which equals zero when $x=-2$, $x=0$, and $x=5$. Also note that $f^{\prime }$ exists for all $x$ in the domain of $f$. Since the domain of $f$ is $(-\mathrm{\infty },\mathrm{\infty })$, all of these values of $x$ are critical points.

Evaluate $f^{″}(x)$ at each critical point where $f^{\prime }(x)=0$ and use the Second Derivative Test to classify each critical point, if possible.

• $x=-2$: Since $f^{\prime }(-2)=0$ and $f^{″}(-2)=60(-2)(16+18-20)<0$, $f$ has a relative maximum at $x=-2$.

• $x=0$: Since $f^{\prime }(0)=0$ and $f^{″}(0)=60(0)(-20)=0$, the Second Derivative test is inconclusive. Instead, using the First Derivative test, we see there is no change in sign of $f^{\prime }(x)$ at $x=0$, and therefore, $f$ does not have a relative extreme at $x=0$.

• $x=5$: Since $f^{\prime }(5)=0$ and $f^{″}(5)=60(5)(100-45-20)>0$, $f$ has a relative minimum at $x=5$.

The graph of the second derivative of a continuous function on the interval (-4,5). The graph of the second derivative is continuous and looks like a cubic polynomial. From left to right, the graph starts below the x-axis and increases until it crosses the x-axis at x = -1.81. It continues to increase, then decreases and crosses the x-axis again at x = 1.61. It continues to decrease, then increases until it crosses the x-axis at x = 4, where the graph increases for all x > 4. The second derivative is negative at x = -3, positive at x = 0, and equal to zero at x = 4.

Since $f^{\prime }$ is continuous for all $x$, and therefore defined for all $x$, the only critical points of $f$ occur where $f^{\prime }(x)=0$. And we were specifically told that the derivative is equal to zero only when $x=-3$, $x=0$, and $x=4$, so these are the three critical points of $f$.

From the given graph of $f^{″}(x)$, we can see that $f^{″}(-3)$ is negative. Therefore, since $f^{\prime }(-3)=0$ and $f^{″}(-3)<0$, by the Second Derivative Test, $f$ has a relative maximum at $x=-3$.

From the given graph of $f^{″}(x)$, we can see that $f^{″}(0)$ is positive. Therefore, since $f^{\prime }(0)=0$ and $f^{″}(0)>0$, by the Second Derivative Test, $f$ has a relative minimum at $x=0$.

Since $f^{\prime }(4)=0$ and $f^{″}(4)=0$, the Second Derivative Test is inconclusive and we proceed to using the First Derivative Test:

• From the graph of $f^{″}(x)$, note that $f^{″}(x)<0$ for $x$ just before 4 and $f^{″}(x)>0$ for $x$ just after 4.

• This means that $f^{\prime }(x)$ is decreasing just before $x=4$ and increasing just after $x=4$.

• But since $f^{\prime }(4)=0$, $f^{\prime }(x)$ must then be positive just before $x=4$ and positive just after $x=4$. A graph of $f^{\prime }(x)$ is shown below. Notice how $f^{\prime }(x)$ is decreasing just before 4 and increasing just after 4.

• In particular, $f^{\prime }(x)$ does not change sign at $x=4$.

• Therefore, by the First Derivative Test, $f$ does not have a relative extrema at $x=4$.

The graph of the second derivative of a continuous function on the interval (-4,5). The graph of the second derivative is continuous and looks like a quartic polynomial. From left to right, the graph starts above the x-axis and decreases until it crosses the x-axis at x = -2.31. It continues to decrease, then increases until it touches (but does not cross) the x-axis at x = 0. It then decreases, then increases until it crosses the x-axis at x = 3.11, where the graph increases for all x > 3.11. The second derivative is positive at x = -3, equal to zero at x = 0, and positive at x = 4.

Since $f^{\prime }(x)$ is continuous for all $x$, and therefore defined for all $x$, the only critical points of $f(x)$ occur where $f^{\prime }(x)=0$. And we were specifically told that the derivative is equal to zero only when $x=-3$, $x=0$, and $x=4$, so these are the three critical points of $f(x)$.

From the given graph of $f^{″}(x)$, we can see that $f^{″}(-3)$ is positive. Therefore, since $f^{\prime }(-3)=0$ and $f^{″}(-3)>0$, by the Second Derivative Test, $f$ has a relative minimum at $x=-3$.

Since $f^{\prime }(0)=0$ and $f^{″}(0)=0$, the Second Derivative Test is inconclusive and we proceed to using the First Derivative Test:

• From the graph of $f^{″}(x)$, note that $f^{″}(x)<0$ for $x$ just before 0 and $f^{″}(x)<0$ for $x$ just after 0.

• This means that $f^{\prime }(x)$ is decreasing just before $x=0$ and decreasing just after $x=0$.

• But since $f^{\prime }(0)=0$, $f^{\prime }(x)$ must then be positive just before $x=0$ and negative just after $x=0$. A graph of $f^{\prime }(x)$ is shown below. Notice how $f^{\prime }(x)$ is decreasing just before 0 and decreasing just after 0.

• In particular, $f^{\prime }(x)$ changes from positive to negative at $x=0$.

• Therefore, by the First Derivative Test, $f$ has a relative maximum at $x=0$.

From the given graph of $f^{″}(x)$, we can see that $f^{″}(4)$ is positive. Therefore, since $f^{\prime }(4)=0$ and $f^{″}(4)>0$, by the Second Derivative Test, $f$ has a relative minimum at $x=4$.