Question

Verify Rolle's theorem for the function:
$f\left(x\right)=\log \left\{\frac{x^2+ab}{(a+b)x}\right\}$ in the interval $[a,b]$ where, $0\notin [a,b]$.

Answer 1

Given: $f\left(x\right)=\log \left(\frac{x^2+ab}{(a+b)x}\right)$
$=\log (x^2+ab)-\log (a+b)-\log x$ ............ (i)
Since, $a>0$ and $\log x$ is continuous for all $x>0$, therefore $f\left(x\right)$ is continuous in $[a,b].$
$f\left(x\right)$ is differentiable in $(a,b)$, and
$f\left(a\right)=\log \left(\frac{a^2+ab}{(a+b)a}\right)=\log 1=0$
$f\left(b\right)=\log \left(\frac{b^2+ab}{(a+b)b}\right)=\log 1=0$
$\Rightarrow f\left(a\right)=f\left(b\right)$.
Thus, all the three conditions of Rolle's theorem are satisfied. Therefore, there exists at least one real number $c$ in $(a,b).$ such that $f^{\prime }\left(c\right)=0$.
Differentiating (i) w.r.t. $x$, we get
$f^{\prime }\left(x\right)=\frac{1}{x^2+ab}.2x-0-\frac{1}{x}$
$=\frac{2x}{x^2+ab}-\frac{1}{x}$
$=\frac{x^2-ab}{x(x^2+ab)}$

Now,

$f^{\prime }\left(c\right)=0$
$\Rightarrow \frac{c^2-ab}{c(c^2+ab)}=0$
$\Rightarrow c^2-ab=0$
$\Rightarrow c^2=ab$
$\Rightarrow c=\pm \sqrt{ab}$
$0\notin [a,b]\Rightarrow -ab\notin [a,b]$
$\therefore c=\sqrt{ab}$
So, there exists a real number

$c=\sqrt{ab}$ in (a, b) such that
$f^{\prime }\left(c\right)=0$
Hence, Rolle's theorem is verified and $c=\sqrt{ab}$.