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
Find the derivative of f when
f(x) = 2x cos 4x 4 sin 4x .
1. f (x) = 8x sin 4x 16 cos 4x
patel (kvp267) HW01 ben-zvi (54740) 2
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
patel (kvp267) HW01 ben-zvi (54740) 3
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
4. limit does not exist
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 .
patel (kvp267) HW01 ben-zvi (54740) 4
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
Determine if the limit
limx
x (x+ 9x 6)
exists, and find its value when it does.
1. limit = 0
2. limit =3
2
3. limit does not exist
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
Determine if the limit
limx
(1 e
x
3ex + 5
)
exists, and if it does, compute its value.
1. limit does not exist
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 ,
patel (kvp267) HW01 ben-zvi (54740) 5
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
Find the derivative of f when
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
Find the derivative of f when
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.
patel (kvp267) HW01 ben-zvi (54740) 6
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
Find the derivative of f when
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
.
patel (kvp267) HW01 ben-zvi (54740) 7
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
patel (kvp267) HW01 ben-zvi (54740) 9
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 .
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