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Calculus help !

  1. Jul 11, 2005 #1
    Calculus help plz!

    The function f(x)= ax3 + bx2 + cx has a point of inflection at the origin and a local maximum at the point (2,4). Find the values of a, b and c.

    I understand that the point of inflection is (0,0) and the local maximum at (2,4) but how can u find a, b & c using these values?
     
  2. jcsd
  3. Jul 11, 2005 #2
    I would start by differentiating what you have.
     
  4. Jul 11, 2005 #3

    LeonhardEuler

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    You need three equations for your three unkowns. The point of inflection indicates that the second derivitive is zero at (0,0). The local maximum at (2,4) gives you two pieces of information which can give you equations. First of all, f(2) must be 4 in order for the graph to pass through this point. Secondly the derivitive is zero at two. This gives you three equations in three unkowns, so you can solve them.
     
  5. Jul 11, 2005 #4

    James R

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    Hint:

    What do f'(x) and f''(x) tell you about points of inflection, local maxima etc.?

    Can you find f'(x) and f''(x) here, at the points x=0 and x=2?
     
  6. Jul 11, 2005 #5
    so is this how it is done:

    f(x)= ax3 + bx2 + cx

    4=a(2)^3 + b2^2 + 2c
    4=8a + 4b +2c
    4=2(4a + 2b + c)
    2=(4a + 2b + c)

    0=3ax^2 + 2b^2 +c0
    0=3a(4) + (4)b + c
    c= -12b - 4b

    6ax + 2b = 0
    b=-3ax
    b=0

    f(x)= ax3 + bx2 + cx
    2=(4a + 2b + c)
    2=4a
    a=1/2
     
  7. Jul 11, 2005 #6
    what am i doing wrong here, is b=0?
     
  8. Jul 11, 2005 #7
    since the inflection point is x=0 b should be 0 if your work is correct. ur second derivative is right.
     
  9. Jul 11, 2005 #8
    im just really being hesitant on the fact that since b=0 then c must also equal 0
     
  10. Jul 11, 2005 #9
    but does a=1/2
     
  11. Jul 11, 2005 #10
    well if b and c equal 0 then a = 1/2
     
  12. Jul 11, 2005 #11
    ok b does equal 0 but what about c
     
  13. Jul 11, 2005 #12
    k nelson....if b=o then c=0 as well right according to the calculations...or is that wrong too?
     
  14. Jul 11, 2005 #13

    dextercioby

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    I get

    [tex] \left\{\begin{array}{c} a=-\frac{1}{4} \\ b=0 \\ c=3 [/tex],

    therefore

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

    Daniel.
     
  15. Jul 11, 2005 #14
    yes, the calculation you may have made a minor error


    0=3a(4) + (4)b + c
    c= -12b - 4b

    yes c would equal zero according to this statement

    but i think you meant
    c=-12a - 4b
     
  16. Jul 11, 2005 #15
    thx for ur help but dexter was write...thx guys
     
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