This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Solutions to Quiz 10A
www.math.uﬂ.edu/˜harringt
April 6, 2008
1. Find the limit. Use l’Hosptial’s Rule where appropriate. If there is a more elementary method,
consider using it.
(a) limx→0 1−cos(x)
x2 lim x→0 1 − cos(x) L
sin x L
cos x
cos(0)
1
= lim
= lim
=
=
x→0 2x
x→0
x2
2
2
2 (b) limx→∞ x1/x
1
ln x L
1
= lim x = lim
=0
x→∞ x
x→∞ 1
x→∞ x ln( lim x1/x ) = lim ln(x1/x ) = lim
x→∞ x→∞ lim x1/x = e0 = 1 x→∞ (c) limx→π sin(x)
1−cos(x) lim x→π sin(π )
0
sin(x)
=
=
=0
1 − cos(x)
1 − cos(π )
1 − (−1) Notice, here we did not use l’Hospital’s rule at all.
2. Let f (x) = −x3 + 9x2 − 15x + 2. Using interval notation, determine where f is concave down. Also,
ﬁnd the absolute maximum.
First we will ﬁnd when the f is concave down. So we look at the second derivative.
f (x) = −3x2 + 18x − 15 = −3(x − 5)(x − 1)
f (x) = −6x + 18 (1)
(2) The function f is concave down when f (x) < 0, so we have −6x + 18 < 0 implies that x > 3.
So f is concave down on (3, ∞). What about the absolute maximum? Well, notice that f is a
polynomial of odd degree, so f does not have an absolute maximum!
Most of my students found the relative maximum (instead of the absolute maximum), so if you
were curious, here is how to ﬁnd the relative maximum. Notice that f (x) = −3x2 + 18x − 15 =
−3(x − 5)(x − 1), so our critical numbers are x = 5 and x = 1. Since, f (1) > 0, then the point at
x = 1 is a local minimum and f (5) < 0, then the point at x = 5 is a local maximum. The graph
below is of f . Notice it has a local max, but as x → −∞, f goes oﬀ to inﬁnity. Figure 1: f (x) = −x3 + 9x2 − 15x + 2 1 ...
View
Full
Document
This note was uploaded on 07/10/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.
 Fall '08
 ALL
 Math, Calculus, Geometry

Click to edit the document details