The maximum value of f(x) = (x + 1)^{frac{1}{ | vedclass

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(B) Given $f(x) = (x + 1)^{\frac{1}{3}} - (x - 1)^{\frac{1}{3}}$.First,find the derivative $f'(x)$:$f'(x) = \frac{1}{3}(x + 1)^{-\frac{2}{3}} - \frac{

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$2$

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(B) Given $f(x) = (x + 1)^{\frac{1}{3}} - (x - 1)^{\frac{1}{3}}$.First,find the derivative $f'(x)$:$f'(x) = \frac{1}{3}(x + 1)^{-\frac{2}{3}} - \frac{1}{3}(x - 1)^{-\frac{2}{3}} = \frac{1}{3} \left[ \frac{1}{(x + 1)^{\frac{2}{3}}} - \frac{1}{(x - 1)^{\frac{2}{3}}} \right]$.For critical points,set $f'(x) = 0$:$\frac{1}{(x + 1)^{\frac{2}{3}}} = \frac{1}{(x - 1)^{\frac{2}{3}}} \implies (x + 1)^2 = (x - 1)^2$.$x^2 + 2x + 1 = x^2 - 2x + 1 \implies 4x = 0 \implies x = 0$.Now,evaluate $f(x)$ at the critical point and the endpoints of the interval $[0, 1]$:At $x = 0$: $f(0) = (0 + 1)^{\frac{1}{3}} - (0 - 1)^{\frac{1}{3}} = 1 - (-1) = 2$.At $x = 1$: $f(1) = (1 + 1)^{\frac{1}{3}} - (1 - 1)^{\frac{1}{3}} = 2^{\frac{1}{3}} - 0 = \sqrt[3]{2} \approx 1.26$.Comparing the values $f(0) = 2$ and $f(1) = \sqrt[3]{2}$,the maximum value is $2$.

The maximum value of $f(x) = (x + 1)^{\frac{1}{3}} - (x - 1)^{\frac{1}{3}}$ for $x \in [0, 1]$ is ....

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