Question

Verify the hypothesis and conclusion of Lagrange's man value theorem for the function
f(x) = $\frac{1}{4x-1},$ 1≤ x ≤ 4.

Answer 1

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

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

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

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

Consequently, there exists some $c\in \left(1,4\right)$ such that
$f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(1\right)}{4-1}=\frac{f\left(4\right)-f\left(1\right)}{3}$

Now,
$f\left(x\right)=\frac{1}{4x-1}$$\Rightarrow$$f\text{'}\left(x\right)=\frac{-4}{{\left(4x-1\right)}^2}$, $f\left(4\right)=\frac{1}{15},f\left(1\right)=\frac{1}{3}$

$\therefore$ $f\text{'}\left(x\right)=\frac{f\left(4\right)-f\left(1\right)}{4-1}$
$$\begin{gathered} \Rightarrow f\text{'}\left(x\right)=\frac{{\displaystyle \frac{1}{15}}-{\displaystyle \frac{1}{3}}}{4-1}=\frac{-4}{45} \\ \Rightarrow \frac{-4}{{\left(4x-1\right)}^2}=\frac{-4}{45} \\ \Rightarrow {\left(4x-1\right)}^2=45 \\ \Rightarrow 16x^2-8x-44=0 \\ \Rightarrow 4x^2-2x-11=0 \\ \Rightarrow x=\frac{1}{4}\left(1\pm 3\sqrt{5}\right) \end{gathered}$$

Thus, $c=\frac{1}{4}\left(1+3\sqrt{5}\right)\in \left(1,4\right)$ such that $f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(1\right)}{4-1}$.

Hence, Lagrange's theorem is verified.