Prove Rolle's Theorem.

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Lesson Parent: 
Geometry of Functions III: The Mean-Value Theorem
Problem: 

Prove Rolle's Theorem.

Answer: 

It is true that if $f(x)$ is differentiable on the open interval $(a,b)$, continuous on the closed interval $[a,b]$, and $f(a)=f(b)=0$, then there is a least one number $e$ in $(a,b)$ such that $f\,'(e)=0$.

SAMPLE SOLUTION

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Solution: 

Recall Rolle's Theorem: if $f(x)$ is differentiable on the open interval $(a,b)$, continuous on the closed interval $[a,b]$, and <Sign in to see all the formulas>, then there is a least one number $e$ in $(a,b)$ such that <Sign in to see all the formulas>.

Proving Rolle's Theorem depends upon the Extreme-Value Theorem. In Rolle's Theorem we assume that the function is continuous on $[a,b]$ so the Extreme-Value Theorem can be used. Let's denote the maximum and minimum values of the function by $U$ and $L$, respectively. The proof of Rolle's Theorem is now broken into two pieces. The first is when $U=L=0$ and the second is when either $U$ or $L$ or both are not zero.

Let's take the trivial case where $U=L=0$. By the extreme value theorem, <Sign in to see all the formulas> for all values of $x$ in $[a,b]$. Since $U=L=0$, then <Sign in to see all the formulas> for all values of $x$ in $[a,b]$ and therefore <Sign in to see all the formulas>. Since $f(x)$ is a constant, its derivative is zero and any number in $(a,b)$ can be chosen for $e$. Therefore, <Sign in to see all the formulas> and the theorem is proved for the case $U=L=0$.

Now let's assume that $U$ or $L$ or both are not zero. Let's first look at the case where <Sign in to see all the formulas>. Since $U$ is the maximum value and <Sign in to see all the formulas> and <Sign in to see all the formulas>. Because <Sign in to see all the formulas> and <Sign in to see all the formulas> then <Sign in to see all the formulas>. In other words, $U$ has to be positive. Now let's denote the point where $f(x)$ takes on its maximum value by $e$. In other words, <Sign in to see all the formulas>. We know by the assumption of the problem that <Sign in to see all the formulas> exists. Now recall that

<Sign in to see all the formulas>



Furthermore, recall that

<Sign in to see all the formulas>



Since $f(e)$ is the maximum value, <Sign in to see all the formulas> is always less than or equal to zero. Now the ratio

<Sign in to see all the formulas>


when $h$ is positive and

<Sign in to see all the formulas>


when $h$ is negative. This means

<Sign in to see all the formulas>


Therefore, <Sign in to see all the formulas> and the only way this can be true is if <Sign in to see all the formulas>.

Now the proof where <Sign in to see all the formulas> is almost identical to the case where <Sign in to see all the formulas>. When <Sign in to see all the formulas>, $L$ is negative, <Sign in to see all the formulas> is always positive, and

<Sign in to see all the formulas>


Again, <Sign in to see all the formulas> and the only way this can be true is if <Sign in to see all the formulas>.