Find the minimum and maximum values of f(x) = x^(1- ln x)

Find the minimum and maximum values of $f(x) = x^{\,1-\ln x}$.

Step 1: Domain, and a change of form

The base must be positive, so the domain is $\bigl(0,+\infty\bigr)$. Writing the power as an exponential: \[f(x) = e^{\bigl(1-\ln x\bigr)\ln x} = e^{\,g(x)}, \qquad g(x) = \ln x-\ln^{2}x. \tag{1}\]

Since $e^{t}$ is strictly increasing, $f$ and $g$ have the same critical points, and extrema of the same type.

Step 2: The derivative

\[ g'(x) = \frac1x-\frac{2\ln x}{x} = \frac{1-2\ln x}{x}, \tag{2} \]

\[ f'(x) = e^{g(x)}g'(x) = x^{1-\ln x}\cdot\frac{1-2\ln x}{x} = \frac{1-2\ln x}{x^{\ln x}}. \tag{3} \]

Step 3: The critical point

As $x^{\ln x}\gt 0$ throughout the domain, $f'$ vanishes only where

\[ 1-2\ln x = 0 \;\Longrightarrow\; \ln x = \frac12 \;\Longrightarrow\; x = \sqrt{e}. \]

And $1-2\ln x\gt 0$ for $x\lt \sqrt{e}$, $\lt 0$ for $x\gt \sqrt{e}$: the derivative passes from $+$ to $-$, so this is a maximum.

Step 4: The behaviour at both ends

Putting $t = \ln x$, the exponent is $g = t-t^{2}$, which tends to $-\infty$ as $t\to\pm\infty$. Hence

\[ \lim_{x\to0^{+}}f(x) = 0, \qquad \lim_{x\to+\infty}f(x) = 0, \] while $f\gt 0$ everywhere. The function rises from $0$, reaches one peak and falls back to $0$: the maximum is global, and the value $0$ is an infimum that is never attained.

Step 5: The maximum value

At $x = \sqrt{e}$, where $\ln x = \tfrac12$: \[g\bigl(\sqrt{e}\bigr) = \frac12-\frac14 = \frac14 \;\Longrightarrow\; f\bigl(\sqrt{e}\bigr) = e^{1/4} = \sqrt[4]{e}.\]

Step 6: Verification

\[f(1) = 1; \qquad f\bigl(\sqrt{e}\bigr) = 1.2840254; \qquad f(2) = 1.2370063 \;\;\checkmark\]

\[f(0.5) = 0.3092516; \qquad f(5) = 0.3749913 \;\;\checkmark\ \text{(falling away on both sides)}\]

\[\text{Symmetry: } g\!\left(\frac{e}{x}\right) = g(x),\ \text{so } f\!\left(\frac{e}{x}\right) = f(x);\ \text{the peak sits at } x = \sqrt{e} \;\;\checkmark\]