How do you use the Extreme Value Theorem to determine the global extrema of the function f(x)=5+ (x^2)/(x+2) on the closed interval [-1,3]?

Sep 12, 2015

Use what some authors call the Closed Interval Method.

Explanation:

The Extreme Value Theorem does not really tell us how to find extrema, it only guarantees that for a function that is continuous on a closed interval, there are extrema.

Nevertheless, we can find the extrema.

They must occur at either a critical number (in the interval) or at an endpoint of the interval.

So the method is: find critical number in the interval, then evaluate $f$ at these critical numbers and at the endpoints.

For $f \left(x\right) = 5 + \frac{{x}^{2}}{x + 2}$ on $\left[- 1 , 3\right]$, we get

$f ' \left(x\right) = \frac{{x}^{2} + 4 x}{x + 2} ^ 2$ So the critical numbers are $- 4 , 0$.

The only critical number in the interval is $0$

Evaluate:

$f \left(- 1\right) = 6$
$f \left(0\right) = 5$
$f \left(3\right) = 6 \frac{4}{5} = 6.8$

The maximum is $6 \frac{4}{5}$ (at $x = 3$)
The minimum is $5$ (at $x = 0$)