The minimum value of the function f(x) = x lo | vedclass

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(D) Given the function $f(x) = x \log x$. To find the minimum value,we first find the derivative $f'(x)$. Using the product rule: $f'(x) = x \cdot \fr

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$-\frac{1}{e}$

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(D) Given the function $f(x) = x \log x$. To find the minimum value,we first find the derivative $f'(x)$. Using the product rule: $f'(x) = x \cdot \frac{d}{dx}(\log x) + \log x \cdot \frac{d}{dx}(x) = x \cdot \frac{1}{x} + \log x = 1 + \log x$. For critical points,set $f'(x) = 0$: $1 + \log x = 0 \implies \log x = -1 \implies x = e^{-1} = \frac{1}{e}$. Now,find the second derivative $f''(x)$: $f''(x) = \frac{d}{dx}(1 + \log x) = \frac{1}{x}$. Evaluate $f''(x)$ at $x = \frac{1}{e}$: $f''(\frac{1}{e}) = \frac{1}{1/e} = e$. Since $e > 0$,the function has a local minimum at $x = \frac{1}{e}$. The minimum value is $f(\frac{1}{e}) = \frac{1}{e} \log(\frac{1}{e}) = \frac{1}{e} \log(e^{-1}) = \frac{1}{e} (-1) = -\frac{1}{e}$.

The minimum value of the function $f(x) = x \log x$ is

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