Explanation

(d)
$f(x) = \frac{x^{2} - 1}{x^{2} + 1}$ (for every real number $x$)
$f^{'}(x) = \frac{\left( x^{2} + 1 \right)(2x) - (x^{2} - 1)2x}{\left( x^{2} + 1 \right)^{2}}$
$\Rightarrow \ \ \ \ \ \ \ f^{'}(x) = \frac{4x}{\left( x^{2} + 1 \right)^{2}}$ ...(i)
For minima or maxima, put $f^{'}(x) = 0$
$\therefore\ \ \ \frac{4x}{\left( x^{2} + 1 \right)^{2}} = 0\ \ \ \ \ \ \Rightarrow \ \ \ \ \ \ \ \ x = 0$
Now, $f^{''}(x) = \frac{\left( x^{2} + 1 \right)^{2}4 - (4x)2(x^{2} + 1)2x}{\left( x^{2} + 1 \right)^{4}}$
$f^{''}(x) = \frac{12x^{2} + 4}{\left( x^{2} + 1 \right)^{3}}$
At $x = 0$
$f^{''}(0) = \frac{- 12 \times 0^{2} + 4}{\left( 0^{2} + 1 \right)^{3}} > 0$, minima
Then, minimum value of $f(x)$ at $x = 0$ is
$f(0) = \frac{0^{2} - 1}{0^{2} + 1} = - 1$