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).

1. sometimes

2. never true

3. sometimes

step-by-step explanation:

these are pretty self explanatory if you know what supp right and com angles are