# How do you find the global extreme values for f(t) = tsqrt(4 - t²) on [-1,2]?

Jul 1, 2015

Maximum is $\left(\sqrt{2} , 2\right)$ and minimum is $\left(- 1 , - \sqrt{3}\right)$.

#### Explanation:

The global extreme values of a function in a set $\left[a , b\right]$ are to be searched between the local extreme values in the set $\left[a , b\right]$ and in $a$ or $b$.

The domain of the function is $D = \left[- 2 , 2\right]$, because the radical argument has to be not-negative.

Let's search the local extreme:

$y ' = 1 \cdot \sqrt{4 - {t}^{2}} + t \cdot \frac{1}{2 \sqrt{4 - {t}^{2}}} \cdot \left(- 2 t\right) =$

$= \sqrt{4 - {t}^{2}} - {t}^{2} / \sqrt{4 - {t}^{2}} = \frac{4 - {t}^{2} - {t}^{2}}{\sqrt{4 - {t}^{2}}} = 2 \cdot \frac{2 - {t}^{2}}{\sqrt{4 - {t}^{2}}}$.

Now:

$y ' \ge 0$

if

$2 - {t}^{2} \ge 0$ (the radical is positive or zero in its domain!),

$- \sqrt{2} \le t \le \sqrt{2}$.

$f \left(- \sqrt{2}\right) = - \sqrt{2} \cdot \sqrt{4 - 2} = - 2$

$f \left(\sqrt{2}\right) = \sqrt{2} \cdot \sqrt{4 - 2} = 2$.

So the point $A \left(- \sqrt{2} , - 2\right)$ is a local minimum and it is not in the set $\left[- 1 , 2\right]$ and the point $B \left(\sqrt{2} , 2\right)$ is a local maximum that is in the set.

$f \left(- 1\right) = - \sqrt{3}$

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

SO:

The point $B \left(\sqrt{2} , 2\right)$ is the global maximum and the point $C \left(- 1 , - \sqrt{3}\right)$ is the global minumum.

graph{xsqrt(4-x^2) [-10, 10, -5,5]}