Math 220
March 5
I. Find the absolute maximum and absolute minimum values of f on the
given interval.
1. f (x) = x2 , [−1, 2]
2. f (x) = sin(x), [−π/4, π/4]
3. f (x) = x2 + 10x + 25, [−1, 4]
4. f (x) = x + sin(x), [0, 2π]
5. f (x) = sec(x), [0, π/4]
6. f (x) = xex , [−1, 0]
7. f (x) = 4x − x1 , [−1, −.1]
8.
f (x) =
9. f (x) =
x
, [0, 2]
x+1
√
x(16 − x), [0, 16]
10. f (x) = ln(x2 + 2x + 2), [−2, 2]
11. f (x) = ln(x3 − 2x2 + x + 1), [0, 2]
1
II.
(a) Find the vertical and horizontal asymptotes
(b) Find the intervals of increase and decrease
(c) Find the local maximum and minimum values
(d) Find the intervals of concavity and the inflections points
1.
f (x) = arctan(x)
2.
f (x) = e1−x
2
3.
f (x) = x3 + 3x3 + 3x + 5
4.
f (x) = earctan(x)
5.
f (x) =
6.
f (x) =
ex+1
1 − ex+1
x2 − 4
x2 + x − 5
III. Find the point of the curve y = 2 − x2 closest to (0,0)
2
1
Solutions
I. Find the absolute maximum and absolute minimum values of f on the
given interval.
1. f (x) = x2 , [−1, 2]
Answer:
f 0 (x) = 2x
f 0 (x) = 0 = 2x
x=0
Since f is continuous on the interval we need to check the critical points
and the endpoints.
f (−1) = 1
f (0) = 0 Absolute Minimum
f (2) = 4 Absolute Maximum
2. f (x) = sin(x), [−π/4, π/4]
Answer:
f 0 (x) = cos(x)
f 0 (x) = 0 = cos(x)
π
x = + kπ, k ∈ Z
2
Since f is continuous on the interval we need to check the critical points
and the endpoints.
3
√
f (−π/4) =√ − 22 Absolute Minimum
f (π/4) = 22 Absolute Maximum
3. f (x) = x2 + 10x + 25, [−1, 4]
Answer:
f 0 (x) = 2x + 10
f 0 (x) = 0 = 2x + 10
−2x = 10
x = −5
Since f is continuous on the interval we need to check the critical points
and the endpoints.
f (−1) = 36 Absolute Minimum
f (4) = 81 Absolute Maximum
4. f (x) = x + sin(x), [0, 2π]
Answer:
f 0 (x) = 1 + cos(x)
f 0 (x) = 0 = 1 + cos(x)
cos(x) = −1
x = π + k2π, k ∈ Z
Since f is continuous on the interval we need to check the critical points
and the endpoints.
4
f (0) = 0 Absolute Minimum
f (π) = π
f (2π) = 2π Absolute Maximum
5. f (x) = sec(x), [0, π/4]
Answer:
f 0 (x) = sec(x) tan(x)
f 0 (x) = 0 = sec(x) tan(x)
sin(x)
0=
cos2 (x)
0 = sin(x)
x = kπ, k ∈ Z
f 0 (x)does not exist
0 = cos2 (x)
π
x = + kπ, k ∈ Z
2
Since f is continuous on the interval we need to check the critical points
and the endpoints.
f (0) = 1 √Absolute Minimum
f (π/4) = 2 Absolute Maximum
6. f (x) = xex , [−1, 0]
Answer:
5
f 0 (x) = ex + xex
f 0 (x) = 0 = (1 + x)ex
0=1+x
x = −1
Since f is continuous on the interval we need to check the critical points
and the endpoints.
f (−1) = −1/e Absolute Minimum
f (0) = 0 Absolute Maximum
7. f (x) = 4x − x1 , [−1, −.1]
Answer:
f 0 (x) = 4 +
1
x2
f 0 (x) = 0 = 4 +
1
x2
No Solution
f 0 (x) does not exist when x = 0
Since f is continuous on the interval we need to check the critical points
and the endpoints.
f (−1) = −3
f (−.1) = 9.6
Absolute Minimum
Absolute Maximum
6
8.
f (x) =
x
, [0, 2]
x+1
Answer:
(x + 1) − x
(x + 1)2
1
=
(x + 1)2
1
f 0 (x) = 0 =
(x + 1)2
No solutions
f 0 (x) does not exist when x = −1
f 0 (x) =
Since f is continuous on the interval we need to check the critical points
and the endpoints.
f (0) = 0
f (2) = 32
Absolute Minimum
Absolute Maximum
√
9. f (x) = x(16 − x), [0, 16]
Answer:
16 − x √
√ − x
2 x
16 − 2x
√
=
2 x
8−x
= √
x
8
−
x
f 0 (x) = 0 = √
x
x=8
0
f (x) does not exist when x = 0
f 0 (x) =
7
Since f is continuous on the interval we need to check the critical points
and the endpoints.
f (0) = 0 √Absolute Minimum
f (8) = 16 2 Absolute Maximum
f (16) = 0 Absolute Minimum
10. f (x) = ln(x2 + 2x + 2), [−2, 2]
Answer:
f 0 (x) =
x2
2x + 2
+ 2x + 2
2x + 2
+ 2x + 2
0 = 2x + 2
2x = −2
x = −1
f 0 (x) = 0 =
x2
Since f is continuous on the interval we need to check the critical points
and the endpoints.
f (−2) = ln(2)
f (−1) = ln(1) = 0 Absolute Minimum
f (2) = ln(10) Absolute Maximum
11. f (x) = ln(x3 − 2x2 + x + 1), [0, 2]
Answer:
8
3x2 − 4x + 1
f (x) = 3
x − 2x2 + x + 1
3x2 − 4x + 1
=
x(x − 1)(x − 1) + 1
0
f 0 (x) = 0 = 3x2 − 4x + 1
0 = (x − 1)(3x − 1)
1
x = 1,
3
Since f is continuous on the interval we need to check the critical points
and the endpoints.
f (0) = ln(1) = 0 Absolute Minimum
f (1) = ln(1) = 0 Absolute Minimum
f (2) = ln(3) Absolute Maximum
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II.
(a) Find the vertical and horizontal asymptotes
(b) Find the intervals of increase and decrease
(c) Find the local maximum and minimum values
(d) Find the intervals of concavity and the inflections points
1.
f (x) = arctan(x)
Answer:
(a)
Vertical Asymtotes: None
Horizontal Asymtotes:y = −π/2, π/2
lim arctan(x) = π/2
x→∞
lim arctan(x) = −π/2
x→−∞
(b)
f 0 (x) =
1
1 + x2
f 0 (x) > 0 for all real numbers:
f is increasing on the interval (−∞, ∞)
(c)
f has no local maximums and minimums
(d)
f 00 (x) =
f 00 (x) = 0 when x = 0
f 00 (x) < 0 when x > 0
f 00 (x) > 0 when x < 0
10
−2x
(1 + x2 )2
f has has an inflection point at x = 0 and is concave up when x < 0
and concave down when x > 0.
2.
f (x) = e1−x
2
Answer:
(a)
Vertical Asymtotes:None
Horizontal Asymtotes: y = 0
2
lim e1−x = 0
x→∞
2
lim e1−x = ∞
x→−∞
(b)
f 0 (x) = −2xe1−x
2
f 0 (x) = 0 when x = 0
f 0 (x) < 0 when x > 0
f 0 (x) > 0 when x < 0
f is increasing on the interval (−∞, 0)
f is decreasing on the interval (0, ∞)
(c)
f has a local maximum at x = 0
(d)
2
2
f 00 (x) = 4x2 e1−x − 2e1−x = (4x2 − 2)e1−x
−1 √1
, 2
f 00 (x) = 0 when x = √
2
−1
f 00 (x) < 0 when x < √
or x > √12
2
−1
f 00 (x) > 0 when √
< x < √12
2
−1 √1
f has inflections points at x = √
, 2
2
11
2
−1
< x < √12
f is concave up when √
2
−1
f is concave down when x < √
or x >
2
√1
2
3.
f (x) = x3 + 3x3 + 3x + 5
Answer:
(a)
Vertical Asymtotes:None
Horizontal Asymtotes:
lim x3 + 3x3 + 3x + 5 = ∞
x→∞
lim x3 + 3x3 + 3x + 5 = −∞
x→−∞
(b)
f 0 (x) = 12x2 + 3
f (x) > 0 for all real numbers.
f is increasing on the interval (−∞, ∞)
(c)
f has no local maximums and minimums
(d)
f 00 (x) = 24x
f 00 (x) = 0 when x = 0
f 00 (x) < 0 when x < 0
f 00 (x) > 0 when x > 0
f has inflections points at x = 0
f is concave up when x > 0
f is concave down when x < 0
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4.
f (x) = earctan(x)
Answer:
(a)
Vertical Asymtotes:None
Horizontal Asymtotes:
lim earctan(x) = eπ/2
x→∞
lim earctan(x) = e−π/2
x→−∞
(b)
f 0 (x) =
earctan(x)
1 + x2
f (x) > 0for all real numbers.
f is increasing on the interval (−∞, ∞)
(c)
f has no local maximums and minimums
(d)
f 00 (x) =
earctan(x) (1 − 2x)
earctan(x) − 2xearctan(x)
=
(1 + x2 )2
(1 + x2 )2
f 00 (x) = 0 when x = 1/2
f 00 (x) > 0 when x < 1/2
f 00 (x) < 0 when x > 1/2
f has inflections points at x = 1/2
f is concave up when x < 1/2
f is concave down when x > 1/2
5.
f (x) =
13
ex+1
1 − ex+1
Answer:
(a)
Vertical Asymtotes:x = −1
Horizontal Asymtotes: y = 0
ex+1
=0
x→∞ 1 − ex+1
ex+1
lim
=0
x→−∞ 1 − ex+1
lim
(b)
f 0 (x) =
ex+1
ex+1 (1 − ex+1 ) + e2x+1
=
(1 − ex+1 )2
(1 − ex+1 )2
f 0 (x) > 0 for all real numbers.
f is increasing on the interval (−∞, ∞)
(c)
f has no local maximums and minimums
(d)
f 00 (x) =
ex+1 (1 − ex+1 )2 − ex+1 2(1 − ex+1 )(−ex+1 )
(1 − ex+1 )4
f 00 (x) =
ex+1 (1 − ex+1 ) ((1 − ex+1 ) + 2ex+1 )
(1 − ex+1 )4
f 00 (x) =
ex+1 (1 − ex+1 ) (1 + ex+1 )
(1 − ex+1 )4
f 00 (x) =
ex+1 (1 + ex+1 )
(1 − ex+1 )3
f 00 (x) does not exist when x = −1
f 00 (x) < 0 when x > −1
f 00 (x) > 0 when x < −1
f has inflections points at x = −1
f is concave up when x < −1
14
f is concave down when x > −1
6.
f (x) =
x2 − 4
x2 + x − 5
Answer:
(a)
Vertical Asymtotes:x = −3
x2 − 4
x−2
=
=0
2
x→2 x + x − 5
x+3
x2 − 4
x−2
lim 2
=
= Does not exist
x→−3 x + x − 5
x+3
lim
Horizontal Asymtotes:
x−2
x2 − 4
=
=1
x→∞ x2 + x − 5
x+3
x2 − 4
x−2
lim 2
=
=1
x→−∞ x + x − 5
x+3
lim
(b)
f 0 (x) =
(x + 3) − (x − 2)
5
=
(x + 3)2
(x + 3)2
f 0 (x) > 0 for all real numbers.
f is increasing on the interval (−∞, ∞)
(c)
f has no local maximums and minimums
(d)
f 00 (x) = −10(x + 3)−3
f 00 (x) does not exist when x = −3
f 00 (x) < 0 when x > −3
15
f 00 (x) > 0 when x < −3
f has inflections points at x = −3
f is concave up when x < −3
f is concave down when x > −3
III. Find the point of the curve y = 2 − x2 closest to (0,0)
Answer:
p
(x − 0)2 + (y − 0)2
p
d = x2 + (2 − x2 )2
√
d = x2 + x4 − 4x2 + 4
√
d = x4 − 3x2 + 4
d=
4x3 − 6x
d0 = √
2 x4 − 3x2 + 4
d0 (x) = 0 = 4x3 − 6x
0 = 2x(2x2 − 3)
p
p
x = 0, − 2/3, 2/3
p
p
f 0 (x) > 0 when −p2/3 < x < 0 and x > p2/3
f 0 (x) < 0 when − 2/3 > x and 0 < x < 2/3
p
p
f has a maximum when x p
= 0 and f has a minimum
when
x
=
−
2/3,
2/3
p
The points closest are (− 2/3, 4/3) and ( 2/3, 4/3)
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