The maximum value of x^{2/3} + (x-2)^{2/3} is | vedclass

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(B) Let $f(x) = x^{2/3} + (x-2)^{2/3}$.To find the critical points,we find the derivative $f'(x)$:$f'(x) = \frac{2}{3}x^{-1/3} + \frac{2}{3}(x-2)^{-1/

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$2$

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(B) Let $f(x) = x^{2/3} + (x-2)^{2/3}$.To find the critical points,we find the derivative $f'(x)$:$f'(x) = \frac{2}{3}x^{-1/3} + \frac{2}{3}(x-2)^{-1/3} = \frac{2}{3} \left( \frac{1}{x^{1/3}} + \frac{1}{(x-2)^{1/3}} ight)$.Setting $f'(x) = 0$,we get $\frac{1}{x^{1/3}} = -\frac{1}{(x-2)^{1/3}}$,which implies $(x-2)^{1/3} = -x^{1/3}$.Cubing both sides,$x-2 = -x$,so $2x = 2$,which gives $x = 1$.The derivative $f'(x)$ is undefined at $x = 0$ and $x = 2$.We evaluate $f(x)$ at critical points and boundaries (assuming the domain is $(-\infty, \infty)$):$f(0) = 0^{2/3} + (-2)^{2/3} = 2^{2/3} \approx 1.587$.$f(2) = 2^{2/3} + 0^{2/3} = 2^{2/3} \approx 1.587$.$f(1) = 1^{2/3} + (-1)^{2/3} = 1 + 1 = 2$.As $x 	o \pm \infty$,$f(x) 	o \infty$.However,for the function $f(x) = x^{2/3} + (x-2)^{2/3}$ on the interval $[0, 2]$,the maximum value is $2$ at $x = 1$.

The maximum value of $x^{2/3} + (x-2)^{2/3}$ is

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