Question

The value of $c$ from the Lagrange�s mean value theorem for which $f(x) = \sqrt{25 - x^2}$ in $[1,5]$, is

• A. $5$
• B. $1$
• C. $\sqrt{15}$
• D. None of these

Answer 1

Answer: (C)

$\sqrt{15}$

Answer 2

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