On the interval [1, e],the greatest value of | vedclass

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(A) Let $f(x) = x^2 \log x$.To find the critical points,we find the derivative $f'(x)$:$f'(x) = \frac{d}{dx}(x^2) \cdot \log x + x^2 \cdot \frac{d}{dx

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$e^2$

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(A) Let $f(x) = x^2 \log x$.To find the critical points,we find the derivative $f'(x)$:$f'(x) = \frac{d}{dx}(x^2) \cdot \log x + x^2 \cdot \frac{d}{dx}(\log x) = 2x \log x + x^2 \cdot \frac{1}{x} = 2x \log x + x = x(2 \log x + 1)$.Setting $f'(x) = 0$,we get $x(2 \log x + 1) = 0$.Since $x \in [1, e]$,$x 
eq 0$. Thus,$2 \log x + 1 = 0 \implies \log x = -\frac{1}{2} \implies x = e^{-1/2} = \frac{1}{\sqrt{e}}$.Since $\frac{1}{\sqrt{e}} \approx 0.606$,which is not in the interval $[1, e]$,there are no critical points in the given interval.Therefore,the maximum value must occur at the endpoints $x = 1$ or $x = e$.$f(1) = 1^2 \cdot \log(1) = 1 \cdot 0 = 0$.$f(e) = e^2 \cdot \log(e) = e^2 \cdot 1 = e^2$.Comparing the values $0$ and $e^2$,the greatest value is $e^2$.

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

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