Let f (x) = 2-|4x-2|. Show that there is no value of c such that f (3) - f (0) = f' (c) (3-0). Why does this not contradict the Mean Value Theorem?

The mean value theorem is valid if f (x) is continuous in the interval (0,3) and differentiable in the interval (0,3), the problem is that f (x) = 2-|4x-2| is not differentiable in x = 1/2 because 4*1/2 - 2 = 0 and the function |x| is not differentiable in x = 0.

f' (x) = (-4) * (4x-2) / |4x-2|

f (3) = 2-|4*3-2| = 8

f (0) = 2-|4*0-2| = 0

Replacing in f' (c) = f (3) - f (0) / (3-0)

(-4) * (4c-2) / |4c-2| = (8 - 0) / 3

(-4) * (4c-2) * 3/8 = |4c-2|

-3/2*4c + 3/2*2 = |4c-2|

-6c + 3 = |4c-2|

That gives us two options

-6c + 3 = 4c-2

5 = 10c

1/2 = c

or

6c - 3 = 4c-2

-1 = - 2c

1/2 = c

But f' (1/2) is not defined, therefore there is no value of c such that f (3) - f (0) = f' (c) (3-0).