In the interval (0, 1),the maximum value of t | vedclass

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(D) For $x \in (0, 1)$,$\ln x < 0$,so $x \ln x < 0$. Thus,$f(x) = |x \ln x| = -x \ln x$. To find the critical points,we calculate the derivative: $f'(

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$e^{-1}$

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(D) For $x \in (0, 1)$,$\ln x Thus,$f(x) = |x \ln x| = -x \ln x$. To find the critical points,we calculate the derivative: $f'(x) = -(1 \cdot \ln x + x \cdot \frac{1}{x}) = -(1 + \ln x)$. Setting $f'(x) = 0$,we get $1 + \ln x = 0$,which implies $\ln x = -1$,so $x = e^{-1} = \frac{1}{e}$. Since $f'(x) > 0$ for $x  \frac{1}{e}$,the function has a local maximum at $x = \frac{1}{e}$. The maximum value is $f(\frac{1}{e}) = -(\frac{1}{e}) \ln(\frac{1}{e}) = -(\frac{1}{e})(-1) = \frac{1}{e} = e^{-1}$.

In the interval $(0, 1)$,the maximum value of the function $f(x) = |x \ln x|$ is:

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