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Find the Values of a and b in a piecewise function. Help please!

  1. Sep 16, 2011 #1
    Find the values of a and b that make the function f(x) differentiable.



    f(x)={ ax^3 + 1 , for x<2
    { b(x-3)^2 +10, for x is greater than or equal to 2



    ax^3 + 1 = b(x-3)^2 +10
    3ax^2 = 2b(x-3)
    From here I'm unsure of where to go.
     
    Last edited by a moderator: Sep 16, 2011
  2. jcsd
  3. Sep 16, 2011 #2

    Mark44

    Staff: Mentor

    The two separate definitions are both continuous and differentiable on their separate domains, so you want to find a and b so that these functions are continuous and differentiable at the point where the definitions change. Where is that?
     
  4. Sep 16, 2011 #3
    The point where the definitions change is at x=2. Correct?
     
  5. Sep 16, 2011 #4

    Mark44

    Staff: Mentor

    Yes, correct.
     
  6. Sep 16, 2011 #5

    HallsofIvy

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    That's certainly a good start! You have used the fact that the function must be continuous at x= 2 and the derivatives must "match up". (A derivative is not necessarily continuous but it must satisfy the "intermediate value property" so that is correct. But the point is that must be true at x= 2. Set x= 2 in your equations and you will have two linear equations to solve for a and b.
     
  7. Sep 16, 2011 #6
    This is my work for setting x=2 in the two given equations.
    a(2)^3 +1 = b(2-3)^2 +10
    8a+1 = b+10
    b=8a-9

    What equation do I substitute "b=8a-9" to solve for the value "a" ?
     
  8. Sep 16, 2011 #7

    SammyS

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    f '(2) must exist. That means that [itex]\lim_{x\to2^+}f'(x)=\lim_{x\to2^-}f'(x)=f'(2)\,.[/itex]

    Differentiate f(x) in both regions, x > 2, and x < 2 .
     
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