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)=-(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.