## Calculus 8th Edition

To see whether $h=f \circ g$ is even or not, let us find $h(-x)$:$$h(-x)=(f \circ g)(-x)=f(g(-x)).$$Since by the assumption $g$ is even, we have$$g(-x)=g(x).$$Thus, we obtain $$h(-x)=f(g(x))=(f \circ g)(x)=h(x),$$so $h$ is even.