Prove that if $f(x)$ is continuous on $[a,b]$ and $f\,'(x)=0$ on $(a,b)$, then $f(x)$ is a constant.

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

Prove that if $f(x)$ is continuous on $[a,b]$ and $f\,'(x)=0$ on $(a,b)$, then $f(x)$ is a constant.

Answer: 

It is true that if $f(x)$ is continuous on $[a,b]$ and $f\,'(x)=0$ on $(a,b)$, then $f(x)$ is a constant.

SAMPLE SOLUTION

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Solution: 

Recall from the Mean-Value Theorem that there there exists a point $e$ where

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But we are told that <Sign in to see all the formulas> for every point in the open interval $(a,b)$. Therefore,

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So now we know that the <Sign in to see all the formulas>.

Now let's take any arbitrary value in $(a,b)$ and call it $x_i$. Again, from the assumptions of the problem the requirements of the Mean-Value Theorem hold. So between $a$ and $x_i$ there is a point (let's call it $e_i$) such that

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But again, we are told that <Sign in to see all the formulas> so that <Sign in to see all the formulas>.

The proof is now complete! We know that <Sign in to see all the formulas> and that for every $x_i$ in $(a,b)$ <Sign in to see all the formulas>. Since $f(x)$ is the same value for every point in $[a,b]$ it is a constant.