Question

The value of $C$ in Mean value theorem for the function $f(x)=x^2$ in $[2,4]$ is

• A. 3
• B. 2
• C. 4
• D. 7/2

Answer 1

Answer: (A)

For mean value theorem
$f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$
$\therefore f(x)=x^2$
$\Rightarrow f^{\prime }(x)=2x$
$\Rightarrow 2c=\frac{f(4)-f(2)}{4-2}$
$\Rightarrow 2c=\frac{4^2-2^2}{4-2}$
$\Rightarrow 2c=4+2$
$\Rightarrow 2c=6$
$c=3$