Verify the hypothesis and conclusion of Lagrange's man value theorem for the function

Solution:

The given function is $f(x)=\frac{1}{4 x-1}$.

Since for each $x \in[1,4]$, the function attains a unique definite value, $f(x)$ is continuous on $[1,4]$.

Also, $f^{\prime}(x)=\frac{-4}{(4 x-1)^{2}}$ exists for all $x \in[1,4]$

Thus, both the conditions of Lagrange's mean value theorem are satisfied.

Consequently, there exists some $c \in(1,4)$ such that

$f^{\prime}(c)=\frac{f(4)-f(1)}{4-1}=\frac{f(4)-f(1)}{3}$

Now,

$f(x)=\frac{1}{4 x-1} \Rightarrow f^{\prime}(x)=\frac{-4}{(4 x-1)^{2}}, f(4)=\frac{1}{15}, f(1)=\frac{1}{3}$

$\therefore f^{\prime}(x)=\frac{f(4)-f(1)}{4-1}$

$\Rightarrow f^{\prime}(x)=\frac{\frac{1}{15}-\frac{1}{3}}{4-1}=\frac{-4}{45}$

$\Rightarrow \frac{-4}{(4 x-1)^{2}}=\frac{-4}{45}$

$\Rightarrow(4 x-1)^{2}=45$

$\Rightarrow 16 x^{2}-8 x-44=0$

$\Rightarrow 4 x^{2}-2 x-11=0$

 

$\Rightarrow x=\frac{1}{4}(1 \pm 3 \sqrt{5})$

Thus, $c=\frac{1}{4}(1+3 \sqrt{5}) \in(1,4)$ such that $f^{\prime}(c)=\frac{f(4)-f(1)}{4-1}$.

Hence, Lagrange's theorem is verified.