Solutions

Since $f$ is continuous and $f(x)\ne0$ for all $x$, $f$ must be either positive or negative. Otherwise, by the IVT, there would be an $x$ such that $f(x)=0$. For contradiction, if for $x_1$ and $x_2$ we have $f(x_1)=g(x_1)$ and $f(x_2)=-g(x_2)$, then the signs of $g(x_1)$ and $g(x_2)$ must be different. By the IVT, there would exist $x_3$ such that

$$0=g(x_3)=\pm f(x_3),$$

which clearly contradicts $f(x)\ne0$.