Calculus: Early Transcendentals 8th Edition

To check this we evaluate $f$ at $-x$ and see if it is equal to $f(x)$ (even), $-f(x)$ (odd), or something else (neither even nor odd). (a) $f(-x) = 2(-x)^5-3(-x)^2+2 = -2x^5-3x^2+2$ where we used the fact that the odd power of $(-x)$ are the same as minus the same power of $x$ and that the even power of $(-x)$ is the same as the same even power of $x$. We see that $f(-x)$ is neither equal to $f(x)$ nor to $-f(x)$ so it is neither even nor odd. (b) $$f(-x)=(-x)^3-(-x)^7=-x^3+x^7=-(x^3-x^7)=-f(x)$$ so the function is odd. (c) $$f(-x)=e^{-(-x)^2}=e^{-x^2}=f(x)$$ so the function is even (d) $$f(-x) = 1+\sin(-x) = 1+(-\sin x) = 1-\sin x,$$ where we used that $\sin(-x) = -\sin x$. We see that $f(x)$ is neither equal to $f(x)$ nor to $-f(x)$ so $f$ is neither even nor odd.