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Unformatted text preview: Math 1110 Name: Homework 10 Due 11/10 or 11/11 in class Please print out these pages. Write your answers to the “text exercises” on separate paper and staple it to these pages. You should include
computational details. These problems will be assessed for completeness.
Always write neatly and legibly.
Please answer the “presentation problems” in the spaces provided.
Include full explanations and write your answers in complete, mathematically and grammatically correct sentences. Your answers will be assessed
for style and accuracy, and you will be given written feedback on these
problems. GRADES
Text exercises / 20 Pres probs / 20 Staple /1 Text exercises. Please do the following problems from the book.
§4.4 #2, 9, 12, 24, 43, 53, 63, 68, 74, 81, 88, 103, 104, 107, 114
§4.5 #2, 3, 6, 13, 20, 23, 30, 38, 43, 53, 66, 68, 75, 79
§4.6 #1, 3, 8, 11, 12, 17, 20, 38, 45, 52, 64
Question 1. Evaluate the following limits. Do not forget to justify your answer!
xa − 1
, where a and b are positive numbers.
x→ 1 xb − 1 (a) lim First, note that 1a − 1 = 0 and 1b − 1 = 0. So we may apply L’Hopital’s rule, to see that
xa − 1
axa−1
lim b
= lim b−1 . Substituting the value x = 1 into the limit, we see that the solution
x→ 1 x − 1
x→ 1 bx
is a .
b
(b) lim xnex, where n > 0 is an integer. x→ −∞ xn
xn
Notice ﬁrst that xnex = −x , so that lim xnex = lim −x . Also note that lim e−x = ∞,
x→ −∞
x→ −∞ e
x→ −∞
e
xn
−∞ x odd
and that lim xn =
. In both cases we have that limx→ −∞ −x takes an inx→ −∞
∞
x even
e
n
x
nxn−1
determinate form, so we may apply L’Hopital’s rule, to get that lim −x = lim
.
x→ −∞ e
x→ −∞ −e−x
xn
n!
Repeating this argument n times, we see that lim −x = lim
= lim n!(−1)nex = 0.
x→ −∞ e
x→ −∞ (−1)ne−x
x→ −∞ Math 1110 (Fall 2011) HW5 Presentation Problems 2 Question 2. Consider the function 1 − x2
.
x2 − 4
We will get you started by computing the ﬁrst two derivatives:
f(x) = f (x) = 6x
(x2 − 4)2 and f (x) = − 18x2 + 24
.
(x2 − 4)3 (a) Find all critical and singular points of f.
We want to ﬁnd all points where f (x) is 0 or undeﬁned. This function is undeﬁned when
the denominator is 0. The points where this occurs are x = 2 or −2. These points are not
in the domain of f, either, so although f is not deﬁned at these points, they are not critical
points. Note that f is equal to 0 when the numerator is equal to 0 and the denominator is
deﬁned and nonzero. The only such point is x = 0. So, the only critical point of f occurs at
x = 0.
(b) Find all asymptotes of f.
To ﬁnd the horizontal asymptotes we compute lim f(x) and lim f(x). Since f is a rational
x→∞ x→∞ function, we can compute these limits by comparing the coefﬁcients and exponents on the
lead terms of the polynomials in the numerator and denominator. We see that the numerator
and denominator have the same degree, so lim f(x) = lim f(x) = −1. In particular, f has
x→∞ x→∞ horizontal asymptotes both as x → −∞ and as x → ∞, and they are both equal to −1. To ﬁnd the vertical asymptotes we compute lim f(x), lim f(x), lim − f(x), lim + f(x), as
+
−
x→ 2 x→ 2 x→ −2 x→ −2 2 and −2 are the values where the function is not deﬁned, and it is continuous at all other
values. Although, strictly speaking, these computations give us more information than is
asked for, it will be useful to have this information in part (e) of this problem. Notice that
for all of these limits, as x approaches the value in question, the numerator approaches −3.
In the ﬁrst and fourth limits, the denominator approaches 0 and is always positive, so the
value of these limits is −∞. In the second and third limits, the denominator approaches 0
and is always negative, so the value of these limits is ∞. In particular, vertical asymptotes
for f occur at x = 2 and x = −2.
(c) Where is f increasing and decreasing?
To determine where f is increasing and decreasing, we need to determine the sign of f on
the intervals (−∞, −2), (−2, 0), (0, 2), and (2, ∞). We can do this by determining the sign
of f at one point in each of these intervals. Observing the following table, we see that f is
increasing on (0, 2), and (2, ∞), and decreasing on (−∞, −2), and (−2, 0).
x
f’(x) 3
18/25 1
2/3 1
2/3 3
18/25 Math 1110 (Fall 2011) HW5 Presentation Problems 3 (d) Where is f concave up and concave down?
Before we can determine where f is concave up and concave down, we need to determine
where f is not deﬁned, and where it is equal to 0. Since the numerator of f has no real
solutions we know that f is never 0. The denominator of f is 0 only when x = 2 or
−2, and since f is a rational function, these are the only places where it is undeﬁned. To
determine where f is concave up and concave down, we need to determine the sign of f on
the intervals (−∞, −2), (−2, 2), and (2, ∞). We can do this by determining the sign of f at
one point in each of these intervals. Observing the following table, we see that f is concave
down on (−∞, −2) and (2, ∞), and f is concave up on (−2, 2).
x
f”(x) 3
186/125 0
24/64 3
186/125 Math 1110 (Fall 2011) HW5 Presentation Problems 4 (e) Using what you have calculated in the ﬁrst four parts of this problem, please graph the
function
1 − x2
f(x) = 2
x −4
on the axes given below. !"#$%&
$'( !'(
$"#$%& Homework 10 Book Problem Answers:
Section 4.4: Section 4.5: Section 4.6: ...
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