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Prove: If the function $f$ is continuous on $[a,b]$, differentiable on $(a,b)$ and $f'(x) = 0$ on $(a,b)$, then $f$ must be a constant function on $[a,b]$.

I need to select some $x_1$ and $x_2$ in $[a,b]$ such that $x_1$ is not equal to $x_2$ therefore by the mean value theorem.. This is where I am getting a bit confused how to apply MVT to this proof.. Hints/Tips appreciated. I may be over thinking things. Thanks.

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Just a note: You should start with $x_1\in[a,b]$ fixed, and $x_2\in[a,b]$ arbitrary. Then show $f(x_2)=f(x_1)$. –  David Mitra Apr 12 '12 at 17:55

1 Answer 1

$$ \begin{align} \frac{f(x_1) - f(x_2)}{x_1-x_2} & = f'(c) \text{ (for some }c\text{ between }x_1\text{ and }x_2\text{)} \\ \\ \\ & = 0 \end{align} $$ Therefore $f(x_1)-f(x_2)=0$.

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