Solutions

Let

$$h(x)=g(b)f(x)+f(a)g(x)-f(x)g(x).$$

Notice that $h$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Moreover,

$$h(a)=h(b)=g(b)f(a).$$

Hence, we can apply Rolle's theorem which gives

$$h'(x)=g(b)f'(x)+f(a)g'(x)-f'(x)g(x)-f(x)g'(x)=0$$

for some $x\in(a,b)$. Since $g'(y)\ne 0$ for all $y\in(a,b)$, we can rearrange the terms above and obtain

$$\begin{align*} f'(x)\big(g(b)-g(x)\big) &= g'(x)\big(f(x)-f(a)\big) \implies \frac{f'(x)}{g'(x)}\big(g(b)-g(x)\big)=f(x)-f(a). \end{align*}$$

We now hope to divide both sides by $g(b)-g(x)$ but this requires that $g(b)\ne g(x)$ for all $x\in(a,b)$. Suppose for contradiction that $g(y)=g(b)$ for some $y\in(a,b)$. Then, by Rolle's theorem, $g'(z)=0$ for some $z\in(y,b)$ which is contradictory.