The greatest value of the real-valued functio | vedclass

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(C) To find the greatest value of $f(x) = (x+1)^{1/3} - (x-1)^{1/3}$ on $[0, 1]$,we find the derivative $f'(x)$.$f'(x) = \frac{1}{3}(x+1)^{-2/3} - \fr

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) To find the greatest value of $f(x) = (x+1)^{1/3} - (x-1)^{1/3}$ on $[0, 1]$,we find the derivative $f'(x)$.$f'(x) = \frac{1}{3}(x+1)^{-2/3} - \frac{1}{3}(x-1)^{-2/3} = \frac{1}{3} \left[ \frac{1}{(x+1)^{2/3}} - \frac{1}{(x-1)^{2/3}} \right]$.Setting $f'(x) = 0$,we get $(x-1)^{2/3} = (x+1)^{2/3}$,which implies $|x-1| = |x+1|$.Squaring both sides,$(x-1)^2 = (x+1)^2$,so $x^2 - 2x + 1 = x^2 + 2x + 1$,which gives $4x = 0$,so $x = 0$.Now,we evaluate $f(x)$ at the critical point $x=0$ and the endpoints $x=0$ and $x=1$.At $x=0$,$f(0) = (0+1)^{1/3} - (0-1)^{1/3} = 1 - (-1) = 2$.At $x=1$,$f(1) = (1+1)^{1/3} - (1-1)^{1/3} = 2^{1/3} - 0 = 2^{1/3}$.Comparing $f(0) = 2$ and $f(1) = 2^{1/3} \approx 1.26$,the greatest value is $2$.

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

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