Find the absolute extrema of the function on the domain

Given function is
[tex]K(z)=\sqrt[]{16-z^2}[/tex]Differentiating K(z) w.r.t z, we have,
[tex]K^{\prime}(z)=-\frac{z}{\sqrt[]{16-z^2}}[/tex]To find the stationary point, let us solve the equation K'(z)=0
[tex]\begin{gathered} K^{\prime}(z)=0 \\ -\frac{z}{\sqrt[]{16-z^2}}=0 \\ z=0 \end{gathered}[/tex]Now, we again differentiate K'(z) w.r.t z.
[tex]K^{\doubleprime}(z)=-\frac{1}{\sqrt[]{16-z^2}}+\frac{z^2}{\sqrt[]{16-z^2}}[/tex]At z=0,
[tex]K^{\doubleprime}(z)=-\frac{1}{4}<0[/tex]Therefore, at z=0, the function attains maximum value.
The maximum value of the function is
[tex]K(0)=4[/tex]In the domain (-4,4), there is no absolute minima for the given function.