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Finding unknown constants

  1. Apr 19, 2015 #1
    1. The problem statement, all variables and given/known data
    determine the constants a,b,c, and d so that the function f(x)=ax^3+bx^2+cx+d has its first derivative equal to 4 at the point (1,0) and its second derivative equal to 5 at the point (2,4)

    2. Relevant equations


    3. The attempt at a solution
    I found the first and second derivative
    f'(x)=3ax^2+2bx+c
    f''(x)=6ax+2b

    I set
    5=12a+2b
    I also set
    4=3a+2b+c
    I get stuck trying to find the two different variables when working with the second derivative first. I know I have to substitute for the unknowns, but I don't know where to start.
     
  2. jcsd
  3. Apr 19, 2015 #2
    Tricky problem but if a function f(x) has a derivative at some specific point then what can you say about f(x)?
     
  4. Apr 19, 2015 #3
    That f(x) is continuous at that specific point?
     
  5. Apr 19, 2015 #4
    Yes, that's true, but I am thinking even more obviously than that?
     
  6. Apr 19, 2015 #5

    SammyS

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    @MrJamesta

    What is f(1) ?
     
  7. Apr 19, 2015 #6
    You tell me.
     
  8. Apr 20, 2015 #7
    Oh, it's an x intercept.
     
  9. Apr 20, 2015 #8
    No, it's not.

    f(1)=?

    You have 4 unknowns so you need 4 equations to solve this. You came up with two of them. What are the other two? It's pretty obvious.
     
  10. Apr 20, 2015 #9

    Mark44

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    Just to be clear, and to reduce confusion on the part of the OP, 1 is an x-intercept.
     
  11. Apr 20, 2015 #10
    Well, what is an x-intercept? I think it is the value of the function when x=0. But f(1) is the value of the function when x=1. So it is not an x-intercept.

    Regardless, it is irrelevant to solve the problem. What are the other two equations?
     
  12. Apr 20, 2015 #11

    Mark44

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    No, that's the y-intercept, a point on the y-axis.
    But your reply to the OP was incorrect, possibly steering him/her in the wrong direction.
     
  13. Apr 20, 2015 #12

    mfb

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    To get back to the original question:
    Right. Can you use it to find another equation?
     
  14. Apr 20, 2015 #13
    I found
    0=a+b+c+d
    4=8a+4b+2c+d
     
  15. Apr 20, 2015 #14

    SteamKing

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    Now, use the other two equations and solve for the unknown coefficients.
     
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