Explanation

(d)
Given, $f(x) = \frac{x}{4 + x + x^{2}}$
Let $f(x) = \frac{1}{u}$, then $u = \frac{4 + x + x^{2}}{x}$
$= \frac{4}{x} + 1 + x$
$\therefore\ \ \frac{du}{dx} = - \frac{4}{x^{2}} + 1,\ \ \ \frac{d^{2}u}{dx^{2}} = \frac{8}{x^{3}}$
For maximum or minimum, put $\frac{du}{dx} = 0$
$\Rightarrow \ \ \ \ 1 - \frac{4}{x^{2}} = 0\ \ \ \ \Rightarrow \ \ x = \pm 2$
$\therefore$ At $x = - 2,\ \frac{d^{2}u}{dx^{2}} = - \frac{8}{(2)^{3}} < 0$, maximum
At $x = - 2,\ \ \frac{d^{2}u}{dx^{2}} = 1 > 0$, minima
$\therefore$ At $x = 2,\ \ f(x)$ is maxima
And at $x = - 2,\ f(x)$ is minima
It is increasing function in the given interval
$\therefore$ The maximum value at $x = 1$ is
$f(1) = \frac{1}{4 + 1 + 1} = \frac{1}{6}$