Intervals of increase/decrease and local extrema of q(x) = x·e^(-x)

Using the product rule, $q'(x) = e^{-x} + x(-e^{-x}) = e^{-x}(1-x)$. Since $e^{-x}$ is always positive, the sign of $q'(x)$ matches the sign of $(1-x)$. Setting $q'(x)=0$ gives the only critical number $x = 1$.

For $x < 1$, $(1-x) > 0$ so $q'(x) > 0$, meaning $q$ is increasing on $(-\infty, 1)$. For $x > 1$, $(1-x) < 0$ so $q'(x) < 0$, meaning $q$ is decreasing on $(1, \infty)$. Since $q'$ changes from positive to negative at $x=1$, the First Derivative Test tells us $q$ has a local maximum there, with value $q(1) = 1\cdot e^{-1} = 1/e$.

The derivative changes from positive to negative at the critical number x equals one. Before x equals one, q prime is positive since q is increasing, and after x equals one, q prime is negative since q is decreasing. This sign change from positive to negative is exactly what identifies a local maximum by the First Derivative Test.