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

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(D) Let $f(x) = {x^{25}}{(1 - x)^{75}}$.To find the maximum value,we differentiate $f(x)$ with respect to $x$:$f'(x) = {x^{25}} \cdot 75{(1 - x)^{74}}

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$1/4$

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(D) Let $f(x) = {x^{25}}{(1 - x)^{75}}$.To find the maximum value,we differentiate $f(x)$ with respect to $x$:$f'(x) = {x^{25}} \cdot 75{(1 - x)^{74}} \cdot (-1) + 25{x^{24}} \cdot {(1 - x)^{75}}$$f'(x) = 25{x^{24}}{(1 - x)^{74}} [ -3x + (1 - x) ]$$f'(x) = 25{x^{24}}{(1 - x)^{74}} (1 - 4x)$For critical points,set $f'(x) = 0$:$25{x^{24}}{(1 - x)^{74}} (1 - 4x) = 0$This gives $x = 0, x = 1,$ or $x = 1/4$.Since $f(0) = 0$ and $f(1) = 0$,and $f(x) > 0$ for $x \in (0, 1)$,the maximum must occur at the critical point $x = 1/4$.Thus,the function takes 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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