Question

Is Rolle's theorem applicable to the function $f\left(x\right)=\log \left[\frac{(x^2+ab)}{(a+b)x}\right]$in $(a,b)?$

Answer 1

We have $f\left(x\right)=log\left\{\frac{x^2+ab}{(a+b)x}\right\}.$
$f\left(a\right)=log\left\{\frac{a^2+ab}{(a+b)a}\right\}=log\left(1\right)=0$ and $f\left(b\right)=log\left\{\frac{b^2+ab}{(a+b)b}\right\}$
$=log\left(1\right)=0\therefore f\left(a\right)=f\left(b\right)=0$
Also $R{f}^{{}^{\prime }}\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$
$=\underset{h\to 0}{lim}\frac{1}{h}[\log \left\{\frac{{(x+h)}^2+ab}{(a+b)(x+h)}\right\}-\log \left\{\frac{x^3+ab}{(a+b)x}\right\}]$
$=\underset{h\to 0}{lim}\frac{1}{h}\left[log\frac{(x^2+2xh+h^2+ab)(a+b)x}{(a+b)(x+h)(x^2+ab)}\right]$
$=\underset{h\to 0}{lim}\frac{1}{h}\left[log\{\frac{(x^2+2h+h^2+ab)}{(x^2+ab)}\times \frac{x}{x+h}\}\right]$
$=\underset{h\to 0}{lim}\frac{1}{h}\left[log\{1+\frac{2hx+h^2}{x^2+ab}-log\{1+\frac{h}{x}\}\}\right]$
$=\underset{h\to 0}{lim}\frac{1}{h}[\frac{2hx}{x^2+ab}-\frac{h}{x}+\cdot \cdot \cdot ]$
$[\because log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}\cdot \cdot \cdot ]$ $=\left[\frac{2x}{(x^2+ab)}\right]-\left(\frac{1}{x}\right).$
Also $L{f}^{\prime }\left(x\right)=\underset{h\to 0}{lim}\left\{\frac{f(x-h)-f\left(x\right)}{-h}\right\}=\underset{h\to 0}{lim}\frac{1}{(-h)}\{\frac{-2hx}{x^2+ab}-\frac{(-h)}{x}+...\}$
$\therefore L{f}^{\prime }\left(x\right)=R{f}^{\prime }\left(x\right)$
Thus $\therefore L{f}^{\prime }\left(x\right)=R{f}^{\prime }\left(x\right).$
$\therefore f^{\prime }\left(x\right)$ exists for all values of x in (a, b).
Also f(x) is continuous for all values of x in (a, b) since f(x) is differentiable for all values of x in (a, b).
Hence, all the three conditions of Rolle's theorem are satisfied.
$\left[\frac{2x}{(x^2+ab)}\right]-\left(\frac{1}{x}\right)=0$
or $2x^2-(x^2+ab)=0$ or $x^2=ab$
or $\therefore f\left(x\right)=0$ for at least one value of x where a< x< b. Equating f'(x) to zero, we have $x=\sqrt{ab}$
which clearly lies in (a, b) as $\sqrt{ab}$ is the G.M. of a and b.