$f(x) = \frac{1}{{4x - 1}}$ in [1,4]

We have, $f(x) = \frac{1}{{4x - 1}}$ in [1,4]

For mean value theorem follow the following algorithm :

If $f$ be a real function such that

(i) $f(x)$ is continuous on [a, b]

(ii) $f(x)$ is differentiable on ]a, b[

Then, there exists a real number $c \in ]a,b[$such that ${f^\prime }(c) = \frac{{f(b) - f(a)}}{{b - a}}$, thus we can verify it for given function.

(i) $f(x)$ is continuous in [1,4] .

Also, at $x = \frac{1}{4},f(x)$ is discontinuous.

Hence, $f(x)$ is continuous in [1,4] .

(ii) ${f^\prime }(x) = - \frac{4}{{{{(4x - 1)}^2}}},$ which exists in (1,4) .

Since, conditions of mean value theorem are satisfied.

Hence, there exists a real number $c \in ]1,4[$ such that
${f^\prime }(c) = \frac{{f(4) - f(1)}}{{4 - 1}}$

$\Rightarrow$ $\frac{{ - 4}}{{{{(4c - 1)}^2}}} = \frac{{\frac{1}{{16 - 1}} - \frac{1}{{4 - 1}}}}{{4 - 1}} = \frac{{\frac{1}{{15}} - \frac{1}{3}}}{3}$

$\Rightarrow$ $\frac{{ - 4}}{{{{(4c - 1)}^2}}} = \frac{{1 - 5}}{{45}} = \frac{{ - 4}}{{45}}$

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

$\Rightarrow$ $4c - 1 = \pm 3\sqrt 5$

$\Rightarrow$ $c = \frac{{3\sqrt 5 + 1}}{4} \in (1,4)$ [neglecting (-ve) value]

Hence, mean value theorem has been verified.