HW02 Solutions-Calc408K UT

8/9/2019 HW02 Solutions-Calc408K UT

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lin (ll25879) HW02 schultz (55325) 1

This print-out should have 21 questions.Multiple-choice questions may continue onthe next column or page nd all choicesbefore answering.

001 10.0 points

If x 1, x 2 are the solutions of the equation

73x2

= 1493x+1

,

compute the value of |x 1 x 2|.

1. |x 1 x 2| = 33

2.

|x 1

x 2

|=

2 3

3

3. |x 1 x 2| = 2 3

3 correct

4. |x 1 x 2| = 2

5. |x 1 x 2| = 2Explanation:

Since1493x+1 = 7

6x2,

the equation can be written as

73x2

= 7 6x2,

which in turn can be rewritten as

3x 2 = 6x 2by taking logs to the base 7 of both sides. Bythe quadratic formula, therefore,

x 1, x 2 = 3 33 .Thus

|x 1 x 2| = 2 3

3 .

002 10.0 points

Find y when

2x = 32 y, 32x = 16 y+1 .

1. y = 421

correct

2. y = 221

3. y = 27

4. y = 17

5. y = 5

21Explanation:

The equations

2x = 32 y, 32x = 16 y+1

are equivalent to

2x = 2 5y, 25x = 2 4(y+1) .

Now set v = 2x

and u = 2y. Then

v = u5, v5 = 16 u 4 .

Substituting for v in the second equation wethus get

u 25 = 16 u 4, i .e ., u = 2 421 .

Consequently, since u = 2 y,

y = 421

.

003 10.0 points

Which of the following is the graph of

f (x ) = 2 2x1 ?


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lin (ll25879) HW02 schultz (55325) 2

1.

2 424

2

4

2

4

2. 2 424

2

4

2

4

3.2 424

2

4

2

4

4.2 424

2

4

2

4

5.2 424

2

4

24

correct

6.2 424

2

4

24

Explanation:Since

limx

2x = 0 ,

we see that

limx

f (x ) = 2 ,

in particular, f has a horizontal asymptotey = 2. This eliminates all but two of thegraphs. On the other hand, f (0) =

32

, so thy-intercept of the given graph must occur aty =

32

.Consequently, the graph is of f is

2 424

2

4

2

4

004 10.0 points


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lin (ll25879) HW02 schultz (55325) 3

Suppose f is a 1-1 function with domain Aand range B.

What is the range of f 1?

1. Cant be determined.

2. B3. A correct

Explanation:The domain of f is the range of its inverse

f 1 , while the domain of f 1 is the range of f .

005 10.0 points

If the graph of f is

4 8

4

8

4

8

4

8

which of the following is the graph of f 1(x )?

1.

4 848

4

8

4

8

2.

4 848

4

8

48

3.

4 848

4

8

4

8

4.

4 8484

8

4

8

5.

4 848

4

8

4

8


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lin (ll25879) HW02 schultz (55325) 4

6.

4 848

4

8

48

correct

Explanation:The graph of f 1(x ) is obtained by reect-

ing the graph of f over the line y = x:

4 8484

8

4

8

006 10.0 points

If f 1 is the inverse of f , determine thevalue of f (f 1(5)).

1. f (f 1(5)) = 25

2. f (f 1(5)) = 125

3. Need to know f

4. f (f 1(5)) =

15

5. f (f 1(5)) = 5 correct

Explanation:The inverse, f 1, of f has the property that

f (f 1(x )) = x, f 1(f (x )) = x .Consequently,

f (f 1(5)) = 5 .

007 10.0 points

Determine the inverse function, f 1, of fwhen

f (x ) = 5 + 4 x , x

5

4 .

1. f 1(x ) = x2 + 4

5 , x 0

2. f 1(x ) = x2 4

5 , x

54

3. f 1(x ) = 5 x 2

4 , x 0

4. f 1(x ) = x2 5

4 , x 0 correct

5. f 1(x ) = 4 x 2

5 , x

54

6. f 1(x ) = x2 5

4 , x

45

Explanation:

Since f is dened on 54 , and is increasing on this interval, the inverse function,

f 1, exists and has range 54

, ; further-more, since f has range [0 , ), the inverse of has domain [0 , ).To determine f 1 we rst solve for x in

y = 5 + 4 x ,

and then interchange x, y . Solving rst for we see that

4x = y2 5 .Consequently, the inverse of f is dened on[0, ) by

f 1(x ) = x2 5

4 .


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lin (ll25879) HW02 schultz (55325) 5

008 10.0 points

Find the inverse of

f (x ) = 1 + 3x9 2x

.

1. f 1(x ) = 9x 12x + 3

correct

2. f 1(x ) = 9x 13x + 2

3. f 1(x ) = 9x + 12x + 3

4. f 1(x ) = 3x 12x + 9

5. f 1(x ) = 9x + 13x + 2

Explanation:To determine the inverse of f we rst solve

for x iny =

1 + 3x9 2x

.

In this case

x = 9y 12y + 3

.

The inverse f 1 is now obtained by inter-changing x, y . Thus

y = f 1(x ) = 9x 12x + 3

.

009 10.0 points

Rewrite

5log3 x = 3in equivalent exponential form.

1. x5 = 27

2. x5 = 10

3. x3 = 127

4. x3 = 10

5. x5 = 127

correct

Explanation:By exponentiation to the base 3,

35log3 x = 127

.

But

35log3 x = 3 log3 x5

= x5.

Hence the exponential form of the given equa-tion is

x5

= 127 .

keywords: LogFunc, LogFuncExam,

010 10.0 points

Use properties of logs to simplify the ex-pression

log7(x x 2 63 ) + log 7(x + x 2 63 ).1. log9 7

2. 1 + log9 7

3. 7 + log7 9

4. 1 + log7 9 correct

5. log7 9

Explanation:By properties of logs the given expressioncan be rewritten as

log7 (x x 2 63)(x + x 2 63 )= log 7 x 2 x 2 63

2 = log 7 63

Butlog7 63 = log7 7 + log7 9 .


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lin (ll25879) HW02 schultz (55325) 6

Thus the given expression reduces to

1 + log 7 9 .


011 10.0 points

Find the value of x when

x = 23

4log4 4 7 7 log7 4

without using a calculator.

1. x = 1112

correct

2. x = 1

3. x = 1

4. x = 1312

5. x = 1112

Explanation:Since

4log4 x = x, 7 log7 x = 1x

,

we see that

4log4 4 = = 4 ,

while7 log7 4 = =

14

.

Consequently,

x = 83

74

= 1112

.

012 10.0 points

Find the value of x when

x = 3log 6 4 63 6log3

3 34 .

1. x = 7112

2. x = 356

3. x = 64. x =

234

correct

5. x = 173

Explanation:Since

r loga z = log a zr , loga a = 1 ,

we see that

log6 4 63 = 3

4 ,

whilelog3

3 34 = 43

.

Consequently,

x = 234 .

013 10.0 points

Use properties of logs to simplify the ex-pression

loga 16 + 25

loga 8 25

loga 4 + loga1

275

as much as possible.

1. loga 8 correct

2. loga 25

3. 8

4. loga 16


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lin (ll25879) HW02 schultz (55325) 7

5. loga 32

Explanation:By properties of logs the given expression

can be rewritten as

loga 24 8 254

25 2 2

25

= log a 8 .

014 10.0 points

Simplify the expression

f (x ) = e6(ln6) log 6 x

as much as possible.

1. f (x ) = e6

2. f (x ) = x6 correct

3. f (x ) = e36

4. f (x ) = x36

5. f (x ) = 6 x

Explanation:By the laws of exponents,

f (x ) = e6(ln 6)log 6 x = ( e6ln6 )log6 x .

So by using the identities

eln y = y , 6log6 y = y

we see that

e6ln6 = eln 66

= 6 6 .

Consequently,

f (x ) = 6 6log6 x = 6 log6 x6

= x6 .

015 10.0 pointsUse the properties of logarithms to expand

ln a (b4 + c3) .1. 2 ln a + 2ln b4 + c3

2. 2 ln a 2 ln b4 + c3

3. ln a

2 +

ln(4b + 3 c)2

4. ln a

2 +

ln b4 + c3

2 correct

5. ln a + ln b4 + c3

2

Explanation:

ln a (b4 + c3) = ln a b4 + c31/ 2

= 12

ln a b4 + c3

= 12

ln a + ln( b4 + c3)

= 12

ln a + 12

ln(b4 + c3)

016 10.0 points

Which one of the following could be thegraph of

f (x ) = log2(3 x )

when dashed lines indicates asymptotes?

1.

2 4

2

2


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lin (ll25879) HW02 schultz (55325) 8

2.

2 4

2

2

correct

3.

2 4

2

2

4.

2 4

2

2

5.

2 4

2

2

6.

2 4

2

2

Explanation:Lets rst review some properties of log 2(x

and log2(x ). Since log2(1) = 0, the graphof log2(x ) has x-intercept at x = 1. On theother hand, log 2(x ) is dened only on (0 , and

limx 0

+log2(x ) = , limx log2(x ) = ,

so x = 0 is a vertical asymptote. Thus thegraph of log2(x ) is

2 4

2

2

To get the graph of log2(x ) we simply ipthe one for log2(x ) over both the x-axis andthe y-axis, producing


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lin (ll25879) HW02 schultz (55325) 9

24

2

2

To obtain the graph of y = log2(3 x )from this last one all we have to do now istranslate horizontally to the right by 3, pro-ducing

2 4

2

2


017 10.0 points

Solve for x when

e2x = 116

.

1. x = 8 ln4

2. x = 4 ln4

3. x = 2 ln2

4. x = 2ln2 correct

5. x = 8ln26. x = 4ln4

Explanation:

Taking logs of both sides we see that

ln(e2x ) = ln 116

.

Thusln

116

= 2x ln e = 2x .

Now

ln 116

= ln2 4 = 4ln2 .

Consequently,

x = 2ln2 .


018 10.0 points

Determine if

f (x ) = ln(5 x + 2) , x > 25 ,has an inverse, and if it does, nd this inverse.

1. f 1(x ) = ex 2

5 , x >

25

2. f 1(x ) = 1

ln(5x + 2) , x >

25

3. f 1(x ) = 1

ln(5x

2)

, x > 25

4. f 1(x ) = ex 2

5 , < x < cor-

rect

5. f 1(x ) = ex + 2

5 , < x <

6. f 1(x ) does not exist

Explanation:


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lin (ll25879) HW02 schultz (55325) 10

Since

ddx

[ln(5x + 2)] = 5

5x + 2 > 0 , x >

25

,

f is strictly increasing on 25

, , so f hasan inverse.

To determine this inverse we solve for y in

x = ln(5 y + 2) .

But then

5y + 2 = ex , i .e ., y = ex 2

5 ,

so

f 1

(x ) =

ex

2

5 .

019 10.0 points

Find the inverse function of

f (x ) = e4 x , x 0 .

1. f 1(x ) =14

ln x2, x > 0

2. f 1(x ) = (4ln x )2, x 13. f 1(x ) =

14

ln x2, x 1 correct

4. f 1(x ) = (ln 4 x )2, x > 1

5. f 1(x ) = ln x4

2, x > 0

6. f 1(x ) = (4ln x )2, x > 0

Explanation:The restriction x > 0 on the domain of f is needed so that the term x is well-dened.But this puts a restriction on the range of f and hence on both the domain and range of the inverse f 1 of f . Indeed, since ex 1when x 0, we see that

domain( f ) = [0 , ),range( f ) = [1 , ) ,

which in turn means that

domain( f 1) = [1 , ),range( f 1) = [0 , ) .

Already this eliminates three of the answers,

so we now need to determine f 1

explicitlyFirst sety = e4 x

and solve for x :

x = 14

ln y, i.e., x =14

ln y2

.

Consequently, after interchanging x and we see that

f 1(x ) =1

4 ln x

2, x

1 .

020 10.0 points

Find the value of f (1) whenf (x ) = tan 1 x 4sin1 x .

1. f (1) = 34

2. f (1) = 94

3. f (1) = 74

correct

4. f (1) = 11

4

5. f (1) = 54

Explanation:Since

tan 1(1) = 4

, sin1(1) = 2

,

we see that

f (1) = 2 14

= 74

.

021 10.0 points


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lin (ll25879) HW02 schultz (55325) 11

Simplify the expression

y = sin tan 1 x 3

by writing it in algebraic form.

1. y = x

x 2 32. y =

xx 2 + 3

3. y = x x 2 + 3 correct

4. y = 3

x 2 + 3

5. y =

x 2 + 3 3

Explanation:The given expression has the form y = sin

where

tan = x 3 ,

2

< < 2

.

To determine the value of sin given the valueof tan , we can apply Pythagoras theoremto the right triangle

3

x

x 2 +

3

From this it follows that

y = sin = x x 2 + 3 .

Alternatively, we can use the trig identity

csc2 = 1 + cot 2

to determine sin .

keywords: TrigFunc, TrigFuncExam,

HW02 Solutions-Calc408K UT

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