Question

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?​

Answer 1

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).