`f(x) = |2x + 1|, [-1,3]` Determine whether the Mean Value Theorem can be applied to `f` on the closed interval `[a,b]`. If the Mean Value Theorem can be applied, find all values of `c` in the open interval `(a,b)` such that `f'(c) = (f(b) - f(a))/(b - a)`. If the Mean Value Theorem cannot be applied, explain why not.
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NO, it isn't.
The function is continuous on [-1, 3] and is differentiable on (-1, 3) but the point where 2x+1=0, or x=-1/2 in [-1, 3]. To the left of this point f(x)=-2x-1, to the right f(x)=2x+1, the graph has an "angle".
Formally, `lim_(h->0)[(f(-1/2+h)-f(-1/2))/h]` doesn't exist (-2 when h<0 and +2 when h>0).
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