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Math 1A: Quiz 6
July 31, 2015
1. (4 + 4 points) Evaluate the following limits.
(a) limx→0
cos(5x)−cos(2x)
.
3x2
We have by L’Hopital’s Rule
cos(5x) − cos(2x)
−5 sin(5x) + 2 sin(2x)
lim
= lim
2
x→0
x→0
3x
6x
−25 cos(5x) + 4 cos(2x)
= lim
x→0
6
−21
−7
−25 + 4
=
=
.
=
6
6
2
0
∵
0
0
∵
0
(b) limx→∞ x1/x .
Since this is of the form ∞0 , we instead consider
1
ln(x)
x→∞ x
∞
1/x
= lim
∵
x→∞ 1
∞
= 0.
lim ln(x1/x ) =
x→∞
lim
Hence, we conclude that limx→∞ x1/x = e0 = 1.
2. (5 points) The sum of two nonnegative numbers is 36. Find them if the sum of their
square roots is to be as large as possible.
Let x and y be the
numbers. We are given that x + y = √
36, x, √
y ≥ 0 and we want
√two √
to maximize S = x + y. Since y = 36 − x, we get S(x) = x + 36 − x; note that
x must necessarily be in [0, 36]. Next, note that
√
√
1
1
S 0 (x) = √ − √
= 0 ⇒ x = 36 − x ⇒ x = 18.
2 x 2 36 − x
Note that S is differentiable everywhere on (0, 36) √
and hence√the only critical points
are x = 0, 18, 36. Note that S(0) = 6, S(18) = 2 18 = 6 2 and S(36) = 6. We
conclude that x = 18 and (correspondingly) y = 18 are the two required numbers.
1
3. (7 points) Sketch the graph of f (x) = x2x−9 . Make sure your work includes the domain,
intercepts, symmetry, asymptotes, slopes and concavity.
• The domain is R−{−3, 3} = (−∞, −3)∪(−3, 3)∪(3, ∞) since f is only undefined
at x = ±3.
• The y-intercept is f (0) = 0. For the x-intercept, set f (x) = 0 to get x = 0.
• Since f (−x) = − x2x−9 = −f (x), we conclude that f is odd.
• Note that limx→∞
−∞.
x
x2 −9
= 0. Also observe that limx→3+
x
x2 −9
= ∞ and limx→3−
x
x2 −9
• We also have
(x2 − 9)(1) − x(2x)
(x2 − 9)2
−x2 − 9
.
=
(x2 − 9)2
f 0 (x) =
Observe that f 0 (x) < 0 whenever f is defined.
• Also note that
(x2 − 9)2 (−2x) − (−x2 − 9)(4x)(x2 − 9)
(x2 − 9)4
2x(x2 − 9)(−(x2 − 9) + 2(x+ 9))
=
(x2 − 9)4
2x(x2 − 9)(x2 + 27)
=
.
(x2 − 9)4
f 00 (x) =
Observe that the sign of f 00 is determined by x(x2 − 9) since all other factors are
always positive. We can write this as x(x−3)(x+3) and conclude that for x < −3
and 0 < x < 3, f 00 (x) < 0 while it is positive for −3 < x < 0 and x > 3.
Gluing this together gives the sketch in Figure 1.
10
8
6
4
2
0
-2
-4
-6
-8
-10
-6
-4
-2
0
Figure 1
2
2
4
6
=