The function f(x) = x^x for x 0 is strictly | vedclass

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(C) Let $f(x) = x^x$. Taking the natural logarithm on both sides,we get $\ln f(x) = x \ln x$. Differentiating both sides with respect to $x$,we have $

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$x > \frac{1}{e}$

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(C) Let $f(x) = x^x$. Taking the natural logarithm on both sides,we get $\ln f(x) = x \ln x$. Differentiating both sides with respect to $x$,we have $\frac{1}{f(x)} f'(x) = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$. Thus,$f'(x) = x^x(1 + \ln x)$. For the function to be strictly increasing,we require $f'(x) > 0$. Since $x > 0$,$x^x$ is always positive. Therefore,we must have $1 + \ln x > 0$,which implies $\ln x > -1$. This is equivalent to $\ln x > \ln(\frac{1}{e})$. Hence,$x > \frac{1}{e}$.

The function $f(x) = x^x$ for $x > 0$ is strictly increasing at

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