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(A) Given function is $f(x) = x^2 \log x$ on the interval $[1, e]$.First,find the derivative $f'(x)$:$f'(x) = x^2 \cdot \frac{1}{x} + (\log x) \cdot 2

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$e^2$

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(A) Given function is $f(x) = x^2 \log x$ on the interval $[1, e]$.First,find the derivative $f'(x)$:$f'(x) = x^2 \cdot \frac{1}{x} + (\log x) \cdot 2x = x + 2x \log x = x(1 + 2 \log x)$.For $x \in [1, e]$,$x$ is always positive and $\log x \ge 0$.Since $1 + 2 \log x > 0$ for all $x \in [1, e]$,$f'(x) > 0$.Thus,$f(x)$ is a strictly increasing function on the interval $[1, e]$.The maximum value occurs at the right endpoint of the interval,which is $x = e$.$f(e) = e^2 \log e = e^2 \cdot 1 = e^2$.

The maximum value of $f(x) = x^2 \log x$ on the interval $[1, e]$ is:

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