Question
Verify mean value theorem for the function: $\text{f(x)}=\sqrt{25-\text{x}^2}\text{ in }[1,5].$

Answer

We have, $\text{f(x)}=\sqrt{25-\text{x}^2}\text{ in }[1,5]$
Since $25 - x^2$ and square root function are continuous and differentiable in their domain.
given function $f(x)$ is also continuous and differentiable.
So, conditions of mean value thecorem are satisfied.
Hence, there exists atleast one $\text{c}\in(1,5)$ such that,
$\text{f}\ '(\text{c})=\frac{\text{f}(5)-\text{f}(1)}{5-1}$
$\Rightarrow\ \frac{-\text{c}}{\sqrt{25-\text{c}^2}}=\frac{0-\sqrt{24}}{4}$
$\Rightarrow\ 16\text{c}^2=24(25-\text{c}^2)$
$\Rightarrow\ 40\text{c}^2=600$
$\Rightarrow\ \text{c}^2=15$
$\Rightarrow\ \text{c}=\sqrt{15}\in(1,5)$
Hence, mean value theorem has been verified.

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