Question

Testify the mean value theorem in the interval $[a,b]$, $f\left(x\right)=\frac{1}{4x-1}$ where $a=1$ and $b=4$.

Answer 1

$f\left(x\right)$ is non-differentiable at $4x-1=0⟹x=\frac{1}{4}$ which does not lie in $[1,4]$.

Hence, $f\left(x\right)$ is differentiable and continuous in given domain.

Also, $f\left(a\right)=\frac{1}{4-1}=\frac{1}{3}$ and $f\left(b\right)=\frac{1}{16-1}=\frac{1}{15}$

Now, consider $\frac{f\left(b\right)-f\left(a\right)}{b-a}=\frac{\frac{1}{15}-\frac{1}{3}}{4-1}$

$\therefore \frac{f\left(b\right)-f\left(a\right)}{b-a}=-\frac{\frac{4}{15}}{3}$

$\therefore \frac{f\left(b\right)-f\left(a\right)}{b-a}=-\frac{4}{45}$

Also, $f^{\prime }\left(x\right)=(-1)\times \frac{1}{(4x-1{)}^2}\times 4=-\frac{4}{(4x-1{)}^2}$

Now, according to Mean Value Theorem, there should be a $c$ such that $f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

$\therefore -\frac{4}{(4c-1{)}^2}=-\frac{4}{45}$

$\therefore \frac{1}{(4c-1{)}^2}=\frac{1}{45}$

$\therefore (4c-1{)}^2=45$

$\therefore c=\frac{1+\sqrt{45}}{4}$

Now, $\sqrt{36}=6$ and $\sqrt{49}=7$.

$\therefore \frac{1+\sqrt{36}}{4}\le \frac{1+\sqrt{45}}{4}\le \frac{1+\sqrt{49}}{4}$

$\therefore \frac{7}{4}\le \frac{1+\sqrt{45}}{4}\le \frac{8}{4}$

$\therefore \frac{7}{4}\le \frac{1+\sqrt{45}}{4}\le 2$

$\therefore \frac{7}{4}\le c\le 2$

$\therefore c\in (1,4)$

Hence, the Mean Value Theorem is satisfied.