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

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.

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

$f\left(x\right)=-x^3+2x^2+4x-8$ $-3\le x\le 3$

Answer:

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

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

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

$f^{\text{'}}\left(x\right)=-3x^2+4x+4$

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

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

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

$f\left(2\right)=0$

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

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

$f\left(-3\right)=25$

$f\left(3\right)=-5$

Check the values of $f\left(x\right)$ 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

$f$ is continuous on $\left[-3,3\right]$ so the candidates test for global extrema shows that:

the global maximum is 25 and occurs at the endpoint $\left(-3,25\right)$

the global minimum is $-\frac{256}{27}$ and occurs at the critical point $\left(-\frac{2}{3},-\frac{256}{27}\right)$