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

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(D) Let $f(x) = x^{25}(1-x)^{75}$.To find the maximum,we calculate the derivative $f'(x)$ using the product rule:$f'(x) = 25x^{24}(1-x)^{75} - 75x^{25

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$1/4$

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(D) Let $f(x) = x^{25}(1-x)^{75}$.To find the maximum,we calculate the derivative $f'(x)$ using the product rule:$f'(x) = 25x^{24}(1-x)^{75} - 75x^{25}(1-x)^{74}$$f'(x) = 25x^{24}(1-x)^{74} [ (1-x) - 3x ]$$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$$f(1) = 0$$f(1/4) = (1/4)^{25}(3/4)^{75} > 0$Since $f(x) > 0$ for $x \in (0, 1)$,the function attains its maximum 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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