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).

not enough information

step-by-step explanation:

we cannot know this, because the other students could have handed out 10 flyers, or they could have handed out 100. one way, the mean would go up, and another way, the mean would go down.

step-by-step explanation:

this is not possible to solve with the info given.   you need at least an one angle somewhere.