Find critical points of $f$, which occur at $f^{\prime }(x)=0$. Compute the derivative:

Critical points of $f$ are at $x=-2$ and $x=3$, so $f^{\prime }(x)$ crosses the $x$-axis at these $x$-values. Test the sign of $f^{\prime }$ on each interval:

For $x<-2$, $f^{\prime }(x)>0$, so $f$ is increasing here, meaning $f^{\prime }(x)$ is positive (above the $x$-axis)

For $-2<x<3$, $f^{\prime }(x)<0$, so $f$ is decreasing here, meaning $f^{\prime }(x)$ is negative (below the $x$-axis)

For $x>3$, $f^{\prime }(x)>0$, so $f$ is increasing here, meaning $f^{\prime }(x)$ is positive (above the $x$-axis)

Since $f$ is cubic, $f^{\prime }$ is quadratic with a positive leading coefficient. The final sketch is an upward-opening parabola crossing the $x$-axis at $x=-2$ and $x=3$, matching our sign intervals.

Concavity of $f$ directly translates to the sign of $f^{\prime \prime }$. Since $f$ is concave down for $x<1$, $f^{\prime \prime }(x)<0$ for all $x<1$, so $f^{\prime \prime }$ lies below the $x$-axis on $(-\infty ,1)$

For $x>1$, $f$ is concave up, so $f^{\prime \prime }(x)>0$ for all $x>1$, meaning $f^{\prime \prime }$ lies above the $x$-axis on $(1,\infty )$

$f$ has an inflection point at $x=1$, so $f^{\prime \prime }(1)=0$, so $f^{\prime \prime }$ crosses the $x$-axis at $(1,0)$

Since $f$ is cubic, its second derivative is linear. The sign of $f^{\prime \prime }$ goes from negative left of 1 to positive right of 1, so the slope of the line is positive. The final graph is a straight line with positive slope crossing the $x$-axis at $x=1$.

First analyze $f(x)$ from $f^{\prime }(x)$: Zeros of $f^{\prime }$ are at $x=-1$ and $x=2$, so $f$ has critical points at these $x$-values

For $x<-1$, $f^{\prime }(x)<0$, so $f$ is decreasing. For $-1<x<2$, $f^{\prime }(x)>0$, so $f$ is increasing. For $x>2$, $f^{\prime }(x)<0$, so $f$ is decreasing. Thus $f$ has a local minimum at $x=-1$ and a local maximum at $x=2$

Next analyze $f^{\prime \prime }(x)$ from $f^{\prime }(x)$: $f^{\prime }$ is a downward-opening quadratic, so its derivative ($f^{\prime \prime }(x)$) is linear

The vertex (maximum) of $f^{\prime }$ is at $x=0.5$, the midpoint of $-1$ and $2$. For $x<0.5$, $f^{\prime }$ is increasing, so $f^{\prime \prime }(x)>0$, meaning $f$ is concave up on $(-\infty ,0.5)$

For $x>0.5$, $f^{\prime }$ is decreasing, so $f^{\prime \prime }(x)<0$, meaning $f$ is concave down on $(0.5,\infty )$. $f^{\prime }$ has a maximum at $x=0.5$, so $f^{\prime \prime }(0.5)=0$, which means $f$ has an inflection point at $x=0.5$

Final summary: $f^{\prime \prime }$ is a straight line with negative slope crossing the $x$-axis at $x=0.5$, and $f$ has key points at $x=-1$ (local min), $x=0.5$ (inflection point), and $x=2$ (local max) with the correct behavior in each interval.

Part (a): Find zeros of $f^{\prime }(x)$: $2\sin x=0$ at $x=0$, $x=\pi$, $x=2\pi$

$f^{\prime }(x)>0$ on $(0,\pi )$, so $f$ is increasing on $(0,\pi )$. $f^{\prime }(x)<0$ on $(\pi ,2\pi )$, so $f$ is decreasing on $(\pi ,2\pi )$. By the First Derivative Test, $f$ has a local maximum at $x=\pi$.

Part (b): Calculate $f^{\prime \prime }(x)$:

Zeros of $f^{\prime \prime }$ are at $x=\frac{\pi }{2}$ and $x=\frac{3\pi }{2}$. $f^{\prime \prime }(x)>0$ on $\left(0,\frac{\pi }{2}\right)\cup \left(\frac{3\pi }{2},2\pi \right)$, so $f$ is concave up here. $f^{\prime \prime }(x)<0$ on $\left(\frac{\pi }{2},\frac{3\pi }{2}\right)$, so $f$ is concave down here. Both points have a sign change, so they are inflection points.

Part (c): Find the antiderivative of $f^{\prime }(x)$:

Substitute $f(0)=2$: $-2(1)+C=2⟹C=4$, so $f(x)=-2\cos x+4$. Key points are $(0,2)$, $\left(\frac{\pi }{2},4\right)$ (inflection), $(\pi ,6)$ (local max), $\left(\frac{3\pi }{2},4\right)$ (inflection), $(2\pi ,2)$.