The hypotheses required are:
(1) f(x) is continuous on [-2,2], and
(2) f(x) is differentiable on (-2,2)
The first statement is easy to prove: |x-1| can be piecewise defined, as 1-x if x < 1, 0 if x=1, and x-1 if x > 1. Since x-1 is a polynomial, and all polynomials are continuous over R, f(x) is continuous over [-2,1) and (1,2].
Next, we must check for continuity at x=1.
f(1) = 0
All 3 numbers match, so f(x) is continuous at x=1.
The second statement can be seen as false graphically: there is a corner point at x=1. However, we must prove this rigorously (no graphical methods are acceptable).
Start with the definition of derivative, calculate the left-hand derivative:
Noting that f(1) = 0, and using the piecewise definition of f(x) as described earlier.
Then calculate the right-hand derivative:
Since they are not the same, f(x) is not differentiable at x=1, so (2) is false.
One of the conditions was not satisfied, so no values of
f'(c) are guaranteed. (1 - 3)/(2 - (-2)) = -0.5.
The only values of f' at the points where it is differentiable are 1 or -1.
p.s. I'm pretty sure the last part of the question is worded wrongly, because:
1) "not satisfied" = theorem doesn't count for anything
2) MVT involves f'(c), and does not state exactly which, or how many, values c could take.