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AP Calculus BC: Differentiability and continuity

  1. Mar 6, 2017 #1
    1. The problem statement, all variables and given/known data
    The function h is differentiable, and for all values of x, h(x)=h(2-x) Which of the following statements must be true?

    1. Integral (from 0 to 2) h(x) dx >0
    2. h'(1)=0
    3.h'(0)=h'(2)=1

    A. 1 only
    B.2 only
    C. 3 only
    D. 2 &3 only
    E. 1,2 &3

    2. Relevant equations

    None that I am aware of

    3. The attempt at a solution
    If h(x) = h(2-x) then h'(x) must also = h'(2-x)
    Therefore, when x=0, h'(0)=h'(2)
    However, we cannot say for sure that both = 1. Thus I eliminate any options with '3'.
    Have been stuck here for a while, please help.
     
  2. jcsd
  3. Mar 6, 2017 #2

    BvU

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    What if you look at statement 2 with this conclusion in mind ?
     
  4. Mar 6, 2017 #3

    Dick

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    I'm having a hard time agreeing with the statement that ##h'(x)=h'(2-x)##. Why don't you try and construct a function that satisfies ##h(x)=h(2-x)## and differentiate it? Then substitute ##2-x## for ##x## and see what happens to the derivative?
     
  5. Mar 6, 2017 #4

    BvU

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    Agree with DIck. My mistake. Re-think ##h'(x) = h'(2-x)## (e.g using the chain rule) That way you get something much more usable for x = 1 :smile:
     
  6. Mar 6, 2017 #5
    I tried using the chain rule. What i get is h'(x) = h'(2-x) (-1) = -h'(2-x)
    When evaluated at x=1 thos gives h(1)= -h(1) which makes no sense to me
     
  7. Mar 6, 2017 #6
    Also, I'm a bit lost as to how i would test the first condition, Inorder to do so wouldnt i need to know that h(x) lies mostly or wholly above the x axis?
     
  8. Mar 6, 2017 #7

    Ray Vickson

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    No, that is not what that equation gives. Try again.
     
  9. Mar 6, 2017 #8
    There is an easy counter-example to rule out the first and third condition.
     
  10. Mar 6, 2017 #9
    You don't get h(1) = -h(1) from the previous equation. Be careful.
     
  11. Mar 6, 2017 #10

    BvU

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  12. Mar 6, 2017 #11
    I'm confused. According to the chain rule its the derivative of the outer function X the derivative of the inner function. Doesn't that give h'(x) = h'(2-x)(-1)?
     
  13. Mar 6, 2017 #12
    Yes. Now plug in x = 1. What do you get?
     
  14. Mar 6, 2017 #13
    h'(1)= -h'(1) ?
    How does this make sense?
     
  15. Mar 6, 2017 #14
    Good. Now solve for h'(1)
     
  16. Mar 6, 2017 #15
    Unless this always implies that h'(x) = 0? Is this reasoning correct?
     
  17. Mar 6, 2017 #16
    It doesn't imply that h'(x) = 0. It implies h'(1) = 0. Can you show me why?
     
  18. Mar 6, 2017 #17
    Phew, alright i think i get that part. Consequentially I understand we can eliminate condition 3 because h'0 will be -h'(2) and not h'(2) and also there is no way to say they are equal to 1 (Hope I'm right). How do you deal with the first condition though? I see no way to test the value of the integral
     
  19. Mar 6, 2017 #18
    Okay so h'(1)=-h'(1)

    2h'(1) = 0

    h'(1) = 0

    Is this correct?
     
  20. Mar 6, 2017 #19
    This is correct. You can eliminate conditions 1 and 3 by using a counter example. There is a certain function, whose derivative is zero, that you can use. It satisfies condition 2 but contradicts condition 1 and 3.
     
  21. Mar 6, 2017 #20
    I'm afraid I don't follow. Wouldn't it have to be a constant function for the derivative to be zero? Why is important that the derivative of this counter example must be zero?
     
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