## Calculus: Early Transcendentals (2nd Edition)

We have that $E(-x)=E(x)$ and $O(-x)=-O(x)$. Let $\Phi=E\circ E$. Then $$\Phi(-x)=E\circ E(-x)=E(E(-x))=E(E(x))=E\circ E(x)=\Phi(x)$$ which means that the function is even.