Question

Find the absolute extrema of the function on the domain

Answer 1

Given function is

$K(z)=\sqrt[]{16-z^2}$

Differentiating K(z) w.r.t z, we have,

$K^{\prime}(z)=-\frac{z}{\sqrt[]{16-z^2}}$

To find the stationary point, let us solve the equation K'(z)=0

$\begin{gathered} K^{\prime}(z)=0 \\ -\frac{z}{\sqrt[]{16-z^2}}=0 \\ z=0 \end{gathered}$

Now, we again differentiate K'(z) w.r.t z.

$K^{\doubleprime}(z)=-\frac{1}{\sqrt[]{16-z^2}}+\frac{z^2}{\sqrt[]{16-z^2}}$

At z=0,

$K^{\doubleprime}(z)=-\frac{1}{4}<0$

Therefore, at z=0, the function attains maximum value.

The maximum value of the function is

$K(0)=4$

In the domain (-4,4), there is no absolute minima for the given function.