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Use the Mean Value Theorem to prove that there is a point $c$ in the interval $[4,6]$ such that the line tangent to the graph of $f(x)=x^2-10x+25$ at $c$ is horizontal. Can we use the same theorem to say the same about the function $g(x)=|x-5|$? Yes or No?

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$f(4)=f(6)$. Using the mean value theorem, we know that there exists $c$ in $(4,6)$ such that $$f'(c)=\frac{f(6)-f(4)}{6-4}=0$$ This does not work for the function $g$, because $g$ is not differentiable at every point in the interval [4,6].

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Thanks for the answer – Dara Nov 9 '12 at 11:12
Consider one small change: "... because $g$ is not differentiable at every point on that interval." – chharvey Nov 9 '12 at 11:53
OK I will change it – Amr Nov 9 '12 at 11:58

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