Intervals of increase/decrease for f(x) = xe^{-x}

Applying the product rule, $f'(x) = e^{-x} + x(-e^{-x}) = e^{-x}(1-x)$. Since $e^{-x} > 0$ for all $x$, the sign of $f'(x)$ matches the sign of $(1-x)$.

When $x < 1$, $(1-x) > 0$ so $f'(x) > 0$, meaning $f$ is increasing on $(-\infty, 1)$. When $x > 1$, $(1-x) < 0$ so $f'(x) < 0$, meaning $f$ is decreasing on $(1, \infty)$.

The critical number is $x=1$, since $f'(1) = 0$. From part (a), $f'(x) > 0$ for $x<1$ and $f'(x) < 0$ for $x>1$, so $f'$ changes from positive to negative at $x=1$.

By the First Derivative Test, this sign change (positive to negative) indicates a local maximum at $x=1$. Evaluating, $f(1) = 1\cdot e^{-1} = e^{-1} = 1/e$.

Since $f$ is decreasing on $(1,\infty)$ and $f(x) = x/e^x$, the exponential $e^x$ in the denominator grows much faster than the numerator $x$ as $x\to\infty$.

Therefore the function values continue to decrease toward zero without becoming negative, so $\lim_{x\to\infty} f(x) = 0$, and $y=0$ acts as a horizontal asymptote.