Finding the critical number of f(x) = x ln(x)

First differentiate using the product rule: $f'(x) = 1\cdot \ln(x) + x\cdot \frac{1}{x} = \ln(x) + 1$. Setting $f'(x) = 0$ gives $\ln(x) = -1$, so $x = e^{-1} = 1/e$. This value lies in $(0, e]$, so it is the only critical number (the derivative is undefined at $x=0$, but that point is not in the domain).

For $0 < x < 1/e$, $\ln(x) < -1$ so $f'(x) < 0$; for $x > 1/e$, $\ln(x) > -1$ so $f'(x) > 0$. Since $f'$ changes from negative to positive at $x = 1/e$, this is a local minimum, with $f(1/e) = \frac{1}{e}\ln(1/e) = -\frac{1}{e}$. Since $f$ decreases then increases on the whole interval, this local minimum is also the absolute minimum on $(0,e]$.

As $x \to 0^+$, although $\ln(x) \to -\infty$, the factor $x \to 0$ dominates, and $x\ln(x) \to 0$. Thus $f$ has no minimum approached there (it approaches a finite limit of 0, not $-\infty$), confirming that the true absolute minimum occurs at the critical point $x=1/e$ rather than near the open end of the interval.