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

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(B) Step $1$: Find the derivative $f'(x)$ using the product rule.$f'(x) = 25x^{24}(1 - x)^{75} + x^{25} \cdot 75(1 - x)^{74}(-1)$Step $2$: Simplify $f

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$1/4$

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(B) Step $1$: Find the derivative $f'(x)$ using the product rule.$f'(x) = 25x^{24}(1 - x)^{75} + x^{25} \cdot 75(1 - x)^{74}(-1)$Step $2$: Simplify $f'(x)$.$f'(x) = 25x^{24}(1 - x)^{74} [(1 - x) - 3x] = 25x^{24}(1 - x)^{74} (1 - 4x)$Step $3$: Set $f'(x) = 0$ to find critical points.$25x^{24}(1 - x)^{74} (1 - 4x) = 0$The critical points are $x = 0$, $x = 1$, and $x = 1/4$.Step $4$: Evaluate $f(x)$ at critical points and endpoints.$f(0) = 0$, $f(1) = 0$, $f(1/4) = (1/4)^{25}(3/4)^{75} > 0$.Since $f(x) \ge 0$ on $[0, 1]$ and the value at $x = 1/4$ is positive, the maximum occurs at $x = 1/4$.

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

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