In these lessons, we will learn

- the definition of local maximum and local minimum
- the definition of global maximum and global minimum
- Fermat's Theorem
- definition of critical number
- the Extreme Value Theorem
- the Closed Interval Method

Related Topics:

More Lessons on Calculus

The local maxima are the largest values (maximum) that a function takes in a point within a given neighborhood.

The local minima are the smallest values (minimum), that a function takes in a point within a given neighborhood.

Definition of local maximum and local minimum

A function

fhas a local maximum (or relative maximum) atc, iff(c) ≥f(x) wherexis nearc.Similarly,

fhas a local minimum atciff(c) ≤f(x) whenxis nearc.

Definition of global maximum or global minimum

A function

fhas a global maximum (or absolute maximum) atciff(c) ≥f(x) for allxinD, whereDis the domain off. The numberf(c) is called the maximum value offonD.Similarly,

fhas a global minimum (or absolute minimum) atciff(c) ≤f(x) for allxinDand the numberf(c) is called the minimum value offonD.The maximum and minimum values of

fare called the extreme values off

Fermat’s Theorem

If

fhas a local maximum or minimum atc, and iff ‘(c) exists then(

f ‘c) = 0

Definition of critical number

A critical number of a function

fis a numbercin the domain offsuch that eitherf‘(c) = 0 off‘(c) does not exists.

This video gives the definition of critical numbers and review several examples.

If

fis continuous on a closed interval [a, b], thenfattains an absolute maximum valuef(c) and an absolute minimum valuef(d) at some numbercanddin [a,b]

This video explains the Extreme Value Theorem.

Find the critical numbers of

**Solution: **

Using the Product Rule, we get

and *f* ‘(*x*) does not exist when *x* = 0.

So, the critical numbers are and 0.

These are the steps to find the absolute maximum and minimum values of a continuous function *f* on a closed interval [*a*, *b*]:

* Step 1:* Find the values of

* Step 2: *Find the values of

* Step 3:* The largest of the values from Steps 1 and 2 is the absolute maximum value and the smallest of these values is the absolute minimum value.

**Example**

Find the absolute maximum and minimum value of the function

**Solution:**

Since *f* is continuous on , we can use the Closed Interval Method.

* Step 1: *Find the values of

When

So, the critical numbers are *x* = 0 and *x* = 2

The values of *f* at these critical numbers are

*f*(0) = 1 and *f*(2) = –3

* Step 2:* Find the values of

The values of *f* at the endpoints of the interval are

* Step 3:* The largest of the values from Steps 1 and 2 is the absolute maximum value and the smallest of these values is the absolute minimum value.

Comparing the four numbers, we see that the absolute maximum value is *f*(4) = 17 and the absolute minimum is *f*(2) = –3.

The basic idea of finding intervals of increase/decrease as well as finding local maximums and minimums.

This video gives a graph and discusses intervals where the function is increasing and decreasing. It also discuss local maximums and minimums.

How to use the second derivative test to find local maximums and local minimums.

This video gives the procedure, gives a short geometric idea as to why this works, and then do a specific example.

You can use the Mathway widget below to practice Algebra or other math topics. Try the given examples, or type in your own problem. Then click "Answer" to check your answer.

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