Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find a cubic function

  1. Mar 26, 2006 #1
    Find a cubic function [tex] f(x) = ax^3 - bx^2 + cx - d [/tex] that has a local maximum value of 40 at x = 1 and a local minimum valud of -68 at x = 4

    Since x = 1 and x = 4 are the max and min, respectively, then f'(x) must equal 0 at x = 1 and x = 4. Therefore [tex] f'(x) = (x-1)(x-4) = x^2 - 5x +4 [/tex]

    Using the original function [tex] f'(x) = 3ax^2 - 2bx +c = x^2 - 5x + 4[/tex]

    Setting the terms equal to each other we get
    [tex] 3ax^2 = x^2 \Rightarrow a = \frac{1}{3} [/tex]
    [tex] -2bx = -5x \Rightarrow b = \frac{5}{2} [/tex]
    [tex] c = 4 [/tex]

    Substituting these values back into the original equation I get

    [tex] f(x) = \frac{x^3}{3} - \frac{5x^2}{2} + 4 - d [/tex]

    When I solve for d using x = 1 and f(x) = 40 I get

    [tex] d = \frac{1}{3} - \frac{5}{2} - 36 = \frac{-229}{6} [/tex]

    However, when I solve for d using x = 4 and f(x) = -68 I get

    [tex] d = \frac{4^3}{3} - \frac{5(4^2)}{2} + 84 = \frac{226}{3} [/tex]

    Any ideas where I went wrong, or if I was down the wrong track to start with?

    Thanks
    Jeff
     
  2. jcsd
  3. Mar 26, 2006 #2

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Eeh, from what I can see, we must have:
    [tex]f'(x)=K(x-1)(x-4)[/tex], where K is some arbitrary constant.
     
  4. Mar 26, 2006 #3
    So if that was brought into the equation, would I be solving for a, b, c, & d in terms of K?
     
  5. Mar 26, 2006 #4

    StatusX

    User Avatar
    Homework Helper

    No, you need to solve for a value of K and a value for the integration constant which give you the desired values at x=1 and 4. And don't worry about solving a,b,c,d seperately as unknown variables. Just find the right cubic and then you just can read them off.
     
  6. Mar 26, 2006 #5

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Well, there was nothing inherently wrong about Jeff's idea. A bit cumbersome, that's all.
     
  7. Mar 26, 2006 #6
    Sorry, not familiar with the idea of an itegration constant. We just finished sections of optimization problems and Newton's method. The next section is antiderivatives. So I have to solve this without any concept of integration.
     
  8. Mar 27, 2006 #7
    Could someone direct me towards information on finding the integration constant, if that is indeed the most direct way to solve this problem?

    Thanks
    Jeff
     
  9. Mar 27, 2006 #8

    benorin

    User Avatar
    Homework Helper

    Scratch the whole integration constant idea gig: no integration for you.
     
  10. Mar 27, 2006 #9

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    f'= k(x-1)(x- 4)= kx^2- 5kx+ 4k= 3ax^2+ bx+ c so
    a= k/3, b= -5k and c= 4k. Evaluate f(x)= (k/3)x^3- 5kx^2+ 4kx+ d at 1 and 4 to get two equations for k and d.
     
  11. Mar 27, 2006 #10

    benorin

    User Avatar
    Homework Helper

    We have 4 unknowns, hence we need 4 equations to solve: they are f(1)=40, f(4)=-68, f'(1)=0, and f'(4)=0.
     
  12. Mar 27, 2006 #11

    J77

    User Avatar

    Where you went wrong in finding d in your OP was substituting the values at f'(x)=0 into f(x)...

    You may also want to use: f''(1)<0, f''(4)>0
     
  13. Mar 27, 2006 #12
    Thanks to all. I used K(x-1)(x-4) = 0, solved for a, b, and c, then went back to the original equation and used x = 1 and x = 4 to get two equations for d in terms of K. I set them equal, solved for K, and when I plugged it back in I ended up with values for a, b, c, and d that satisfied all conditions.
     
  14. May 15, 2007 #13
    Can anyone explain this further?

    Can anyone explain this further? I have a similar problem, and I'm having trouble following what the correct answer Jeff figured out in the end was.
    How does one make an equation for d in terms of K, for instance?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Similar Discussions: Find a cubic function
Loading...