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Finding constants in a piecewise function that allow the function to be differentiabl

  1. Oct 9, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the values of the constants a and b such that the function f(x) is differentiable on R


    2. Relevant equations
    f(x) = ax2 if x < 2

    f(x) = -4(x-3) + b if x >= 2


    3. The attempt at a solution
    ax2 = -4(x-3) + b
    2xa = -4x
    a = -2


    I believe that I need to equate the equations but with a value of a how do I find b and then prove that it's differentiable?
     
  2. jcsd
  3. Oct 9, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi souldoutt! Welcome to PF! :smile:

    (however did you get 2xa = -4x? :confused:)

    Hint: the only problem is at x = 2.

    So just bung x = 2 in, and check for continuity and differentiability. :smile:
     
  4. Oct 11, 2008 #3
    Re: Finding constants in a piecewise function that allow the function to be different

    I got the 2xa = -4x by differentiating both sides.
    But with 2 separate unknown constants, how would i solve for them? I can plug the value x = 2 into the equations but i wont get an answer to confirm whether the second part of the function is actually starting from x = 2.

    Then wouldn't I still need to have the constants in order to check whether the slopes of the tangents are the same? (therefore differentiable)?



    thanks for the welcome too.
     
  5. Oct 12, 2008 #4

    tiny-tim

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    Hi souldoutt! :smile:

    For continuity, if you put x = 2, you get f(2) = 4a and = 4 + b,

    so your continuity equation is 4a = 4 + b.

    And similarly for differentiability you get f'(2) = 4a = -8.

    So … ? :smile:
     
  6. Oct 12, 2008 #5

    HallsofIvy

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    Re: Finding constants in a piecewise function that allow the function to be different

    Yes, a must be -2. Now, you put a= -2 and x= 2 in the first equation you have -4= 4+ b. Solve for b.


    To prove it is differentiable, with the correct values for a and b, Look at the difference quotient limit.
     
  7. Oct 13, 2008 #6
    Re: Finding constants in a piecewise function that allow the function to be different

    It should be -8 instead of -4 right? Because it is ax2 which = -8 when the numbers are plugged in.

    Then, once I have the values of a and b the difference quotient limit to check for differentiability is the difference quotient limit of the derivatives correct?
     
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