Explanation

(b)
Let the number be $x$, then $f(x) = \frac{x}{x^{2} + 16}$
On differentiating w.r.t. $x$, we get
$f^{'}(x) = \frac{\left( x^{2} + 16 \right).1 - x(2x)}{\left( x^{2} + 16 \right)^{2}}$
$= \frac{x^{2} + 16 - 2x^{2}}{\left( x^{2} + 16 \right)^{2}} = \frac{16 - x^{2}\ }{\left( x^{2} + 16 \right)^{2}}$ …(i)
Put $f^{'}(x) = 0$ for maxima or minima
$f^{'}(x) = 0\ \ \Rightarrow \ \ 16 - x^{2} = 0\ \ \Rightarrow \ \ x = 4,\ - 4$
Again, on differentiating w.r.t. $x$, we get
$f^{''}(x) = \frac{\left( x^{2} + 16 \right)^{2} - ( - 2x) - \left( 16 - x^{2} \right)2(x^{2} + 16)2x}{\left( x^{2} + 16 \right)^{4}}$
At $x = 4,\ f^{''}(x) < 0$
$\therefore\ \ f(x)$ is maximum at $x = 4$
And at $x = - 4,\ f^{''}(x) > 0\ f(x)$ is minimum
$\therefore$ Least value of $f(x) = \frac{- 4}{16 + 16} = - \frac{1}{8}$