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

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(C) Let $f(x) = x^{25}(1-x)^{75}$.To find the critical points,we differentiate $f(x)$ with respect to $x$ using the product rule:$f'(x) = 25x^{24}(1-x

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$\frac{1}{4}$

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(C) Let $f(x) = x^{25}(1-x)^{75}$.To find the critical points,we differentiate $f(x)$ with respect to $x$ using the product rule:$f'(x) = 25x^{24}(1-x)^{75} + x^{25} \cdot 75(1-x)^{74} \cdot (-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 $x = \frac{1}{4}$.Since $f(0) = 0$ and $f(1) = 0$,and $f(x) > 0$ for $x \in (0,1)$,the function must attain its maximum at the critical point $x = \frac{1}{4}$.

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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