Math 234
February 28
I.Find all vertical and horizontal asymptotes of the graph of the given
function.
1. f (x) = 1/(x − 3)
Answer:
V. A. :
x − 3 = 0 when x = 3
Vertical Asymptotes: x = 3
H.A.:
limx→∞ 1/(x − 3) = 0
limx→−∞ 1/(x − 3) = 0
Horizontal asymptotes: y = 0
2. f (x) = (3x − 1)/(x + 2)
Answer:
V. A. :
x + 2 = 0 when x = −2
Vertical Asymptotes: x = −2
H.A.:
limx→∞ (3x − 1)/(x + 2) = 3
limx→−∞ (3x − 1)/(x + 2) = 3
Horizontal Asymptotes: y = 3
3. f (x) = (3 − 2x)/(4 − x)
Answer:
V. A. :
4 − x = 0 when x = 4
Vertical Asymptotes: x = 4
H.A.:
1
limx→∞ (3 − 2x)/(4 − x) = 2
limx→−∞ (3 − 2x)/(4 − x) = 2
Horizontal Asymptotes: y = 2
4. f (x) = (x2 + 3)/(x2 + 4)
Answer:
V. A. :
x2 + 4 = 0 when x = −2, 2
Vertical Asymptotes: x = −2, 2
H.A.:
limx→∞ (x2 + 3)/(x2 + 4) = 1
limx→−∞ (x2 + 3)/(x2 + 4) = 1
Horizontal Asymptotes: y = 1
5. f (x) = (3x2 + 3)/(4x2 − 4)
Answer:
V. A. :
4x2 − 4 = 0 when x = −1, 1
Vertical Asymptotes: x = −1, 1
H.A.:
limx→∞ (3x2 + 3)/(4x2 − 4) = 3/4
limx→−∞ (3x2 + 3)/(4x2 − 4) = 3/4
Horizontal Asymptotes: y = 3/4
6. f (x) = (3x + 4)/x2
Answer:
V. A. :
x2 = 0 when x = 0
Vertical Asymptotes: x = 0
H.A.:
2
limx→∞ (3x + 4)/x2 = 0
limx→−∞ (3x + 4)/x2 = 0
Horizontal Asymptotes: y = 0
7. f (x) = (x3 + 3x + 5)/(6x + 2)
Answer:
V. A. :
6x + 2 = 0 when x = −1/3
Vertical Asymptotes: x = −1/3
H.A.:
limx→∞ (x3 + 3x + 5)/(6x + 2) = ∞
limx→−∞ (x3 + 3x + 5)/(6x + 2) = −∞
Horizontal Asymptotes: none
8. f (x) = (x3 + +3x2 + 3x + 5)/(2x3 + 5x2 + 6x + 2)
Answer:
V. A. :
2x3 + 5x2 + 6x + 2 = 0 when x = −1/2
Vertical Asymptotes: x = −1/2
H.A.:
limx→∞ (x3 + +3x2 + 3x + 5)/(2x3 + 5x2 + 6x + 2) = 1/2
limx→−∞ (x3 + +3x2 + 3x + 5)/(2x3 + 5x2 + 6x + 2) = 1/2
Horizontal Asymptotes: y = 1/2
9. f (x) = 2/x − 1/(3x − 1)
Answer:
V. A. :
2/x = 0 when x = 0
1/(3x − 1) = 0 when x = 1/3
Vertical Asymptotes: x = 0, 1/3
3
H.A.:
2
1
2(3x − 1) − x
5x − 1
−
= lim
= lim
=0
x→∞ x
x→∞ (3x2 − x)
3x − 1 x→∞ x(3x − 1)
lim
2
1
5x − 1
−
= lim
=0
x→−∞ x
3x − 1 x→∞ (3x2 − x)
lim
Horizontal Asymptotes: y = 0
√
10. g(x) = 3x/( x2 − 9)
Answer:
V.A.
√
x2 − 9 = 0 when x = −3, 3
Vertical Asymptotes: x = −3, 3
H.A.
√
3x
3x
(1/ x2 )
3
√
lim √
= lim √
= lim q
x→∞ ( x2 − 9)
x→∞ ( x2 − 9) (1/ x2 )
x→∞
1−
=3
9
x2
√
1
3x −x
3x
−3
3x
(1/ x2 )
√
= lim q
= lim q
= lim √
lim √
2
2
x→−∞ ( x − 9) (1/ x2 )
x→−∞
x→−∞
x→−∞ ( x − 9)
1 − 92
1−
x
Horizontal Asymptotes: y = −3, 3
√
11. f (x) = x2 / x4 + 1
Answer:
V.A.
√
x4 + 1 > 0. Therefore there are no vertical asymptotes.
Vertical Asymptotes: none
H.A.
x2
x2
lim √
= lim √
x→∞
x4 + 1 x→∞ x4 + 1
4
√1
x4
1
√
x4
1
= lim q
x→∞
1+
=1
1
x4
= −3
9
x2
x2
x2
lim √
= lim √
x→−∞
x4 + 1 x→−∞ x4 + 1
√1
x4
1
√
x4
x2 12
1
= lim q x
= lim q
x→−∞
x→−∞
1 + x14
1+
Horizontal Asymptotes: y = 1
II. Sketch the graph of the given function.
1. f (x) = 1/(x − 3)
Answer:
Domain: x 6= 3
y-intercept: f (0) = −1/3
x-intercept: none
Vertical Asymptotes: x = 3
Horizontal Asymptotes: y = 0
f (x) = −1/(x − 3)2
Critical numbers: x = 3
f 0 (x) < 0 for x/neq3
f (x) is decreasing: (∞, 3) ∪ (3, ∞)
There are no extreme.
f 00 (x) = 2/(x − 3)3
f 00 (x) < 0 for x < 3 and f 00 (x) > 0 for x > 3
f (x) is concave down: (−∞, 3)
f (x) is concave up: (3, ∞)
Since f (x) does not exist at x = 3 there is no inflection point to plot.
The graph looks like:
5
=1
1
x4
1.5
1.0
0.5
2
-2
4
6
8
-0.5
-1.0
-1.5
2. f (x) = x5 − 5x4 + 93
Answer:
Domain: all real numbers
y-intercept: f (0) = 93
x-intercept: f (x) = 0 when x = −1.91502, 2.459829, 4.82897 (You have
to use a calculator for this one)
Vertical Asymptotes: none
Horizontal Asymptotes: none
f 0 (x) = 5x4 − 20x3 = 5x3 (x − 4)
Critical numbers: x = 0, 4
f 0 (x) < 0 when 0 < x < 4 and f 0 (x) > 0 when x < 0 or x > 4.
f 0 (x) is increasing: x < 0 or x > 4
f 0 (x) is decreasing: 0 < x < 4
There is a relative max at (0, 93) and a relative min (4, −163).
f 00 (x) = 20x3 −60x2 = 20x2 (x−3) f 00 (x) < 0 when x < 3 and f 00 (x) > 0
when x > 3. f (x) is concave down: (−∞, 3)
f (x) is concave up: (3, ∞)
There is an inflection point at (3, −69)
The graph looks like:
6
50
-2
1
-1
2
3
4
5
-50
-100
-150
-200
3. f (x) = 3x4 − 4x2 + 3
Answer:
Domain: all real numbers
y-intercept: f (0) = 3
x-intercept: none
Vertical Asymptotes: none
Horizontal Asymptotes: none
2
f 0 (x) = 12x3 − 8x = 4x(3x
p − 2) p
Critical numbers: x = −
p 2/3, 0, 2/3 p
0
fp
(x) < 0 when x < − p2/3 or 0 < x < 2/3; and f 0 (x) > 0 when
− 2/3 < x < 0 or x p
> 2/3
p
0
f (x) is increasing: − 2/3p< x < 0 or x > p2/3
f 0 (x) is decreasing: x < − 2/3 or 0 < x < 2/3
p
Therep
is a relative maximum at (0, 3) and relative minimums at (− 2/3, 5/3)
and ( 2/3, 5/3)
f 00 (x) = 36x2 − 8 p
p
p
f 00 (x) <p0 when − 2/9 < x < 2/9 and f 00 (x) > 0 when x < − 2/9
or x > 2/9 .
p
p
f (x) is concave down: (− p2/9, 2/9)
p
f (x) is concave up: (−∞, 2/9)p
∪ ( 2/9, ∞)
p
There are inflection points at (− 2/9, 61/27) and ( 2/9, 61/27)
The graph looks like:
7
3.0
2.8
2.6
2.4
2.2
2.0
1.8
-1.0
0.5
-0.5
1.0
4. f (x) = x3 − 3x4
Answer:
Domain: all real numbers
y-intercept: f (0) = 0
x-intercept:f (x) = 0 when x = 0, 1/3
Vertical Asymptotes: none
Horizontal Asymptotes: none
f 0 (x) = 3x2 − 12x3 = 3x2 (1 − 4x) Critical numbers: x = 0, 1/4
f 0 (x) > 0 when x < 1/4 and f 0 (x) < 0 when x > 1/4
f 0 (x) is increasing: x < 1/4
f 0 (x) is decreasing: x > 1/4
There is a relative maximum at (1/4, 1/256)
f 00 (x) = 6x − 36x2 = 6x(1 − 6x)
f 00 (x) > 0 when x < 0 or x > 1/6 and f 00 (x) < 0 when 0 < x < 1/6
f (x) is concave down: (0, 1/6)
f (x) is concave up: (−∞, 0) ∪ (1/6, ∞)
There are inflection points at (0, 0) and (1/6, 1/432)
The graph looks like:
8
-0.2
0.1
-0.1
0.2
0.3
0.4
0.5
-0.005
-0.010
-0.015
-0.020
-0.025
-0.030
5. f (x) = 1/(2x + 3)
Answer:
Domain: x 6= −3/2
y-intercept: f (0) = −1/3
x-intercept: none
Vertical Asymptotes: x = −3/2
H.A.
1
=0
2x + 3
1
lim
=0
x→−∞ 2x + 3
Horizontal Asymptotes: y = 0
f 0 (x) = −2/(2x + 3)2
Critical numbers: x = −3/2
f (x) < 0 for x 6= −3/2
f 0 (x) is decreasing: x 6= −3/2
There are no extreme.
f 00 (x) = 8/(2x + 3)3
f 00 (x) > 0 for x > −3/2 and f 00 (x) < 0 for x < −3/2
f (x) is concave down: (−∞, −3/2)
f (x) is concave up: (−3/2, ∞)
Since f (x) does not exist at x = −3/2 there is no inflection point to
plot.
lim
x→∞
9
The graph looks like:
1.5
1.0
0.5
-5
-4
-3
-2
1
-1
2
-0.5
-1.0
-1.5
6. f (x) = x2 /(x + 2)
Answer:
Domain: x 6= −2
y-intercept: f (0) = 0
x-intercept: f (x) = 0 when x = 0
Vertical Asymptotes: x = −2
H.A.
x2
=∞
x→∞ x + 2
x2
= −∞
lim
x→−∞ x + 2
Horizontal Asymptotes: none
lim
f 0 (x) =
2x(x + 2) − x2
x2 + 4x
x(x + 4)
=
=
2
2
(x + 2)
(x + 2)
(x + 2)2
Critical numbers: x = −4, −2, 0
f 0 (x) < 0 when −4 < x < −2 or −2 < x < 0
f 0 (x) > 0 when x < −4 or x > 0
f 0 (x) is increasing: (−∞, −4) ∪ (0, ∞)
f 0 (x) is decreasing: (−4, 0)
There is a relative maximum at (−4, −8) and a relative minimum at
10
(0, 0)
f 00 (x) =
=
=
=
(2x+4)(x+2)2 −(x2 +4x)(2(x+2))
(x+2)4
(x+2)[(2x+4)(x+2)−2x2 −8x]
(x+2)4
(x+2)[2x2 +8x+8−2x2 −8x]
(x+2)4
(x+2)8
(x+2)4
f 00 (x) < 0 for x < −2 and f 00 (x) > 0 for x > −2
f (x) concave down: x < −2
f (x) concave up: x > −2
Since f (x) does not exist at x = −2 there is no inflection point to plot.
The graph looks like:
15
10
5
-5
-4
-3
-2
1
-1
-5
-10
-15
-20
7. f (x) = 1/(x2 − 9)
Answer:
Domain: x 6= −3, 3
y-intercept: f (0) = −1/9
x-intercept: None
Vertical Asymptotes: x = −3, 3
H.A.
1
=0
−9
1
lim 2
=0
x→−∞ x − 9
lim
x→∞
x2
11
2
Horizontal Asymptotes: y = 0
f 0 (x) =
(x2
2x
− 9)2
Critical numbers: x = −3, 0, 3
f 0 (x) < 0 when x < −3 or −3 < x < 0; and f 0 (x) > 0 when 0 < x < 3
or x > 3.
f 0 (x) is increasing: (−∞, −3) ∪ (−3, 0)
f 0 (x) is decreasing: (0, 3) ∪ (3, ∞)
There is a relative maximum at (0, −1/9)
f 00 (x) =
=
=
=
2(x2 −9)2 −2x2(x2 −9)2x
(x2 −9)4
(x2 −9)(2(x2 −9)−8x2 )
(x2 −9)4
−6x2 −18
(x2 −9)3
(6)(−x2 −3)
(x2 −9)3
f 00 (x) < 0 when −3 < x < −3 ; and f 00 (x) > 0 when x < −3 or x > 3.
f (x) concave down: (−3, 3)
f (x) concave up: (−∞, −3) ∪ (3, ∞)
Since f (x) does not exist at x = −3 or x = 3 there is no inflection
point to plot.
The graph looks like:
0.4
0.2
-6
-4
2
-2
-0.2
-0.4
-0.6
√
8. f (x) = 1/ 1 − x2
Answer:
12
4
6
Domain: (−1, 1)
y-intercept: f (0) = 1
x-intercept: None
Vertical Asymptotes: x = −1, 1
H.A.
1
=0
lim √
x→∞
1 − x2
1
lim √
=0
x→−∞
1 − x2
Horizontal Asymptotes: y = 0
f 0 (x) =
=
=
√
0( 1−x2 )−1( 12 (1−x2 )−1/2 (−2x))
√
( 1−x2 )2
2
−1/2
x(1−x )
√
( 1−x2 )2
x
(1−x2 )3/2
Critical numbers: x = 0 and x < −1 or x > 1(f 0 (x) does not exist)
f 0 (x) > 0 for 0 < x < 1 and f 0 (x) < 0 for −1 < x < 0
f 0 (x) is increasing: (0, 1)
f 0 (x) is decreasing: (−1, 0)
There is a relative minimum at (0, 1)
f 00 (x) =
=
=
=
(1−x2 )3/2 −(3/2)(1−x2 )1/2 (−2x)
((1−x2 )3/2 )2
(1−x2 )3/2 +3(1−x2 )1/2
((1−x2 )3/2 )2
(1−x2 )1/2 ((1−x2 )+3)
for − 1
(1−x2 )3
2
(4−x )
for − 1 < x < 1
(1−x2 )5/2
f 00 (x) > 0 for −1 < x < 1
f (x) is concave up for the domain of f (x).
There are no inflection points.
The graph looks like:
13
<x<1
2.5
2.0
1.5
-2
1
-1
9. f (x) = (x2 − 9)/(x2 + 1)
Answer:
Domain: all real numbers
y-intercept: f (0) = −9
x-intercept: f (x) = 0 when x = −3, 3
Vertical Asymptotes: none
H.A.
x2 − 9
=1
x→∞ x2 + 1
x2 − 9
=1
lim 2
x→−∞ x + 1
Horizontal Asymptotes: y = 1
lim
f 0 (x) =
=
=
2x(x2 +1)−(x2 −9)(2x)
(x2 +1)2
2x+18x
(x2 +1)2
20x
(x2 +1)2
Critical numbers: x = 0
f 0 (x) < 0 when x < 0 and f 0 (x) > 0 when x > 0
f 0 (x) is increasing: (0, ∞)
f 0 (x) is decreasing: (−∞, 0)
There is a relative minimum at (0, −9)
14
2
f 00 (x) =
=
=
=
20((x2 +1)2 )−20x2(x2 +1)2x)
(x2 +1)4
(x2 +1)(20(x2 +1)−80x2 )
(x2 +1)4
(x2 +1)(−60x2 +20
(x2 +1)4
(x2 +1)(20)(1−3x2
(x2 +1)4
p
p
p
f 00 (x) >p0 when − 1/3 < x < 1/3; and f 00 (x) < 0 when x < − 1/3
or x > 1/3
p
p
f (x) concave up: (− 1/3, p1/3)
p
∪
(
1/3, ∞)
f (x) concave down: (−∞, − 1/3)
p
p
There are inflection points at (− 1/3, −13/2) and ( 1/3, −13/2)
The graph looks like:
-4
2
-2
4
-2
-4
-6
-8
III. Find the absolute maximum and absolute minimum of the given function
on the specified interval.
1. f (x) = x3 + x2 + 1 [−3, 2]
Answer:
f (x) is continuous on [−3, 2]. Therefore the function has the extreme
value property.
First we find the critical numbers.
f 0 (x) = 3x2 + 2x = x(3x + 2)
f 0 (x) = 0 when x = 0, −2/3. Both values are in the given interval
f 0 (x) < 0 when −2/3 < x < 0 and f (x) > 0 when x < −2/3 or x > 0.
15
There is a relative max at (−2/3, 31/27) and a relative min at (0, 1)
At the endpoints, f (−3) = −17 and f (2) = 13
The absolute maximum is 13 and the absolute minimum is -17.
2. f (x) = (x2 − 4)5 [−3, 2]
f (x) is continuous on [−3, 2]. Therefore the function has the extreme
value property.
First we find the critical numbers.
f 0 (x) = 5(x2 − 4)4 (2x)
f 0 (x) = 0 when x = −2, 0, 2. These values are in the given interval.
f 0 (x) < 0 when x < −2 or −2 < x < 0; and f 0 (x) > 0 when 0 < x < 2
or x > 2.
There is a relative minimum at (0, −1024).
At the endpoints, f (−3) = 3125 and f (2) = 0.
The absolute maximum is 3125 and the absolute minimum is -1024.
3. f (x) = x2 /(x − 1) [−2, −1/2]
f (x) is continuous on[−2, −1/2]. Therefore the function has the extreme value property.
First we find the critical numbers.
f 0 (x) =
x2 − 2x
x(x − 2)
2x(x − 1) − x2
=
=
2
2
(x − 1)
(x − 1)
(x − 1)2
f 0 (x) = 0 when x = 0, 2. Neither value is in the given interval, therefore we only have to check the endpoints.
At the endpoints, f (−2) = −4/3 and f (−1/2) = −1/6.
The absolute maximum is -1/6 and the absolute minimum is -4/3.
16