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

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(A) Given function: $f(x) = x \log x$ To find the minimum value,we first find the derivative $f'(x)$: $f'(x) = x \cdot \frac{1}{x} + \log x \cdot 1 =

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$-\frac{1}{e}$

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(A) Given function: $f(x) = x \log x$ To find the minimum value,we first find the derivative $f'(x)$: $f'(x) = x \cdot \frac{1}{x} + \log x \cdot 1 = 1 + \log x$ Set $f'(x) = 0$ for critical points: $1 + \log x = 0 \Rightarrow \log x = -1 \Rightarrow x = e^{-1} = \frac{1}{e}$ Now,find the second derivative $f''(x)$: $f''(x) = \frac{d}{dx}(1 + \log x) = \frac{1}{x}$ Evaluate $f''(x)$ at $x = \frac{1}{e}$: $f''(\frac{1}{e}) = \frac{1}{1/e} = e > 0$ Since $f''(\frac{1}{e}) > 0$,the function has a 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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