Ask Question
25 April, 16:23

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?

+2
Answers (1)
  1. 25 April, 16:37
    0
    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).
Know the Answer?
Not Sure About the Answer?
Find an answer to your 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 ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers