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4.1 Maximum and Minimum Values
Math 1271, TA: Amy DeCelles
1. Overview
Outline:
- Definition of absolute and local maximum and minimum values of a function
- Theorem: Local max/min values will always occur at a critical number, i.e. a number c in the domain of f such that f ′(c) = 0 or f ′(c) does not exist.
- Theorem: A continuous function will always have an absolute max value and an absolute min value on a closed interval.
- Finding abs max/min values (for a continuous function on a closed interval)
- Either the max/min value occurs at an endpoint, or it occurs somewhere in the middle.
- If it occurs in the middle, it has to be a local max/min value as well, and we know that local max/min values always occur at critical numbers.
- So the abs max/min will be either at an endpoint or a critical number.
- You just have to check the value of the function at the endpoints and the critical numbers and see what value is the greatest and which the least; those are your absolute max and min values.
Note: We will learn how to find local max/min values in section 4.3.
You should be able to look at a graph of a function and determine:
- the critical numbers
- the local min/max values
- the absolute min/max values
2. Examples
1.) Find the critical numbers of
g(x) =
1 − x^2 Remember that the critical numbers of g are numbers in the domain of g where g′(x) = 0 or g′(x) DNE. So we start by taking the derivative (have to use the chain rule): g′(x) = 12 (1 − x^2 )−^1 /^2 · (− 2 x) = −
x √ 1 − x^2 When is g′(x) = 0? Just when x = 0 (because the numerator will be zero, and the denominator will be nonzero.)
When does g′(x) DNE? Well, g′(x) DNE if the denominator is zero, i.e. if x = ±1, and g′(x) also DNE if the inside of the square root is negative, i.e. if: 1 − x^2 < 0 1 < x^2 √ 1 < |x| 1 < |x| x > 1 or x < − 1
So g′(x) DNE if x ≥ 1 or x ≤ −1.
To sum up: g′(x) = 0 for x = 0 and g′(x) DNE for x ≥ 1 or x ≤ −1. Now we just need to check which of these numbers are in the domain of g. The domain of g is going to be all real numbers, except those numbers that make the inside of the square root negative. We’ve already figured out that the square root is negative when x > 1 or x < −1, so the domain of g is [− 1 , 1]. So the critical numbers are − 1 , 0 , and 1.
2.) Find the absolute max/min values of f (x) = x^3 − 3 x + 1 on the interval [0, 3].
We know that the absolute max/min values of f (x) will occur either at an endpoint or a critical number. So we start by finding the critical numbers. We need to take the derivative: f ′(x) = 3x^2 − 3 = 3(x^2 − 1) Where is f ′(x) = 0? If we factor f ′(x),
f ′(x) = 3(x − 1)(x + 1)
we can see that f ′(x) = 0 when x = ±1. Where does f ′(x) DNE? Nowhere. Since the numbers ±1 are in the domain of f , ±1 are the critical numbers of f (x).
So we need to check the value of f (x) at each endpoint of [0, 3] and each critical number in [0, 3]. Since −1 is not in the interval, we don’t have to check f (x) there.
f (0) = 0 − 0 + 1 = 1 f (1) = 1 − 3(1) + 1 = − 1 f (3) = 33 − 3(3) + 1 = 27 − 9 + 1 = 19
So the absolute max value is 19 and the absolute min value is −1.
3.) Find the absolute max/min values of f (x) = x
(^2) − 4 x^2 +4 on the interval [−^4 ,^ 4]. We know that the absolute max/min values of f (x) will occur either at an endpoint or a critical number. So we start by finding the critical numbers. We take the derivative using the quotient rule:
f ′(x) =
(2x)(x^2 + 4) − (x^2 − 4)(2x) (x^2 + 4)^2
= 2 x((x^2 + 4) − (x^2 − 4)) (x^2 + 4)^2
= 2 x(x^2 + 4 − x^2 + 4) (x^2 + 4)^2
=
16 x (x^2 + 4)^2
We can see that f ′(x) = 0 just when x = 0, and that there are no values of x where f ′(x) DNE. Since x = 0 is in the domain of f , it is a critical number.
