Mathematics - Application of Derivative Question with Solution | TestHub

It is clear that $\mathrm{f}(\mathrm{x})$ has a definite and unique value for each $x\in [1,5].$ Thus, for every point in the interval $[1,5]$, the value of $\mathrm{f}(\mathrm{x})$ exists. So, $f(x)$ is continuous in the interval $[1,5]$. Also, $f^{\prime }(x)=\frac{-x}{\sqrt{25-x^2}}$, which clearlyexists for all $x$ in an open interval $(1,5)$. So, ${\mathrm{f}}^{\prime }(\mathrm{x})$ is differentiable in $(1,5)$. So, there must be a value $c\in [1,5]$ such that $\begin{array}{rl}& {\mathrm{f}}^{\prime }(\mathrm{c})=\frac{\mathrm{f}(5)-\mathrm{f}(1)}{5-1}=\frac{0-\sqrt{24}}{4}\\ & =\frac{0-2\sqrt{6}}{4}=\frac{-\sqrt{6}}{2}\\ & \text{ But }{\mathrm{f}}^{\prime }(\mathrm{c})=\frac{-\mathrm{c}}{\sqrt{25-{\mathrm{c}}^2}}\\ \Rightarrow & \frac{-\mathrm{c}}{\sqrt{25-{\mathrm{c}}^2}}=-\frac{\sqrt{6}}{2}\\ \Rightarrow & 4{\mathrm{c}}^2=6\left(25-{\mathrm{c}}^2\right)\\ \Rightarrow & 4{\mathrm{c}}^2=150-6{\mathrm{c}}^2\Rightarrow 10{\mathrm{c}}^2=150\\ \Rightarrow & {\mathrm{c}}^2=15\Rightarrow \mathrm{c}=\pm \sqrt{15}\\ \therefore & \mathrm{c}=\sqrt{15}\in [1,5]\end{array}$