A relative maximum on an odd polynomial

An odd function satisfies $p(-x)=-p(x)$, so its graph is unchanged by a half turn about the origin. Every point $(a,b)$ on the graph has a partner $(-a,-b)$.

The value first: $p(-3)=-p(3)=-(-4)=4$.

Now the kind of point. Near $x=3$ the graph rises to $(3,-4)$ and then falls, a peak. Turn that picture 180 degrees about the origin and the peak becomes a valley sitting at $(-3,4)$. In symbols, if $p(3)\ge p(x)$ for $x$ near 3, then for $x$ near $-3$, $$p(x)=-p(-x)\ge -p(3)=p(-3),$$ so $p(-3)$ is the smallest nearby value. It is a relative minimum.

So $p(-3)=4$ is a relative minimum, choice C.

Each choice

• A. It gets the 4, then forgets that a half turn flips a peak into a valley. Negating the output while assuming the shape carries over unchanged gives this.
• B. Treating the function as even gives this: a mirror image across the $y$-axis keeps both the value and the type of extremum. For an odd function the output changes sign and a maximum becomes a minimum.
• C (correct). Odd symmetry gives $p(-3)=-p(3)=4$, and rotating a relative maximum about the origin turns it into a relative minimum.
• D. Changing a maximum into a minimum is the right idea, but the output keeps its sign here. Odd symmetry negates the output, so $p(-3)$ is $4$, not $-4$.