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

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Solution

The correct option is C 1
$\begin{matrix} f (x) = {(x + 1)}^{1 / 3} - {(x - 1)}^{1 / 3} d i f f e r e n t i a t e w . r . t . x \Rightarrow f^{'} (x) = \frac{1}{3} {(x + 1)}^{- 2 / 3} - \frac{1}{3} {(x - 1)}^{- 2 / 3} \Rightarrow f^{'} (x) \Rightarrow \frac{1}{3} [{(x + 1)}^{- 2 / 3} - {(x - 1)}^{- 2 / 3}] = 0 \Rightarrow {(x + 1)}^{- 2 / 3} = {(x - 1)}^{- 2 / 3} \Rightarrow {(x - 1)}^{2 / 3} = {(x + 1)}^{2 / 3} x = 0 \Rightarrow f^{''} (x) = \frac{1}{3} [- \frac{2}{3} {(x + 1)}^{- 5 / 3} + \frac{2}{3} {(x - 1)}^{- 5 / 3}] p u t x = 0 \Rightarrow f (0) = {(0 + 1)}^{1 / 3} - {(0 - 1)}^{1 / 3} = 1 - {(- 1)}^{1 / 3} G r e a t e s t v a l u e s i s 1 \end{matrix}$

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