Question

If $f\left(x\right) = \frac{x^{2} -1}{x^{2} +1} ,x\in R$ then the minimum value of $f$ is

• A. 0
• B. 44656
• C. 44625
• D. -1

Answer 1

The Correct Option is D

We have, $f\left(x\right) = \frac{x^{2} -1}{x^{2} +1}$
Maximum or minimum value point is given by $f'(x) = 0$
$f'\left(x\right) =\frac{\left(x^{2}+1\right)2x-\left(x^{2}-1\right)2x}{\left(x^{2}+1\right)^{2}} =0 $
$\Rightarrow2x\left(x^{2}+1-x^{2}+1\right)=0 $
$ \Rightarrow 4x =0 \Rightarrow x=0 $
Now, $f''\left(x\right) = \frac{d}{dx} \left(\frac{2x^{3} +2x -2x^{3} +2x}{\left(x^{2} +1\right)^{2}} \right) $
$ =\frac{d}{dx}\left(\frac{4x}{\left(x^{2}+1\right)^{2}}\right) $
$ f''\left(x\right)= \frac{\left(x^{2}+1\right)^{2} 4 -4x\times2\left(x^{2}+1\right)\times 2x}{\left(x^{2}+1\right)^{4}} $
$= \frac{4\left(x^{2}+1\right)^{2}-16x^{2}\left(x^{2}+1\right)}{\left(x^{2}+1\right)^{4}} $
$= \frac{4\left(x^{2}+1\right)\left(x^{2}+1 -4x^{2}\right)}{\left(x^{2}+1\right)^{4}} =\frac{4\left(-3x^{2}+1\right)}{\left(x^{2}+1\right)^{3}}$
Now $f''(0) = \frac{4}{1} > 0$
$ \therefore \:\: x = 0$ is a minimum point and minimum value of $f$ is given by $ f(0) = \frac{0-1}{0+1} = - 1$