> [!Theorem] Cauchy Mean Value Theorem > Let $[a,b]$ be a [[Closed Real Interval|closed real interval]]. Let $f,g:[a,b]\to \mathbb{R}$ [[Continuous Real Function|continuous real functions]] that are [[Fréchet Differentiation|differentiable]] on the [[Open Real Interval|open real interval]] $(a,b).$ If for all $t\in(a,b),$ $g'(t)\neq 0$ then there is a point $c\in(a,b)$ where $\frac{f'(c)}{g'(c)} = \frac{f(b)-f(a)}{g(b)-g(a)}.$ **Proof**: Consider the function $x\mapsto h(x)=f(x)(g(b)-g(a))-g(x)(f(b)-f(a))$. Note that $h(a)= h(b) =f(a)g(b)-f(b)g(a)$. So by $h(a)=h(b)$ and [[Rolle's Theorem]], there is point $t\in(a,b)$ where $h'(t)=0$. This means that $f'(t)(g(b)-g(a))=g'(t)(f(b)-f(a)) \tag{1}$Since $g'$ is non-zero on $(a,b)$, by *Rolle's Theorem* applied to $g$ shows that $g(b)-g(a) \neq 0$. Also we can rearrange $(1)$ to get the desired result.