The maximum value of [x(x-1)+1]^{frac{1}{3}} | vedclass

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

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$1$

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(A) Let $f(x) = [x(x-1)+1]^{\frac{1}{3}} = [x^2-x+1]^{\frac{1}{3}}$.To find the critical points,we calculate the derivative $f'(x)$:$f'(x) = \frac{1}{3}[x^2-x+1]^{-\frac{2}{3}} \cdot (2x-1) = \frac{2x-1}{3(x^2-x+1)^{\frac{2}{3}}}$.Setting $f'(x) = 0$,we get $2x-1 = 0$,which implies $x = \frac{1}{2}$.Since $x = \frac{1}{2}$ lies within the interval $[0, 1]$,we evaluate $f(x)$ at the critical point and the endpoints:$f(0) = [0(0-1)+1]^{\frac{1}{3}} = 1^{\frac{1}{3}} = 1$.$f(1) = [1(1-1)+1]^{\frac{1}{3}} = 1^{\frac{1}{3}} = 1$.$f(\frac{1}{2}) = [\frac{1}{2}(\frac{1}{2}-1)+1]^{\frac{1}{3}} = [\frac{1}{2}(-\frac{1}{2})+1]^{\frac{1}{3}} = [-\frac{1}{4}+1]^{\frac{1}{3}} = (\frac{3}{4})^{\frac{1}{3}}$.Comparing the values $1$,$1$,and $(\frac{3}{4})^{\frac{1}{3}}$,the maximum value is $1$.Thus,the correct answer is $A$.

The maximum value of $[x(x-1)+1]^{\frac{1}{3}}$ for $0 \leq x \leq 1$ is

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