7. Given that the domain of $f(x)$ is symmetric about the origin, $f(x)=g(x)+h(x)$, where $g(x)$ is an odd function, $h(x)$ is an even function, then $g(x)=$ $\qquad$ (express $g(x)$ in terms of $f(x)$)
7. $g(x)=\frac{f(x)-f(-x)}{2}$ Hint: From $f(x)=g(x)+h(x), f(-x)=g(-x)+h(-x)$, $f(-x)=-g(x)+h(x)$, solve the system of equations.