Verify Lagranges mean value theorem for function `f(x)=sqrt(25-x^2)` on `[-3,\ 4]` and find a point `' c '` in the indicated interval:

Answer

∴ C `=pm frac{5}{2 sqrt{2}}`

Explanation

$$ f(x)=\sqrt{25-x^{2}} \text { in }[-3.4] $$
The function is algebraic so it is continuous in `[-3,4]` and differentiable at `(-3,4)`. Then there must exist a value 'c' for which
$$
\begin{aligned}
&f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \\ \\
&f^{\prime}(x)=\frac{1}{2}\left(25-x^{2}\right)^{-1 / 2} \cdot(-2 x) \\ \\
...