =

166 (3 4)

3

=

2 5( 1)

3 3

=

So the answer is B.

30) E.

Solution:

Here we will take the derivative recalling the following rule: ( ) lnx xd

a a adx

= , also keep

in mind we will use the chain rule for the derivative of the inside, 2x 1.

2 1'( ) 2 ln 2(2)xf x =

Again, the last (2) is because of the chain rule.

At this point, plug in x = 0.

2(0) 1 1'(0) 2 ln 2(2) 2 ln 2(2) ln 2f = = =

So the answer is E.


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31) E.

Solution:

For f (0), we need to take the derivative and plug in 0.

2'( ) 12 (2 (2)) cosx x f x x xe e x= + +

You may notice for the second term, I used the product rule.

Now plugging in x = 0 we get:2 0 0'(0) 12(0) (2(0) (2)) cos0 f e e= + + = 0 (0 + 2) + 1 = -2 + 1 = -1 (answer).

We also need f (0), so we will take the derivative again. First, I will rewrite thederivative:

2'( ) 12 (2 (2)) cosx x f x x xe e x= + + 2'( ) 12 2 2 cosx x f x x xe e x= +

Now when taking the derivative again, the second term will require the product rule,giving us:

''( ) 24 (2 (2)) 2 sin x x x f x x xe e e x= +

Plugging in x = 0 gives us:0 0 0''(0) 24(0) (2(0) (2)) 2 sin 0 f e e e= + =0 (0+2) 2 0 = -4

So f (0) = -1, f (0) = -4, the answer is E.


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32) C.

Solution:We can either

I. use a product rule, quotient rule, AND chain rule, OR

II. use logarithmic differentiation.

In this case I will use logarithmic differentiation.We start by taking the ln of both sides:

4ln ln

1

xe x

yx

+=

I will simplify the right side:

4ln

1

xe x

x

+

=1

ln ln 4 ln( 1) ln( 4) ln( 1)2

xe x x x x x+ + = + +

Now we have:

1ln ln( 4) ln( 1)

2 y x x x= + +

Take the derivative to get:

1 1 1 11

2 4 1

dy

y dx x x= +

+

Multiply by y on both sides (which is the ORIGINAL EQUATION) to get:

1 1 1 4

1 2 4 1 1

xdy e x

dx x x x

+

= + +

Now plug in x = 0.01 1 1 0 4

12 0 4 0 1 0 1

dy e

dx

+ = + +

= ( ) ( )1 17 17

1 1 2 28 8 4

+ + = =

So the answer is C.


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33) A.

Solution:To find the equation of a line, we need point and slope.

The point comes from the original.

Plug in x = 0 into f(x) to get:4(0)

2 0 2 1(0)

3 3 1 2f

e

+= = =

+ +, so we have POINT:

1(0, )

2

To find the slope, we need to take the derivative using the quotient rule:

2'( )

LdH HdLf x

L

=

4 4

4 2

(3 )(1) (2 )( (4))

'( ) (3 )

x x

x

e x e

f x e

+ +

= +

You may notice the (4) at the end of the top right of the derivative came from the

derivative of 4xe .

Now we need to plug in x = 0 for the slope (no need to simplify since all we need is theslope).

4(0) 4(0)

4(0) 2

(3 )(1) (2 0)( (4))'(0)

(3 )

e ef

e

+ +=

+=

2

(4)(1) (2)(4) 4 1

(3 1) 16 4

= =

+

Now we have point and slope, we can use point-slope form: 1 1( )y y m x x =

Point:1

(0, )2

Slope:1

4 Line:

1 1( 0)

2 4y x = ,

1 1

4 2y x= +

So the answer is A.


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34) E.

Solution:

We need to take the derivative and plug in 1. To take the derivative we will use the

quotient rule: 2'( ) LdH HdLf x L=

2 3

2

(1 2 )(3 ) ( 2)( 2)'( )

(1 2 )

x x xf x

x

+ =

no need to simplify sine we will plug in x = -1 to

get:

2 3

2

(1 2( 1))(3( 1) ) (( 1) 2)( 2)'(1)

(1 2( 1))f

+ =

=

2

(3)(3) (1)( 2) 11

(3) 9

=

HOWEVER, the question asked for the slope of the NORMAL line, so the slope of the

normal line is the perpendicular slope, which is9

11 , so the answer is E.

35) B.

Solution:To find a horizontal tangent line, we need to find where f (x) = 0.

Taking the derivative we get: 3 2'( ) 4 6 2 f x x x x= +

Now set it equal to zero to find where the slope is zero (hence a horizontal line).3 2

2

4 6 2 0

2 (2 3 1) 0

2 (2 1)( 1) 0

10, ,12

x x x

x x x

x x x

x

+ =

+ =

=

=

So the answer is B.


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36) D.

Solution:To find a horizontal tangent line, we need to find where f (x) = 0.

Taking the derivative, we need to use the product rule:f (x) = (first)(derivative of the second) + (second)(derivative of the first)Note: We will also use the chain rule while taking the derivative.

Now the derivative is:1 2'( ) ( )(2(3 6) (3)) (3 6) (1) f x x x x= +

Since we set it equal to zero, we will need to factor to solve:1 2( )(2(3 6) (3)) (3 6) (1) 0

(3 6)( (2)(3) 3 6) 0

(3 6)(6 3 6) 0(3 6)(9 6) 0

3 6 0, 2

29 6 0,

3

x x x

x x x

x x xx x

x x

x x

+ =

+ =

+ =

=

= =

= =

So the answer is D.


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37) A.

Solution:To find the equation of a line, we need POINT and SLOPE.

The POINT comes from the original equation, so we have 2(3) ln (3) 8 ln1 0f = = = Giving us the POINT: (3,0)

Now the SLOPE will come from the derivative. Before we take the derivative, I will useproperties of logarithms to simplify f(x) as follows:

2 21( ) ln 8 ln( 8)2

f x x x= =

Now taking the derivative gives us:

2

1 1'( ) (2 )

2 8

f x x

x

=

Note: The (2x) came from using the chain rule.

Plug in x = 3 to get2

1 1 1 1'(3) (2(3)) (6) 3

2 3 8 2 1f = = =

Now we have

POINT: (3,0)

Slope: 3

Using point-slope form we have:

y 0 = 3(x 3)y = 3x 9So the answer is A.


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38) A.

Solution:We need to use implicit differentiation; notice the first term will require the product rule.

(1 ) (1) 2 2 3dy dy x y x ydx dx

+ + = +

Now we can just plug in (-1,1) and solve fordy

dx

( 1)(1 ) (1)(1) 2( 1) 2(1) 3

1 1 2 2 3

3 4

4

3

dy dy

dx dx

dy dy

dx dx

dydx

dy

dx

+ + = +

+ = +

=

=

So the answer is A.

39) A.

Solution:

We need to use implicit differentiation; notice the first term will require the product rule.

2 2

2 2

2 2

2

2

(2 ) 9 2

2 9 2

2 9 (2 )

2 9

2

dy dy x y x x

dx dx

dy dy xy x x

dx dx

dy xy x x

dx

dy xy x

dx x

+ + =

+ =

+ =

+=

So the answer is A.


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40) B.

Solution:We are finding a line so we will need POINT and SLOPE.The POINT is already given: (1,3)

We need to use implicit differentiation to find the slope; notice the second term willrequire the product rule.

2 2 23 (2 ) (2 ) 0dy dy

y x y y xdx dx

+ + =

At this point we just need the slope at (1,3), so plug in and solve fordy

dx(aka the slope).

2 2 2

3(3) (1) (2(3) ) (3) (2(1)) 0

27 6 18 0

33 18

18 6

33 11

dy dy

dx dx

dy dy

dx dx

dy

dx

dy

dx

+ + =

+ + =

=

= =

Now we have:POINT: (1,3)

SLOPE:6

11

Using point-slope form of a line we obtain:

63 ( 1)

11

6 63

11 11

6 39

11 11

y x

y x

y x

=

= + +

= +

So the answer is B.


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21

41) D.

Solution:

A(x)= g(f(x))

A (x) will be the chain rule, giving us:A (x) = g (f(x)) f (x)

Plug in x = 1 to obtain:A (1) = g (f(1)) f (1)

We have f(1) and f (1)A (1) = g (-1) (-2)

All we need is g (-1), well we have 2( ) 3g x x x= + , so '( ) 6 1g x x= + , and '( 1) 5g = .

So we have

A (1) = (-5)(-1) = 10So the answer is D.

42) A.

Solution:The first thing we need to look at is where any of the functions are not continuous at

x = 0, and by examining all the functions,2

3 2

3 6

2

xy

x x

=

has a domain restriction at x = 0,

so the answer is A.

43) E.

Solution:P) We have a discontinuity at x = 0, so if f(x) is not continuous at x=0, it is notdifferentiable at x = 0. Thus the given statement is TRUE.

Q) f(0) = 0, so the function is continuous, but that does not imply differentiability. We

need to take the derivative and obtain1

31

3

2 2'( )

3 3

f x x

x

= = , and thus f (0) is undefined

and hence not differentiable. Thus the given statement is FALSE.

R) It may appear that the answer is false, however if you graph f(x) = |x 1|, you will seethe x value that is being asked is NOT the point where it changes, so the statement endsup being TRUE.

So the answer is E.


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22

44) C.

Solution:

Maximum height occurs when the derivative (velocity), is 0.

'( ) 320 32 0

32 320

10

s t t

t

t

= =

=

=

Plug t = 10 into s(t) to get s(10) = 1600 feet, so the answer is C.

45) C.

Solution:

To find where f(x) is negative, we plug in the answer choices and find where f(x) < 0.To find where f(x) is increasing, we plug in the answer choices and find where f (x) > 0.

We have3 2

2

( ) 3 5

'( ) 3 6

f x x x

f x x x

= +

= +

Plug in all the answers until you get a solution with f(x) < 0, and f (x) > 0 and you will

see the answer is C.

46) C. First, to find where the function is decreasing, we used the first derivative. Withthe first derivative, we will find the critical points, draw a number line, and find where itis increasing and decreasing:

f (x) = 9x2 18x 72 = 9(x2 2x -8). Setting this equal to zero gives you criticalpoints of x = -2 and x = 4. Plotting these on a number line and testing points willshow you that f(x) is decreasing on (-2, 4).

Now do the same thing with the second derivative to find where f is concave up andconcave down:

f (x) = 18x 18. Setting this equal to zero gives you an inflection point at x = 1.Plotting this on a number line and testing points will show you that f is concave

down on (-, 1).

So f(x) is BOTH decreasing and concave down on the interval (-2, 1)


23/41


24/41

24

49) B. To find where f(x) is decreasing we need to look at the first derivative (product rule).

1'( ) 2 2(1 ln ) 2 2 2ln 4 2ln

'( ) 4 2ln

f x x x x xx

f x x

= + + = + + = +

= +

Now set this derivative equal to 0 and set up a number line, but remember the domain of the

original function. You cannot take the ln of 0 or a negative number, so the number line actually

starts at 0, not -.

0 = 4 + 2ln x

-2 = ln x

2

1

e= x

- +

0

2

1

e

So the function is decreasing on2

10,

e

50) D. To find the relative max and min first take the first derivative and set it equal to 0.

3'( ) 2 1 f x x

x= using a common denominator of x gives us:

22 3'( )

x xf x

x

=

22 3 (2 3)( 1)'( )

x x x xf x

x x

+= = , setting the numerator and denominator = 0

gives us x = 0, -1, and3

2, however, x = 0 and -1 are not in the domain of f(x)!

The domain is x > 0 so you can only plug in numbers greater than 0.,

- +|-------|------- >

0

3

2

Remember there is a minimum when the derivative goes from negative to positive (when theoriginal graph goes from decreasing to increasing) around a critical number. And there is amaximum when the derivative goes from positive to negative around a critical number. So we

have no maximum and we have a minimum at x =3

2


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51) C. To find the inflection points first take the second derivative and set it equal to zero.f(x) = x6 10x4 6x + 24f (x) = 6x5 40x3 6f (x) = 30x4 120x2 = 30x2(x2 4) =30x2(x 2)(x + 2)

Setting f (x) equal to 0 and solving gives you x = 0, 2, and -2. Now plot these points on a

number line and test points to see where the second derivative changes signs. Remember, thex-value is only an inflection point if the second derivative changes signs, and is in the domain ofthe original (in this case the domain is all reals).

+ - - +< ------|---------|---------|------- >

-2 0 2

So the only inflection points are x = 2 and -2.

52) C.

2

2 2'( )

4 (2 )(2 )

x xf x

x x x

= =

+

Setting the numerator and denominator equal to zero gives you x = 0, 2, and -2.However, x = 2 and -2 are not in the domain of the original, thus the only critical numberis x = 0.

53) B.

A. FALSE. To find out if f has a maximum at x = 0, find the critical numbers bysetting the top and bottom of the first derivative equal to 0. This gives you x = 0

and x =3

5 . Plotting these on a number line and testing points will show that

f(x) is increasing on (-,3

5 ) U (0, ) and decreasing on (

3

5 , 0). This means

that at x = 0, f (x) goes from negative to positive. So x = 0 is a relativeminimum, not a relative maximum.

B. TRUE. Since the denominator of the derivative equals 0 at x = 0. This means

that x = 0 is a vertical tangent since f (x) is undefined AND in the domain of f(x).Since the first derivative goes from a negative slope to a positive slope at x = 0, ithas a cusp.

C. FALSE. See explanation for B.D. FALSE. f(x) is continuous so it does not have a hole.E. FALSE. f(x) does not have a vertical asymptote because f(0) is defined.


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54) B.

Plug each x-value of the horizontal tangent line into the second derivative:f(-2) = -5f(0) = 3f(2) = -5.

Using the second derivative test this tells you that there is a maximum at x = 2 and -2,and there is a minimum at x = 0.

55) B. Use the second derivative test for this problem. Test x = -1, 0, and 4 because those arecritical numbers.

f (-1) = +, so it is a minimumf (0) = + so it is a minimumf (4) = - so it is a maximum

56) C.

Solution:

Maximum height occurs when the derivative (velocity), is 0.

'( ) 320 32 0

32 320

10

s t t

t

t

= =

=

=

Plug t = 10 into s(t) to get s(10) = 1600 feet, so the answer is C.

57) B. First find an equation that relates the information we are given. Here we are talking aboutcircumference and diameter so use c d= .

Take the derivative of this equation with respect to time.

dc dd

dt dt =

Now plug in the values we are given.

4dd

dt=

4 dd

dt=


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58) D.

1

2 A bh= also 2h b=

Since the question gives us 5h = and 2dhdt

= , we should solve the area equation in terms

of height, meaning:

1

2b h= , 2

1 1 1 1

2 2 2 4 A bh h h h

= = =

So 21

4A h=

Taking the derivative with respect to time gives us

1

2

dA dhh

dt dt =

Plug in our knowns to get 5dA

dt= (answer).

59) E.To find the absolute min and max of f(x), take the first derivative (using the quotient rule)and set the numerator and denominator equal to zero.

2 2 2 2

2 2 2 2

( 2)(2 ) 2 4 4 ( 4)'( )

( 2) ( 2) ( 2) ( 2)

x x x x x x x x x xf x

x x x x

= = = =

Set the numerator and denominator to zero to get x = 0, 4, and 2, however x = 0 and 2 arenot the interval so we do not use them. Now plug 3, 4, and 7 into f(x) to obtain themaximum and minimum.

f(3) = 9, f(4) = 8, f(7) = 495

.

The absolute maximum value is49

5, the absolute minimum value is 8.


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63) A.

Start with the equation:34

3V r=

V = 36

So plug into the volume equation find r.

34363

r =

Solve to get r = 3.

Now the change, dr = -0.2

Taking the derivative and solving for dV gives us:

2

2

2

4

4

4 (3) ( 0.2)

7.2

dV rdr

dV r dr

dV

dV

=

=

=

=

So it decreases by 7.23

sec

in.

64) B. Remember the formula for linearization is L(x) = f (a) + f (a) (x a). (or just use point-slope form)

The point is at x = 1, and f(1) = 1 ln 1 = 0. Thus we have the point (1,0)

Now for the derivative (product rule):

1'( ) ln f x x x

x

= +

1'(1) 1 ln1

1f

= +

So the slope is 1.

Now just plug everything else into the linearization formula (or point-slope form of a line)L(x) = 0 + 1 (x 1) = x 1.

y = x 1 (answer).


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71) A.

To find the maximum and minimum values of f(x), take the derivative and set it equal to zero to

find critical numbers. Plug in any critical numbers, that are in the interval [0, 4], and the

endpoints of the interval back into the original function. The highest value is the maximum, the

lowest is the minimum.

2 4'( ) (2 4) 0x x f x e x= =

2 4x xe

will never equal 0, so we only need to look at 2x 4 = 0, which happens at x = 2.

f(0) = 1

f(2) = e-4

f(4) = 1So the maximum value is at 1 and the minimum is at e-4. To find the upper and lower bound,

multiply by 4, the length of the interval of. This gives you:

4e-4 < A < 4 answer.

72) C.

This just means to take the integral and evaluate from 0 to 2. Use substitution.

2

4 2

0

6 xe dx

4 2

62

3

3 3

u

u

u x

due

e du

e e

=

4 2

62

3

3 3

u

u

u x

due

e du

e e

=

Evaluate from 0 to 2.

(-3e0)- (-3e4 )= -3 + 3e4 = 3e4 3 (answer).


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76) D.

Use substitution for this integral

ln

1

u x

du dxx

dx xdu

=

=

=

3

3

2

1

2

xduxu

u du

u

=

2

2

1

2(ln )

e

ex

=2 2 2

1 1 1 1 3

2(ln ) 2(ln ) 8 2 8e e

= =

(answer.)

77) B. In this case, substitution would be used.

u = 2x3 + 4x2 + 10x

du = (6x2

+ 8x + 10)dx

2 26 8 10 2(3 4 5)

du dudx

x x x x= =

+ + + +

1

3 2

12

1

2

1 1ln | | ln | 2 4 10 |

2 2

duu

u du

u C x x x C

+ = + + +


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84) D.

Here you are given that f (x) = sinx cosx. (Remember the slope of the tangent line is the first

derivative). So to find f(x), given f (x), you have to integrate. Use substitution.

sin

cos

cos

u x

du xdx

dudx

x

=

=

=

2

coscos

2

duu x

x

udu

uC+

(Note that this will also work if you chose u = cosx)

2sin( )

2

x f x C = +

To find C, use the condition that f(0) = 1

2sin 01

2C= + , So C = 1 and

2sin( ) 1

2

xf x = +

To find2

f

just plug in /2.

2sin1 32

1 12 2 2 2

f

= + = + =

(answer).

85) D.

Since they are asking you to evaluate the integral at a variable, you know the trick with the chain

rule and FTC. In this problem f(t) = t cos(t2). To use the theorem just do this:

[f(11) * (the derivative of 11)] [f(x) * (the derivative ofx)]

11cos(112) * 0 (x cos(x) * x-1/2)

-x cosx = - cosx (answer)2 x 2


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90) A.

This is an optimization problem. So we need the primary and constraint:Using 2 numbers, call them x and y, we have:Primary: Maximize the product, so P = xy

Constraint: x + 2y = 32

Solve the constraint for one variable, plug into the primary.

x = 32 2yP = (32 2y)y = 32y 2y2

Take the derivative, set equal to zero, and solve.

P= 32 4y = 0y = 8

Plug back into the constraint for the x.

x = 32 2(8) = 16

So the maximum product P = xy = (16)(8) = 128 (answer).

91. C.Here, you need a primary and a constraint:

Primary: Perimeter = P = 2x + 3y (it said to minimize this, and remember theinner part).Constraint: Area = 216= xy (given info).

Solve the constraint for one variable:216/x = y

Plug into the primary:P = 2x + 3(216/x) = 2x + 648/x

Take the derivative, set it equal to zero, and solveP = 0 = 2 648/x22 =648/x2

2x2 = 648x2 = 324x = 18, -18x = 18 since it is a real world problem, and we cant have a dimension of 18.

Now, plug x = 18 into the constraint to get the y value.216 = 18yy = 12


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