Determine the intervals of monotonicity of the function y

Determine the intervals of monotonicity of the function $y = x\,e^{1/x}$.

Step 1: Domain

The exponent $\dfrac1x$ requires $x\neq0$, so the domain is $\bigl(-\infty,0\bigr)\cup\bigl(0,+\infty\bigr)$ — two separate branches.

Step 2: The derivative

By the product rule with $u = x$ and $v = e^{1/x}$, where $v' = e^{1/x}\cdot\left(-\dfrac{1}{x^{2}}\right)$: \[y' = e^{1/x}+x\,e^{1/x}\left(-\frac{1}{x^{2}}\right) = e^{1/x}\left(1-\frac1x\right) = e^{1/x}\cdot\frac{x-1}{x}. \tag{1}\]

Step 3: Sign of the derivative

Since $e^{1/x}\gt 0$ for every $x\neq0$, the sign of $y'$ is the sign of $\dfrac{x-1}{x}$, which vanishes at $x = 1$ and is undefined at $x = 0$: \[x\lt 0:\ \frac{(-)}{(-)} = +\ \Rightarrow y'\gt 0; \qquad 0\lt x\lt 1:\ \frac{(-)}{(+)} = -\ \Rightarrow y'\lt 0; \qquad x\gt 1:\ +\ \Rightarrow y'\gt 0.\]

Step 4: The critical point

At $x = 1$ we have $y' = 0$, and $y'$ changes from $-$ to $+$: a local minimum, with

\[ y(1) = 1\cdot e^{1} = e. \]

The point $x = 0$ is not a critical point but a break in the domain. As $x\to0^{-}$, $\dfrac1x\to-\infty$ and $e^{1/x}\to0$, so $y\to0^{-}$; as $x\to0^{+}$, $e^{1/x}\to+\infty$ and $y\to+\infty$. The two branches behave quite differently.

Step 5: Verification

\[y(-2) = -1.2130613 \lt y(-0.5) = -0.0676676 \lt y(-0.1) = -4.5\cdot10^{-6} \;\;\checkmark\ \text{(rising to } 0^{-})\]

\[y(0.5) = 3.6945280 \gt y(1) = 2.7182818 \lt y(2) = 3.2974425 \lt y(5) = 6.1070138 \;\;\checkmark\]

\[y''(x) = \frac{e^{1/x}}{x^{3}} \;\Longrightarrow\; y''(1) = e \gt 0,\ \text{confirming the minimum} \;\;\checkmark\]