Question

The value of $c$ for which the conclusion of Lagrange's mean value theorem holds for the function $f\left(x\right)=\sqrt{25-x^2}$ on the interval $[1,5]$ is

• A. $\sqrt{3}$
• B. $\sqrt{5}$
• C. $\sqrt{15}$`
• D. $2$

Answer 1

The correct option is C $\sqrt{15}$`

$\begin{array}{c}f\left(x\right)=\sqrt{25-x^2}\\ f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\\ =\frac{0-\sqrt{24}}{4}\\ \frac{1}{2}\frac{1}{\sqrt{25-x^2}}\left(-2x\right)=\frac{-\sqrt{24}}{4}\\ \frac{c}{\sqrt{25-c^2}}=\frac{\sqrt{24}}{4}\\ \frac{c^2}{25-c^2}=\frac{24}{16}\\ 2c^2+3c^2=75\\ 5c^2=75\\ c^2=\frac{75}{5}=\sqrt{15}\end{array}$