Explanation

(c)
$f(x) = (x + 1)^{1/3} - (x - 1)^{1/3}$
$f^{'}(x) = \frac{1}{3}(x + 1)^{- 2/3} - \frac{1}{3}(x - 1)^{- 2/3}$
Now, put $f^{'}(x) = 0$
$\Rightarrow \ \ \ \ \ \frac{1}{3}(x + 1)^{- 2/3} = \frac{1}{3}\ (x - 1)^{- 2/3}$
$\Rightarrow \ \ \ \ \ (x - 1)^{2} = (x + 1)^{2}$
$\Rightarrow \ \ \ x = 0$
At $x = 0,\ \ f''$ is negative, so it is maximum.
$\therefore$ The greatest value of
$f(x) = (0 + 1)^{1/3} - (0 - 1)^{1/3} = 2$