Prove the Mean-Value Theorem.

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

Prove the Mean-Value Theorem.

Answer: 

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

SAMPLE SOLUTION

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Solution: 

Recall the Mean-Value Theorem: if $f(x)$ is differentiable on the open interval $(a,b)$ and continuous on the closed interval $[a,b]$, then there is a least one number $e$ in $(a,b)$ such that

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Let's construct the function

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Note that <Sign in to see all the formulas> and <Sign in to see all the formulas>.

Our method of proof is to show that $F(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$ and therefore Rolle's Theorem can be used.

The function $F(x)$ is really just the sum of two functions - $f(x)$ and a polynomial. Let's define

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Since $f(b)$, $f(a)$, $b$, and $a$ are all constants, $g(x)$ is just a polynomial. Therefore, <Sign in to see all the formulas> is the sum of $f(x)$ and the polynomial $g(x)$.

Because we know that a polynomial is a continuous function, we know that $g(x)$ is continuous on $[a,b]$. Now we know that the sum of two continuous functions is a continuous function. Therefore, $F(x)$ is continuous on $[a,b]$.

Because $g(x)$ is a polynomial and because polynomials are differentiable, $g(x)$ is differentiable on $(a,b)$. Furthermore, since $F(x)$ is the sum of two differentiable functions, $F(x)$ is differentiable on $(a,b)$.

Now $F(x)$ has the following properties as shown above:

Therefore, from Rolle's Theorem we know there exists a point $e$ in $(a,b)$ such that <Sign in to see all the formulas>. Now since

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we know that

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and therefore

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