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## AP®︎/College Calculus BC

### Unit 5: Lesson 5

Using the candidates test to find absolute (global) extrema

# Absolute minima & maxima review

Review how we use differential calculus to find absolute extremum (minimum and maximum) points.

## How do I find absolute minimum & maximum points with differential calculus?

An absolute maximum point is a point where the function obtains its greatest possible value. Similarly, an absolute minimum point is a point where the function obtains its least possible value.
Supposing you already know how to find relative minima & maxima, finding absolute extremum points involves one more step: considering the ends in both directions.

## Finding absolute extrema on a closed interval

Extreme value theorem tells us that a continuous function must obtain absolute minimum and maximum values on a closed interval. These extreme values are obtained, either on a relative extremum point within the interval, or on the endpoints of the interval.
Let's find, for example, the absolute extrema of h, left parenthesis, x, right parenthesis, equals, 2, x, cubed, plus, 3, x, squared, minus, 12, x over the interval minus, 3, is less than or equal to, x, is less than or equal to, 3.
h, prime, left parenthesis, x, right parenthesis, equals, 6, left parenthesis, x, plus, 2, right parenthesis, left parenthesis, x, minus, 1, right parenthesis, so our critical points are x, equals, minus, 2 and x, equals, 1. They divide the closed interval minus, 3, is less than or equal to, x, is less than or equal to, 3 into three parts:
Intervalx-valueh, prime, left parenthesis, x, right parenthesisVerdict
minus, 3, is less than, x, is less than, minus, 2x, equals, minus, start fraction, 5, divided by, 2, end fractionh, prime, left parenthesis, minus, start fraction, 5, divided by, 2, end fraction, right parenthesis, equals, start fraction, 21, divided by, 2, end fraction, is greater than, 0h is increasing \nearrow
minus, 2, is less than, x, is less than, 1x, equals, 0h, prime, left parenthesis, 0, right parenthesis, equals, minus, 12, is less than, 0h is decreasing \searrow
1, is less than, x, is less than, 3x, equals, 2h, prime, left parenthesis, 2, right parenthesis, equals, 24, is greater than, 0h is increasing \nearrow
Now we look at the critical points and the endpoints of the interval:
xh, left parenthesis, x, right parenthesisBeforeAfterVerdict
minus, 39minus\nearrowMinimum
minus, 220\nearrow\searrowMaximum
1minus, 7\searrow\nearrowMinimum
345\nearrowminusMaximum
On the closed interval minus, 3, is less than or equal to, x, is less than or equal to, 3, the points left parenthesis, minus, 3, comma, 9, right parenthesis and left parenthesis, 1, comma, minus, 7, right parenthesis are relative minima and the points left parenthesis, minus, 2, comma, 20, right parenthesis and left parenthesis, 3, comma, 45, right parenthesis are relative maxima.
left parenthesis, 1, comma, minus, 7, right parenthesis is the lowest relative minimum, so it's the absolute minimum point, and left parenthesis, 3, comma, 45, right parenthesis is the largest relative maximum, so it's the absolute maximum point.
Notice that the absolute minimum value is obtained within the interval and the absolute maximum value is obtained on an endpoint.
Problem 1
f, left parenthesis, x, right parenthesis, equals, x, cubed, minus, 3, x, squared, plus, 12
What is the absolute maximum value of f over the closed interval open bracket, minus, 2, comma, 4, close bracket?

Want to try more problems like this? Check out this exercise.

## Finding absolute extrema on entire domain

Not all functions have an absolute maximum or minimum value on their entire domain. For example, the linear function f, left parenthesis, x, right parenthesis, equals, x doesn't have an absolute minimum or maximum (it can be as low or as high as we want).
However, some functions do have an absolute extremum on their entire domain. Let's analyze, for example, the function g, left parenthesis, x, right parenthesis, equals, x, e, start superscript, 3, x, end superscript.
g, prime, left parenthesis, x, right parenthesis, equals, e, start superscript, 3, x, end superscript, left parenthesis, 1, plus, 3, x, right parenthesis, so our only critical point is x, equals, minus, start fraction, 1, divided by, 3, end fraction.
Intervalx-valueg, prime, left parenthesis, x, right parenthesisVerdict
left parenthesis, minus, infinity, comma, minus, start fraction, 1, divided by, 3, end fraction, right parenthesisx, equals, minus, 1g, prime, left parenthesis, minus, 1, right parenthesis, equals, minus, start fraction, 2, divided by, e, cubed, end fraction, is less than, 0g is decreasing \searrow
left parenthesis, minus, start fraction, 1, divided by, 3, end fraction, comma, infinity, right parenthesisx, equals, 0g, prime, left parenthesis, 0, right parenthesis, equals, 1, is greater than, 0g is increasing \nearrow
Let's imagine ourselves walking on the graph of g, starting all the way to the left (from minus, infinity) and going all the way to the right (until plus, infinity).
We will start by going down and down until we reach x, equals, minus, start fraction, 1, divided by, 3, end fraction. Then, we will be forever going up. So g has an absolute minimum point at x, equals, minus, start fraction, 1, divided by, 3, end fraction. The function doesn't have an absolute maximum value.