The function f is defined by f(x) = x^p (1 - | vedclass

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(D) Given the function $f(x) = x^p (1 - x)^q$. To find the critical points,we find the first derivative $f'(x)$ and set it to $0$. Using the product r

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$\frac{p}{p+q}$

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(D) Given the function $f(x) = x^p (1 - x)^q$. To find the critical points,we find the first derivative $f'(x)$ and set it to $0$. Using the product rule: $f'(x) = p x^{p-1} (1 - x)^q + x^p \cdot q (1 - x)^{q-1} (-1)$. $f'(x) = x^{p-1} (1 - x)^{q-1} [p(1 - x) - qx]$. $f'(x) = x^{p-1} (1 - x)^{q-1} [p - px - qx]$. $f'(x) = x^{p-1} (1 - x)^{q-1} [p - (p + q)x]$. Setting $f'(x) = 0$,we get critical points at $x = 0$,$x = 1$,and $x = \frac{p}{p+q}$. Since $p, q$ are positive integers,for $x$ in the interval $(0, 1)$,the sign of $f'(x)$ changes from positive to negative at $x = \frac{p}{p+q}$. Therefore,the function has a maximum value at $x = \frac{p}{p+q}$.

The function $f$ is defined by $f(x) = x^p (1 - x)^q$ for all $x \in R$,where $p, q$ are positive integers. The function has a maximum value for $x$ equal to:

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