exam 2 - Version 033 Exam 2 Radin(56520 This print-out...

Version 033 – Exam 2 – Radin – (56520) 1
This print-out should have 18 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001 10.0 points
Determine if
lim
x → ∞
x
3
− 4 x
5 x 3 + 4 x 2 + 5
exists, and if it does, find its value.
1. limit does not exist
2. limit = 0
3. limit = −
1
5
4. limit =
2
5
5. limit =
3
5
6. limit =
1
5
correct
Explanation:
After division by x
3
in both the numerator
and denominator, we see that
x
3
− 4 x
5 x
3
− 4 x
2
+ 5
=
1 −
4
x 2
5 +
4
x
+
5
x
3
.
On the other hand,
lim
x → ∞
parenleftBig
1 −
4
x 2
parenrightBig
= 1 ,
while
lim
x → ∞
parenleftBig
5 +
4
x
+
5
x 3
parenrightBig
= 5 .
Consequently,
lim
x → ∞
x
3
− 4 x
5 x 3 + 4 x 2 + 5
exists and
limit =
1
5
.
002 10.0 points
Find the rational function f so the figure
below is the graph of y = f ( x ) on [ − 8 , 8].
-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7
2 4 6 − 2 − 4 − 6
2
4
6
− 2
− 4
− 6
1. f ( x ) =
x
2
+ 3 x + 8
x
2
− 4
2. f ( x ) =
x
2
− 3 x + 8
x
2
− 4
3. f ( x ) =
x
2
− 3 x + 8
4 − x 2
4. f ( x ) =
x
2
+ 3 x − 8
x
2
− 4
correct
5. f ( x ) =
x
2
+ 3 x − 8
x − 2
6. f ( x ) =
x
2
+ 3 x − 8
x 2 + 4
Explanation:
The graph has vertical asymptotes at x =
± 2, so a term x
2
− 4 must appear in the
denominator of f ( x ); in particular, therefore,
f ( x ) negationslash =
x
2
+ 3 x − 8
x 2 + 4
and
f ( x ) negationslash =
x
2
+ 3 x − 8
x − 2
,

Version 033 – Exam 2 – Radin – (56520) 2
eliminating two of the possible answers. The
graph also has a horizontal asymptote y = 1,
so the ratio of coefficients of degree 2 terms in
the numerator and denominator of f ( x ) must
be 1. Thus
f ( x ) negationslash =
x
2
− 3 x + 8
4 − x
2
,
eliminating a third answer. Hence one of
f ( x ) =
x
2
+ 3 x − 8
x
2
− 4
or
f ( x ) =
x
2
− 3 x + 8
x 2 − 4
or
f ( x ) =
x
2
+ 3 x + 8
x
2
− 4
must hold. But the y -intercept of the graph is
positive, while the y -intercepts of these possi-
ble f are
f (0) = 2 , or − 2
and only one has the property f (0) = 2. Con-
sequently,
f ( x ) =
x
2
+ 3 x − 8
x
2
− 4
.
003 10.0 points
Determine which of the following graphs
could be that of
f ( x ) =
3 + 2 x
2
1 + x 2
.
1.
2.
3.
4. correct
Explanation:
The line y = 2 is a horizontal asymptote
because
lim
x → ±∞
3 + 2 x
2
1 + x 2
= 2 .
On the other hand,
f
′
( x ) = −
2 x
(1 + x 2 ) 2
.
Thus the critical point of f occurs at x = 0.
In addition, f is increasing on ( −∞ , 0] and
decreasing on [0 , ∞ ); in particular, therefore,
f has a local maximum at x = 0. The only
graph with these properties is

Version 033 – Exam 2 – Radin – (56520) 3
004 10.0 points
If f is the function defined on [ − 4 , 4] by
f ( x ) = x + | x | − 4 ,
which of the following properties does f have?
A. Absolute minimum at x = 0 .
B. Differentiable at x = 0 .
1. B only
2. both of them
3. neither of them
4. A only correct
Explanation:
Since
| x | =
braceleftbigg
x , x ≥ 0 ,
− x , x < 0 ,
we see that
f ( x ) = x + | x |− 4 =
braceleftbigg
2 x − 4 , x ≥ 0 ,
− 4 , x < 0 .
Thus on [ − 4 , 4] the graph of f is
2 4 − 2 − 4
2
4
− 2
− 4
Consequently:
A. True: by inspection.