Maxima and Minima

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 f has a local maximum (or relative maximum) at c, if f(c) ≥ f(x) where x is near c.

Similarly, f has a local minimum at c if f(c) ≤ f(x) when x is near c.

Definition of global maximum or global minimum

A function f has a global maximum (or absolute maximum) at c if f(c) ≥ f(x) for all x in D, where D is the domain of f. The number f(c) is called the maximum value of f on D.

Similarly, f has a global minimum (or absolute minimum) at c if f(c) ≤ f(x) for all x in D and the number f(c) is called the minimum value of f on D.

The maximum and minimum values of f are called the extreme values of f

Fermat’s Theorem

If f has a local maximum or minimum at c, and if f ‘(c) exists then
f ‘(c) = 0

Definition of critical number

A critical number of a function f is a number c in the domain of f such that either f ‘(c) = 0 of f ‘(c) does not exists.

The Extreme Value Theorem

If f is continuous on a closed interval [a, b], then f attains an absolute maximum value f(c) and an absolute minimum value f(d) at some number c and d in [a, b]

Example:
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.

The Closed Interval Method

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 f at the critical numbers of f in (a, b).

Step 2: Find the values of f at the endpoints of the interval.

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 f at the critical numbers of f in
f(x) = x³ – 3x² + 1
f ‘(x) = 3x² – 6x = 3x(x – 2)
When f ‘(x) = 3x(x – 2) we get x = 0 or x = 2

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 f at the endpoints of the interval.

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.

Videos

Finding Intervals of Increase/Decrease Local Max/Mins
The basic idea of finding intervals of increase/decrease as well as finding local maximums and minimums.

Increasing/Decreasing , Local Maximums/Minimums
This video gives a graph and discusses intervals where the function is increasing and decreasing. It also discuss local maximums and minimums.

Finding Local Maximums/Minimums - Second Derivative Test
How to use the second derivative test to find local maximums and local minimums

The Closed Interval Method to Find Absolute Maximums and Minimums
The Closed Interval Method to Find Absolute Maximums and Minimums of continuous functions on closed intervals. Patrick gives the procedure, gives a short geometric idea as to why this works, and then do a specific example.

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