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Questions
Example Find the absolute maximum and absolute minimum values of f (x) =
x x^2 + 1
on the interval [0, 2]
Example Find the absolute maximum and absolute minimum values of f (x) =
ln x x
on the interval [1, 3]
Example If a and b are positive numbers, find the maximum value of f (x) = xa(1 − x)b, 0 ≤ x ≤ 1.
Solutions
Example Find the absolute maximum and absolute minimum values of f (x) = x x^2 + 1
on the interval [0, 2]
Absolute extrema on a closed interval are found using the Closed Interval Method:
- Find the values of f at the critical numbers of f in (a, b).
- Find the values of f at the endpoints of the interval.
- The largest of the values from 1) and 2) is the absolute maximum; the smallest of these values is the absolute minimum.
We need the critical numbers. We need to find where f ′(c) = 0 and where f ′(x) does not exist. Since x^2 + 1 6 = 0 for real valued x, the derivative always exists.
f (x) =
x x^2 + 1
f ′(x) =
d dx
[
x x^2 + 1
]
(x^2 + 1) (^) dxd [x] − x (^) dxd [x^2 + 1] (x^2 + 1)^2
= (x
(^2) + 1)(1) − x(2x) (x^2 + 1)^2
=
x^2 + 1 − 2 x^2 (x^2 + 1)^2
=
−x^2 + 1 (x^2 + 1)^2
For f ′(c) = 0, the numerator must equal zero,
f ′(c) = 0 =
−c^2 + 1 (c^2 + 1)^2 0 = −c^2 + 1 c^2 = 1 c = ± 1
We have shown that f ′(1) = 0 and f ′(−1) = 0. The critical numbers are +1, −1. Only +1 is in the interval (0, 2).
Now, we evaluate the function at the critical numbers in the interval and at the endpoints of the interval:
f (+1) = 1 (1)^2 + 1
=^1
f (0) =
(0)^2 + 1
f (2) = 2 (2)^2 + 1
=^2
The largest number is 1/2, so this is the absolute max and it occurs at x = +1. The smallest number is 0, so this is the absolute min and it occurs at x = 0.
Example Find the absolute maximum and absolute minimum values of f (x) =
ln x x on the interval [1,^ 3]
Absolute extrema on a closed interval are found using the Closed Interval Method:
- Find the values of f at the critical numbers of f in (a, b).
- Find the values of f at the endpoints of the interval.
- The largest of the values from 1) and 2) is the absolute maximum; the smallest of these values is the absolute minimum.
We need the critical numbers, which means we need to find where f ′(c) = 0 and where f ′(x) does not exist. The only place we could have the derivative not defined is for x ≤ 0; luckily, this is outside of the interval (1, 3) so we don’t need to worry about the derivative being undefined.
f (x) = ln^ x x f ′(x) =
d dx
[
ln x x
]
(x) (^) dxd [ln x] − ln x (^) dxd [x] (x)^2
=
(x)
x
− ln x(1) x^2 =
1 − ln x x^2
For f ′(c) = 0, the numerator must equal zero,
f ′(c) = 0 =
1 − ln c c^2 0 = 1 − ln c ln c = 1 c = e
We have shown that f ′(e) = 0. The critical number is e, which lies in the interval (2, 3).
The only critical numbers will be if f ′(c) = 0:
f (x) = xa(1 − x)b
f ′(x) = d dx
[
xa(1 − x)b
]
= xa^
d dx
[
(1 − x)b
]
d dx
[xa]
= bxa(1 − x)b−^1 (−1) + a(1 − x)bxa−^1 = a(1 − x)bxa−^1 − bxa(1 − x)b−^1 = a(1 − x)bxax−^1 − bxa(1 − x)b(1 − x)−^1 = (1 − x)bxa^
ax−^1 − b(1 − x)−^1
= (1 − x)bxa
a x
− b 1 − x
= (1 − x)bxa
a(1 − x) − bx x(1 − x)
= (1 − x)b−^1 xa−^1 (a − (a + b)x)
The critical number is therefore c =
a a + b
Now, we evaluate the function at the critical numbers in the interval and at the endpoints of the interval:
f
a a + b
a a + b
)a ( 1 −
a a + b
))b
a a + b
)a ( b a + b
)b
0 (since a and b are positive)
f (0) = 0 a(1 − 0)b^ = 0 f (1) = 1 a(1 − 1)b^ = 0
Therefore, the maximum value of f (x) = xa(1 − x)b, 0 ≤ x ≤ 1 is
a a + b
)a ( b a + b
)b which occurs at x =
a a + b
