The function f(x) = x^2 log x in the interval | vedclass

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(D) Let $f(x) = x^2 \log x$. To find the critical points,we calculate the first derivative: $f'(x) = 2x \log x + x^2 \cdot \frac{1}{x} = 2x \log x + x

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Neither a point of maximum nor minimum

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(D) Let $f(x) = x^2 \log x$. To find the critical points,we calculate the first derivative: $f'(x) = 2x \log x + x^2 \cdot \frac{1}{x} = 2x \log x + x = x(2 \log x + 1)$. Setting $f'(x) = 0$,we get $x(2 \log x + 1) = 0$. Since $x \in (1, e)$,$x 
eq 0$. Thus,$2 \log x + 1 = 0$,which implies $\log x = -\frac{1}{2}$,or $x = e^{-1/2} = \frac{1}{\sqrt{e}}$. Since $\frac{1}{\sqrt{e}} \approx 0.606$,this value does not lie in the interval $(1, e)$. In the interval $(1, e)$,$f'(x) = x(2 \log x + 1)$ is always positive because for $x > 1$,$\log x > 0$,so $2 \log x + 1 > 1$. Since $f'(x) > 0$ for all $x \in (1, e)$,the function is strictly increasing in this interval. Therefore,the function has neither a point of maximum nor a point of minimum in the interval $(1, e)$.

The function $f(x) = x^2 \log x$ in the interval $(1, e)$ has

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