calc 2 hw 1

patel (kvp267) HW01 ben-zvi (54740) 1

This print-out should have 22 questions.Multiple-choice questions may continue onthe next column or page find all choicesbefore answering.

001 10.0 points

Find the value of

limx1

2

x+ 1

(1

x2 + 2 1

3

).

1. limit =2

9

2. limit =4

9correct

3. limit does not exist

4. limit =4

3

5. limit =2

3

Explanation:

After the second term in the product isbrought to a common denominator it becomes

3 x2 23(x2 + 2)

=1 x2

3(x2 + 2).

Thus the given expression can be written as

2(1 x2)3(x+ 1)(x2 + 2)

=2(1 x)3(x2 + 2)

so long as x 6= 1. Consequently,

limx1

2

x+ 1

(1

x2 + 2 1

3

)

= limx1

2(1 x)3(x2 + 2)

.

By properties of limits, therefore,

limit =4

9.

002 10.0 points

Find the derivative of f when

f(x) =1 + 2 cosx

sinx.

1. f (x) =2 + sinx

cos2 x

2. f (x) = 1 + 2 cosxsin2 x

3. f (x) =1 2 cosxsin2 x

4. f (x) =2 cosxsin2 x

5. f (x) =2 sinx+ 1

cos2 x

6. f (x) =sinx 2cos2 x

7. f (x) = 2 + cosxsin2 x

correct

8. f (x) =2 sinx 1cos2 x

Explanation:

By the quotient rule,

f (x) =2 sin2 x cosx(1 + 2 cosx)

sin2 x

=2(sin2 x+ cos2 x) cosx

sin2 x.

But cos2 x+ sin2 x = 1. Consequently,

f (x) = 2 + cosxsin2 x

.

003 10.0 points


f(x) = 2x cos 4x 4 sin 4x .

1. f (x) = 8x sin 4x 16 cos 4x


2. f (x) = 16 cos 4x 14x sin 4x

3. f (x) = 16 cos 4x+ 8x sin 4x

4. f (x) = 8x sin 4x14 cos 4x correct

5. f (x) = 8x sin 4x 14 cos 4x

Explanation:

Using formulas for the derivatives of sineand cosine together with the Product andChain Rules, we see that

f (x) = 2 cos 4x 8x sin 4x 16 cos 4x= 8x sin 4x 14 cos 4x .

004 10.0 points

Find f (x) when

f(x) =1

8x x2 .

1. f (x) =4 x

(8x x2)3/2

2. f (x) =4 x

(x2 8x)3/2

3. f (x) =x 4

(8x x2)3/2 correct

4. f (x) =x 4

(x2 8x)3/2

5. f (x) =4 x

(x2 8x)1/2

6. f (x) =x 4

(8x x2)1/2

Explanation:

By the Chain Rule,

f (x) = 12(8x x2)3/2 (8 2x) .

Consequently,

f (x) =x 4

(8x x2)3/2 .

005 10.0 points

Find f (x) when

f(x) = 3 sec2 x tan2 x .

1. f (x) = 4 tan2 secx

2. f (x) = 8 sec2 x tanx

3. f (x) = 4 sec2 x tanx

4. f (x) = 4 sec2 x tanx correct

5. f (x) = 4 tan2 sec x

6. f (x) = 8 tan2 sec x

Explanation:

Since

d

dxsecx = secx tanx,

d

dxtanx = sec2 x,

the Chain Rule ensures that

f (x) = 6 sec2 x tanx 2 tanx sec2 x .

Consequently,

f (x) = 4 sec2 x tanx .

006 10.0 points

Determine the third derivative, f (x), of fwhen

f(x) =2x+ 3 .

1. f (x) = (2x+ 3)5/2

2. f (x) = (2x+ 3)3/2

3. f (x) = 3(2x+ 3)5/2


4. f (x) = 3(2x+ 3)5/2 correct

5. f (x) = 3(2x+ 3)3/2

6. f (x) = (2x+ 3)3/2

Explanation:

To use the Chain Rule successively its moreconvenient to write

f(x) =2x+ 3 = (2x+ 3)1/2 .

For then

f (x) =1

2 2 (2x+ 3)1/2

= (2x+ 3)1/2 ,

while

f (x) = 12 2 (2x+ 3)3/2

= (2x+ 3)3/2 ,and

f (x) =3

2 2 (2x+ 3)5/2 .

Consequently,

f (x) = 3(2x+ 3)5/2 .

007 10.0 points

Determine if the limit

limx

x+ 4

x2 3x+ 5exists, and if it does, find its value.

1. limit = 4

2. limit doesnt exist

3. limit = 5

4. limit =4

5

5. limit = 0 correct

6. limit = 13

Explanation:

Dividing in the numerator and denominatorby x2, the highest power, we see that

x+ 4

x2 3x+ 5 =1

x+

4

x2

1 3x+

5

x2

.

On the other hand,

limx

1

x= lim

x1

x2= 0 .

By Properties of limits, therefore, the limitexists and

limit = 0 .

008 10.0 points

Determine if

limx

(2x

x 1 +4x

x+ 1

)

exists, and if it does, find its value.

1. limit = 4

2. limit = 2

3. limit = 3


5. limit = 5

6. limit = 6 correct

Explanation:

Bringing the expression to a common de-nominator, we see that

2x

x 1 +4x

x+ 1=

2x(x+ 1) + 4x(x 1)(x 1)(x+ 1)

=6x2 2xx2 1 .


Thus after dividing through by x2 we see that

limx

(2x

x 1 +4x

x+ 1

)

= limx

6 2x

1 1x2

.

Consequently, the limit exists and

limit = 6 .

009 10.0 points


limx

x (x+ 9x 6)

exists, and find its value when it does.

1. limit = 0

2. limit =3

2


4. limit = 3

5. limit =15

2correct

6. limit = 15

Explanation:

By rationalization,

x+ 9x 6 = (x+ 9) (x 6)

x+ 9 +x 6

=15

x+ 9 +x 6 .

On the other hand,x

x+ 9 +x 6 =

11 +

9

x+

1 6

x.

Since

limx

1 +

9

x= lim

x

1 6

x= 1,

it thus follows by properties of limits that

limx

1 +

9

x+

1 6

x

exists and has value 2. Consequently, againby properties of limits, the limit

limx

x (x+ 9x 6)

exists and

limit =15

2.

010 10.0 points


limx

(1 e

x

3ex + 5

)

exists, and if it does, compute its value.


2. limit = 0

3. limit = 13

4. limit =2

3correct

5. limit = 23

Explanation:

Adding, we see that

1 ex

3ex + 5=

2ex + 5

3ex + 5,

and so

1 ex

3ex + 5=

2 + 5ex

3 + 5ex

after dividing by ex in both numerator anddenominator. But

limx

ex = 0 ,


in which case

limx

(2 + 5ex

)= 2 ,

whilelim

x3 + 5ex = 3 .

Consequently, by properties of limits, thegiven limit exists and

limit =2

3.

011 10.0 points


f(x) = 4e2x+5 + 3e2x+5.

1. f (x) = 2(4e2x 3e2x)

2. f (x) = e5 (4e2x 3e2x)

3. f (x) = e5 (3e2x 4e2x)

4. f (x) = 2e5 (4e2x 3e2x) correct

5. f (x) = e5 (4e2x + 3e2x)

6. f (x) = 2(4e2x + 3e2x)

Explanation:

After differentiation,

f (x) = 8e2x+5 6e2x+5.Consequently,

f (x) = 2e5 (4e2x 3e2x) .

012 10.0 points

Determine f (x) when

f(x) = e4x+5.

1. f (x) = 2e4x+5

4x+ 5

correct

2. f (x) = 4e4x+5

3. f (x) =4e4x+5

4x+ 5

4. f (x) =1

2

e4x+5

4x+ 5

5. f (x) = 2e4x+5

4x+ 5

Explanation:

By the chain rule

f (x) = e4x+5

(d

dx

4x+ 5

)

= 2e4x+5

4x+ 5

.

013 10.0 points


f () = ln (cos 5) .

1. f () =5

cos 5

2. f () = 5 tan 5 correct

3. f () = 5 tan 5

4. f () = 1sin 5

5. f () = cot 5

6. f () = 5 cot 5

Explanation:

By the Chain Rule,

f () =1

cos(5)

d

d(cos 5) = 5 sin 5

cos 5.


Consequently,

f () = 5 tan 5 .

014 10.0 points

Differentiate the function

f(x) = cos(ln 5x) .

1. f (x) = sin(ln 5 x)x

correct

2. f (x) =5 sin(ln 5x)

x

3. f (x) =sin(ln 5 x)

x

4. f (x) =1

cos(ln 5 x)

5. f (x) = 5 sin(ln 5x)x

6. f (x) = sin(ln 5 x)Explanation:

By the Chain Rule

f (x) = sin(ln 5x)x

.

015 10.0 points

Determine f (x) when

f(x) = e(3 ln(x5)) .

1. f (x) =3

x2e3 ln(x

5)

2. f (x) = 14x15

3. f (x) = 15x14 correct

4. f (x) = e15/x

5. f (x) =1

xe3 ln(x

5)

6. f (x) = 15(lnx)e3 ln(x5)

Explanation:

Since

r lnx = lnxr , eln x = x ,

we see that

f(x) = e(ln x15) = x15 .

Consequently,

f (x) = 15x14 .

016 10.0 points


f(x) = 2 (8x) 3 log8 x .

1. f (x) = 2 (8x) ln 8 3x

2. f (x) = 2 (8x) ln 8 3x ln 8

correct

3. f (x) = 2 (8x) ln 8 3log8 x

4. f (x) = 2 (8x) 3x ln 8

5. f (x) = 2 (8x) ln 8 3 ln 8x

Explanation:

Note that

8x = ex ln 8, log8 x =lnx

ln 8.

By the chain rule, therefore,

f (x) = 2ex ln 8 ln 8 3x ln 8

.

Consequently,

f (x) = 2 (8x) ln 8 3x ln 8

.


017 10.0 points

Simplify the expression

y = sin

(tan1

x11

)

by writing it in algebraic form.

1. y =x

x2 + 11correct

2. y =

11

x2 + 11

3. y =x

x2 + 11

4. y =

x2 + 11

11

5. y =x

x2 11Explanation:

The given expression has the form y = sin where

tan =x11

, pi2

< 0 on (5, 3),B. f has exactly 3 critical points,

C. f has exactly 2 local extrema.

1. A and B only

2. A and C only correct

3. C only

4. A only


5. none of them

6. B and C only

7. all of them

8. B only

Explanation:

A. True: the graph of f is concave UP on(5, 3).

B. False: f (x) = 0 at x = 3, 1, while f (x)does not exist at x = 2; in addition,the graph of f has a vertical tangent atx = 1. All of these are critical points.

C. True: f has a local minimum at x = 2and a local maximum at x = 1; the graphof f does have a horizontal tangent at(3, 1), but this is an inflection point.

022 10.0 points

In drawing the graph

P

of f the x-axis and y-axis have been omitted,but the point P = (1, 1) on the graph hasbeen included.Use calculus to determine which of the fol-

lowing f could be.

1. f(x) = 23+ 3x x2 1

3x3

2. f(x) =10

3 5x+ 3x2 1

3x3

3. f(x) =8

3 3x+ x2 + 1

3x3

4. f(x) = 83+ 3x+ x2 1

3x3 correct

5. f(x) =14

3 3x x2 + 1

3x3

6. f(x) = 43+ 5x 3x2 + 1

3x3

Explanation:

From the graph, f(x) as x .Of the six given choices, therefore, f musthave the form

f(x) = a+ bx+ cx2 13x3

with {b, c} being one of

{3, 1}, {5, 3}, {3, 1} .

Now

f (x) = b+ 2cx x2, f (x) = 2(c x) .

Since the graph has an inflection point atx = 1, it follows that

c 1 = 0 ,

i.e., b = 3 and c = 1. On the other hand, todetermine a we use the fact that

f(1) = a+ b+ c 13

= 1 .

Consequently,

f(x) = 83+ 3x+ x2 1

3x3 .

calc 2 hw 1

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