Question

Verify Mean Value Theorem for the function $f\left(x\right)=x^2$ in the interval $[2,4]$.

Answer 1

Verify conditions
Given: $f\left(x\right)=x^2$ in the interval $[2,4]$
Mean Value Theorem is satisfied if following conditions are met.
$\left(i\right)f\left(x\right)$ is continuous in $[a,b];f\left(x\right)=x^2$ is a polynomial of degree 'two'. so, $f\left(x\right)$ is continuous in $[2,4]$

(ii) $f\left(x\right)$is differentiable in $(a,b);f\left(x\right)=x^2$ is a polynomial of degree 'two'. so, $f\left(x\right)$ is differentiable in $(2,4)$
Hence function is satisfying the conditions of Mean Value Theorem.

Finding ${}^{\prime }{c}^{\prime }$
Let ${}^{\prime }{c}^{\prime }$ be any point in the interval $[2,4]$ such that
$f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$
$f\left(x\right)=x^2$
$f^{\prime }\left(x\right)=2x$
Putting $x=c,f^{\prime }\left(c\right)=2c$
$f^{\prime }\left(c\right)=\frac{f\left(4\right)-f\left(2\right)}{4-2}$
$2c=\frac{4^2-2^2}{2}$
$2c=\frac{16-4}{2}$
$2c=\frac{12}{2}$
$2c=6$
$c=3$
Hence $c=3\in (2,4)$
Thus, Mean Value Theorem is verified.