Find the value of `c` prescribed by Lagranges mean value theorem for the function `f(x)=sqrt(x^2-4)` defined on `[2,\ 3]` .

Answer

` C = sqrt{5}`

Explanation

Clearly,

`f(x)=\sqrt{{x}^{2}-4}` is continuous on `[2,3]` and differentiable on `(2,3)`.

So, by Lagrange's mean value theorem, there exists c `\varepsilon(2,3)`
such that `f^{\prime}(c)=\frac{f(3)-f(2)}{3-2}`
$$
\begin{aligned}
&\Rightarrow \frac{\mathrm{c}}{\sqrt{c^{2}-4}}=\sqrt{5}-0 \\
&{\left[\because \mathrm{f}(\mathrm{x})=\sqrt{\mathrm{x}^{2}-4} \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}-4}}\right]} \\
&\Rightarrow \mathrm{c}^{2}=5\left(\mathrm{c}^{2}-4\right) \\
&\Rightarrow 4 \mathrm{c}^{2}=20 \\
...