## Calculus: Early Transcendentals 8th Edition

If $f(x)$ and $g(x)$ are both even, $f(x)+g(x)$ is even. If $f(x)$ and $g(x)$ are both odd, $f(x)+g(x)$ is odd. If $f(x)$ is even and $g(x)$ is odd, $f(x)+g(x)$ is neither even nor odd.
If $f(x)$ and $g(x)$ are both even, $f(x)=f(-x)$ (i) $g(x)=g(-x)$ (ii) Let $S(x)=f(x)+g(x)$ $=f(-x)+g(-x)$ (from i) and ii)) $=S(-x)$ $\therefore f(x)+g(x)$ is even. If $f(x)$ and $g(x)$ are both odd, $-f(x)=f(-x), f(x)=-f(-x)$ (i) $-g(x)=g(-x), g(x)=-g(-x)$ (ii) Let $S(x)=f(x)+g(x)$ $=-f(-x)-g(-x)$ (from i) and ii)) $=-(f(-x)+g(-x))$ $=-S(x)$ $\because S(-x)=-S(x),$ $\therefore f(x)+g(x)$ is odd. If $f(x)$ is even and $g(x)$ is odd, $f(x)=f(-x)$ (i) $-g(x)=g(-x), g(x)=-g(-x)$ (ii) Let $S(x)=f(x)+g(x)$ $=f(-x)-g(-x)$ (from i) and ii)) $S(-x)=f(-x)+g(-x)$ $\ne \pm S(x)$ $\because S(-x)\ne \pm S(x)$ $\therefore f(x)+g(x)$ is neither even nor odd.