Why does function not contradict the mean value theorem
The function f is defined by f(x) = |x -1|. Sketch its graph. Observe that there is no value c such that f(3)-f(0) = f'(c)(3-0). Why does this not contradict the Mean Value Theorem?
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Prove that if f is differentiable at a and f(a) 6= 0, then |f| is also differentiable at a. Give an example to show why the assumption f(a) 6= 0 is necessary.
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