Locate and classify the local extremum using the Second Derivative Test

Using the product rule, $f'(x) = e^x \cdot x + e^x \cdot 1 = e^x(x+1)$. Since $e^x$ is never zero, setting $f'(x)=0$ gives $x+1=0$, so the only critical number is $x=-1$.

Differentiating again, $f''(x) = e^x(x+1) + e^x(1) = e^x(x+2)$. Evaluating at $x=-1$: $f''(-1) = e^{-1}(1) = e^{-1} > 0$, so by the Second Derivative Test, $x=-1$ gives a local minimum. The value is $f(-1) = -1\cdot e^{-1} = -1/e$.

From $f''(x) = e^x(x+2)$, setting $f''(x) = 0$ gives $x = -2$ (since $e^x \neq 0$ for all $x$). This is a candidate inflection point that must be checked for an actual sign change.

For $x < -2$, say $x=-3$: $f''(-3) = e^{-3}(-1) < 0$, so $f$ is concave down. For $x > -2$, say $x=0$: $f''(0) = e^0(2) = 2 > 0$, so $f$ is concave up. Since concavity changes from down to up at $x=-2$, this confirms an inflection point there, located at $(-2, -2e^{-2})$.

The local minimum occurs at x equals negative one, while the inflection point occurs at x equals negative two. Since negative two is farther from zero in the negative direction, the inflection point lies farther to the left. This directly follows from the critical number found in part a and the inflection point found in part b.