On the interval [0, 1],the function f(x) = x^ | vedclass

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(B) Let $f(x) = x^{25}(1-x)^{75}$.To find the maximum value on $[0, 1]$,we find the derivative $f'(x)$ and set it to $0$.Using the product rule: $f'(x

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$1/4$

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(B) Let $f(x) = x^{25}(1-x)^{75}$.To find the maximum value on $[0, 1]$,we find the derivative $f'(x)$ and set it to $0$.Using the product rule: $f'(x) = 25x^{24}(1-x)^{75} + x^{25} \cdot 75(1-x)^{74}(-1)$.$f'(x) = 25x^{24}(1-x)^{74} [ (1-x) - 3x ]$.$f'(x) = 25x^{24}(1-x)^{74} (1 - 4x)$.Setting $f'(x) = 0$,we get critical points at $x = 0$,$x = 1$,and $x = 1/4$.Evaluating $f(x)$ at these points:$f(0) = 0^{25}(1-0)^{75} = 0$.$f(1) = 1^{25}(1-1)^{75} = 0$.$f(1/4) = (1/4)^{25}(3/4)^{75} > 0$.Since $f(x) \geq 0$ on $[0, 1]$ and the values at the endpoints are $0$,the maximum value occurs at $x = 1/4$.

On the interval $[0, 1]$,the function $f(x) = x^{25}(1-x)^{75}$ takes its maximum value at the point

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