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

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

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$1/4$

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(B) Given,$f(x) = x^{25}(1-x)^{75}$.To find the maximum value,we calculate the derivative $f'(x)$ using the product rule:$f'(x) = 25x^{24}(1-x)^{75} - 75x^{25}(1-x)^{74}$.Factor out the common terms:$f'(x) = 25x^{24}(1-x)^{74} [ (1-x) - 3x ]$.$f'(x) = 25x^{24}(1-x)^{74} (1-4x)$.Setting $f'(x) = 0$ gives critical points $x = 0, 1, 1/4$.For $x \in (0, 1/4)$,$f'(x) > 0$ (function is increasing).For $x \in (1/4, 1)$,$f'(x) Since the derivative changes sign from positive to negative at $x = 1/4$,the function attains its maximum value 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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