Question

Prove the Mean Value Theorem: If $f$is continuous on $[a,b]$and differentiable on $(a,b)$, then there is some value $c\in (a,b)$with $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$.

Short answer

We have proved the Mean Value Theorem.

Step-by-step solution

Step 1. Given Information

A function $f$is continuous on $[a,b]$and differentiable on$(a,b)$.

Step 2. Mean Value Theorem.

If a function $f$is continuous on $[a,b]$and differentiable on $(a,b)$and $l\left(x\right)$be the secant line from $(a,f(a\left)\right)$to $(b,f(b\left)\right)$.We rotate the secant line $l\left(x\right)$such that it plays the role of the x-axis. We will consider the function $g\left(x\right)=f\left(x\right)-l\left(x\right)$. The graph of this new function $g\left(x\right)$will have roots at $x=a$and $x=b$,and we will be able to apply Rolle's theorem. Since the secant line $l\left(x\right)$has the slope $\frac{f\left(b\right)-f\left(a\right)}{b-a}$and passes through the point $(a,f(a\left)\right)$, its equation is,

$l\left(x\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}(x-a)+f\left(a\right)$

This means that the function $g\left(x\right)=f\left(x\right)-l\left(x\right)$is equal to,

$g\left(x\right)=f\left(x\right)-\frac{f\left(b\right)-f\left(a\right)}{b-a}(x-a)-f\left(a\right)$

We can apply Rolle's Theorem to $g\left(x\right)$as it satisfies all the hypotheses that are:

(i) it is continuous on $[a,b]$

(ii) it is differentiable on $(a,b)$

(iii) $g\left(a\right)=0$and $g\left(b\right)=0$.

We can thus conclude that there exist some $c\in (a,b)$for which $g\text{'}\left(c\right)=0$. Now, calculating $g\text{'}\left(x\right)$:

$$\begin{gathered} g\text{'}\left(x\right)=\frac{d}{dx}\left[f\right(x)-\frac{f\left(b\right)-f\left(a\right)}{b-a}(x-a)-f(a\left)\right] \\ =f\text{'}\left(x\right)-\frac{f\left(b\right)-f\left(a\right)}{b-a}(1-0)-0 \\ =f\text{'}\left(x\right)-\frac{f\left(b\right)-f\left(a\right)}{b-a} \end{gathered}$$

Hence, through the calculation and by conclusion from Rolle's Theorem we know that, $g\text{'}\left(c\right)=0$,for this value of $c$we have,

$$\begin{gathered} g\text{'}\left(c\right)=f\text{'}\left(c\right)-\frac{f\left(b\right)-f\left(a\right)}{b-a} \\ =0 \end{gathered}$$

Therefore, $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$.

Hence, proved.