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Unformatted text preview: Version 208 – Exam 2 – Radin – (58305)
5. ππ
−,
33 4. limit = − π
π
,
, π correct
3
3
Explanation:
After diﬀerentiation,
6. −π , − 6 1
3 5
5. limit = − correct
2
Explanation:
Dividing by x3 in the numerator and denominator we see that f ′ (x) = 2 cos x − 1 .
Now f will be decreasing on an interval [a, b]
if f ′ (x) < 0 on (a, b). But for x in (−π, π ),
f ′ (x) = 2 cos x − 1 = 0
when x = −π/3, π/3. Thus the graph 2 x − 5 x3 1+
3 + 2 x3 With s = 1 1
2
+ 2 −5
3
x
=x
.
3
+2
x3 1
, therefore,
x 1 + 2 x − 5 x3
x → −∞
3 + 2 x3
lim π
−π − 34 − π − π
2
4 π
4 1 π
2 3π
4 π = 2 limit = − of f ′ on (−π, π ) shows that f ′ (x) < 0 only on
π
3 , π
,π .
3 014 Consequently, f will be decreasing only on
the intervals
−π , − s → 0− s 3 + 2s 2 − 5
.
3s 3 + 2 Consequently, the limit exists, and 3 −π , − lim π
,
3 π
,π
3 . 5
.
2 10.0 points For which function f does its graph have
vertical asymptotes x = − 1 , 5 and horizon5
tal asymptote y = 5.
1. f (x) = 5x2 − 24x − 5
25x2 + 1 2. f (x) = 5 x2 + 2 x − 5
5x2 + 24x − 5 3. f (x) = 25x2 − 2x + 5
correct
5x2 − 24x − 5 5
1. limit =
2 4. f (x) = 5x2 + 24x − 5
25x2 + 1 2. limit does not exist 5. f (x) = 25x2 − 1
5x2 + 24x − 5 1
3. limit =
3 6. f (x) = 5 x2 + 1
5x2 − 24x − 5 013 10.0 points Determine if the limit
lim x → −∞ 1 + 2 x − 5 x3
3 + 2 x3 exists, and if it does, ﬁnd its value. ...
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This note was uploaded on 02/14/2012 for the course MATH 408 K taught by Professor Clark,c.w./hoy,r.r during the Spring '08 term at University of Texas at Austin.
 Spring '08
 CLARK,C.W./HOY,R.R

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