Maximum value of the function f(x) = [x(x-1) | vedclass

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(B) To find the maximum value of $f(x) = [x(x-1) + 1]^{\frac{1}{3}}$ on the interval $[0, 1]$,we first find the derivative $f'(x)$.Let $g(x) = x(x-1)

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$1$

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(B) To find the maximum value of $f(x) = [x(x-1) + 1]^{\frac{1}{3}}$ on the interval $[0, 1]$,we first find the derivative $f'(x)$.Let $g(x) = x(x-1) + 1 = x^2 - x + 1$.Then $f(x) = [g(x)]^{\frac{1}{3}}$.$f'(x) = \frac{1}{3} [g(x)]^{-\frac{2}{3}} \cdot g'(x) = \frac{1}{3} [x^2 - x + 1]^{-\frac{2}{3}} (2x - 1)$.Setting $f'(x) = 0$,we get $2x - 1 = 0$,which implies $x = \frac{1}{2}$.Now,we evaluate $f(x)$ at the critical point and the endpoints:$f(0) = [0(0-1) + 1]^{\frac{1}{3}} = 1^{\frac{1}{3}} = 1$.$f(1) = [1(1-1) + 1]^{\frac{1}{3}} = 1^{\frac{1}{3}} = 1$.$f(\frac{1}{2}) = [\frac{1}{2}(\frac{1}{2}-1) + 1]^{\frac{1}{3}} = [-\frac{1}{4} + 1]^{\frac{1}{3}} = (\frac{3}{4})^{\frac{1}{3}}$.Comparing the values $1$,$1$,and $(\frac{3}{4})^{\frac{1}{3}}$,the maximum value is $1$.

Maximum value of the function $f(x) = [x(x-1) + 1]^{\frac{1}{3}}$ for $0 \leq x \leq 1$ is . . . . . . .

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