The greatest value of
$f (x) = (x + 1)^{1 / 3} - (x - 1)^{1 / 3}$ on $[0, 1]$ is

1

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2

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3

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\frac{1}{3}

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Solution

The correct option is C $2$
We have, $f (x) = (x + 1)^{1 / 3} - (x - 1)^{1 / 3}$
$∴ f^{'} (x) = \frac{1}{3} [\frac{1}{(x + 1)^{2 / 3}} - \frac{1}{(x - 1)^{2 / 3}}]$
$= \frac{(x - 1)^{2 / 3} - (x + 1)^{2 / 3}}{3 (x^{2} - 1)^{2 / 3}}$
Clearly, $f^{'} (x)$ does exist at $x = \pm 1$
$∴ f^{'} (x) = 0$
$\Rightarrow (x - 1)^{2 / 3} = (x + 1)^{2 / 3} \Rightarrow x = 0$
Clearly, $f^{'} (x) \neq 0$ for any other value of $x ϵ [0, 1]$ . The value of $f (x)$ at $x = 0$ is $2$ .
Hence, the greatest value of $f (x)$ is $2$ .

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