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utf14-003
Let $f(x)=|x|-1$. Then $f(-1)=f(1)=0$, but $f'(x) \neq 0$ on $ [-1,1] $. Does this contradict Rolle's Theorem? Explain!
Does the Mean Value Theorem apply to the function $f(x)=\frac{x^2-4x +3 }{ x-3}$ on $ [2,4] $?
Is there a point $c$ on $ [2,4] $ for which $f'(c)= \frac{f(4)-f(2) }{ 4-2}$ where $f(x)$ is the function of part $b)$?
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