Candidates Test for Global Extrema (College Board AP® Calculus AB)

Study Guide

Test yourself

Written by: Jamie Wood

Reviewed by: Dan Finlay

Candidates test

What is the candidates test for global extrema?

For a continuous function on a closed interval, the extreme value theorem guarantees
- that there will be at least one global maximum and one global minimum
The global extrema (minimum or maximum) of a continuous function on a closed interval can only occur at:
- Critical points
- or at endpoints
This fact can be used to find global extrema using the following steps:
First check that $f (x)$ is continuous on the interval $[a, b]$
Then find the critical points of $f (x)$ on the interval $(a, b)$
Find the value of $f (x)$ at:
- The critical points
- The endpoints $x = a$ and $x = b$
Out of these values,
- the largest value of $f (x)$ is the global maximum
- the smallest value of $f (x)$ is the global minimum
This process is known as the candidates test
- The "candidates" for the global extrema are the critical points, and the endpoints

Examiner Tips and Tricks

Sketching a graph (or plotting on your calculator) is usually helpful
- You may be able to spot immediately if an extremum will be at an endpoint or a critical point
A common mistake is to find critical points and assume they are global extrema
- Even if the function is a quadratic, the critical point may not be the global extremum if the domain has been limited

Worked Example

Use the candidates test to find the coordinates of the global minimum and global maximum on the graph of the function $f$ defined by:

$f (x) = - x^{3} + 2 x^{2} + 4 x - 8$ $- 3 \leq x \leq 3$

Answer:

Check that $f (x)$ is continuous on the interval

$f (x)$ is continuous on [-3, 3] as it is a polynomial
(a cubic) defined for all values in the interval

Find the critical points of $f (x)$ on the interval, where $f^{'} (x) = 0$

$f^{'} (x) = - 3 x^{2} + 4 x + 4$

$- 3 x^{2} + 4 x + 4 = 0$

$x = 2$ and $x = - \frac{2}{3}$

Find the values of $f (x)$ at the critical points

$f (2) = 0$

$f (- \frac{2}{3}) = - \frac{256}{27} = - 9.481481 . . .$

Find the values of $f (x)$ at the endpoints of the interval, in this case -3 and 3

$f (- 3) = 25$

$f (3) = - 5$

Check the values of $f (x)$ at the critical points and the endpoints to find the largest and smallest; these will be the global maximum and minimum respectively

The largest is 25, which occurs at an end point

The smallest is $- \frac{256}{27}$ which is -9.481481..., which occurs at a critical point

The global maximum is $(- 3, 25)$

The global minimum is $(- \frac{2}{3}, - \frac{256}{27})$

Last updated: 16 August 2024

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