Q. In the mean value theorem $f(b) - f(a) = (b - a)f^{'}(c)$, if $a = 4,\ b = 9$ and $f(x) = \sqrt{x}$, then the value of $c$ is
A) 8.00
B) 5.25
C) 4.00
D) 6.25
Explanation
(d)
Given, $f(b) - f(a) = (b - a)f^{'}(c)$
$\Rightarrow f^{'}(c) = \frac{f(b) - f(a)}{b - a} = \frac{3 - 2}{9 - 4}$
$\Rightarrow f^{'}(c) = \frac{1}{5}$
$\Rightarrow \frac{1}{2^{\ }}c^{- 1/2} = \frac{1}{5}$
$\Rightarrow c = \left( \frac{5}{2} \right)^{2}$=6.25
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.