Suppose $y=f(x)$ is [[Continuity on a Domain|continuous]] over a closed interval $[a, b]$ and [[Derivative of a function|differentiable]] on the interval's interior $(a, b)$ Then there is at least one point $c$ in $(a, b)$ at which $ \frac{f(b)-f(a)}{b-a}=f^{\prime}(c) $ ![[Screen Shot 2022-04-15 at 7.23.16 PM.png]] #### Proof The [[Secant line]] connecting $A :=(a,f(a))$ and $B:=(b,f(b))$ is a function $g(x)$. This makes $g(a)=f(a)$ and $g(b)=f(b)$. The slope of this line is $g'(x)=\frac{f(b)-f(a)}{b-a}$ The difference between them will be a function $h(x)=f(x)-g(x)$ that satisfies [[Rolle's Theorem]] on $[a,b]$, since $h(a)=h(b)=0$ So there must be a $c\in(a,b) : h^\prime (c)=0$ $\begin{aligned} h^{\prime}(x) &=f^{\prime}(x)-\frac{f(b)-f(a)}{b-a} \\ h^{\prime}(c) &=f^{\prime}(c)-\frac{f(b)-f(a)}{b-a} \\ 0 &=f^{\prime}(c)-\frac{f(b)-f(a)}{b-a} \\ f^{\prime}(c) &=\frac{f(b)-f(a)}{b-a} \end{aligned}$ That same $c$ is the one we sought to prove.