The function f(x) = x^{1/x} is | vedclass

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(C) Let $y = x^{1/x}$.Taking the natural logarithm on both sides,we get $\ln y = \frac{1}{x} \ln x$.Differentiating with respect to $x$ using the prod

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Increasing in $(1, e)$,decreasing in $(e, \infty)$

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(C) Let $y = x^{1/x}$.Taking the natural logarithm on both sides,we get $\ln y = \frac{1}{x} \ln x$.Differentiating with respect to $x$ using the product rule:$\frac{1}{y} \frac{dy}{dx} = \left(-\frac{1}{x^2}\right) \ln x + \frac{1}{x} \left(\frac{1}{x}\right) = \frac{1 - \ln x}{x^2}$.Thus,$\frac{dy}{dx} = x^{1/x} \left(\frac{1 - \ln x}{x^2}\right)$.Since $x^{1/x} > 0$ for all $x > 0$ and $x^2 > 0$,the sign of $\frac{dy}{dx}$ depends on $(1 - \ln x)$.For $x \in (1, e)$,$\ln x 0$,which implies $\frac{dy}{dx} > 0$. Thus,$f(x)$ is increasing in $(1, e)$.For $x \in (e, \infty)$,$\ln x > 1$,so $1 - \ln x < 0$,which implies $\frac{dy}{dx} < 0$. Thus,$f(x)$ is decreasing in $(e, \infty)$.

The function $f(x) = x^{1/x}$ is

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