Respuesta :

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.

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