Find critical points and classify local extrema

Differentiate to get $g'(x) = 3x^2 - 6x = 3x(x-2)$. Setting $g'(x) = 0$ gives critical points at $x = 0$ and $x = 2$.

Compute $g''(x) = 6x - 6$. At $x=0$, $g''(0) = -6 < 0$, so this is a local maximum with $g(0) = 2$. At $x=2$, $g''(2) = 6 > 0$, so this is a local minimum with $g(2) = -2$.

From part (a), $g''(x) = 6x - 6$. Setting $g''(x) = 0$ gives $x = 1$.

For $x < 1$, $g''(x) < 0$ (concave down); for $x > 1$, $g''(x) > 0$ (concave up). Since concavity changes, $x=1$ is an inflection point, and $g(1) = 1 - 3 + 2 = 0$.

As $x \to \infty$, the dominant term $x^3$ grows without bound, so $g(x) \to \infty$.

As $x \to -\infty$, the term $x^3 \to -\infty$ dominates the lower-order terms, so $g(x) \to -\infty$.

Since $g'(x) = 3x(x-2)$ is positive for $x<0$, negative for $0<x<2$, and positive for $x>2$, $g$ is increasing on $(-\infty,0)$ and $(2,\infty)$, and decreasing on $(0,2)$.

Since $g''(x) = 6x-6$ is negative for $x<1$ and positive for $x>1$, $g$ is concave down on $(-\infty,1)$ and concave up on $(1,\infty)$, matching the plotted curve which rises to $(0,2)$, falls through the inflection point $(1,0)$ to the local min $(2,-2)$, then rises again.

The local maximum occurs at x equals zero, which lies in the interval where g double prime is negative. Since g double prime of zero equals negative six, the curve is concave down there. This matches the second derivative test used to classify x equals zero as a maximum. So at the local maximum, the graph is concave down.