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P.O.I Question

  1. Mar 29, 2007 #1
    1. The problem statement, all variables and given/known data
    The slope of the curve y = ax(cubed) + bx(squared) + c is -12 at its point of inflection (2,-11), find a,b,c.


    2. Relevant equations
    f ' (x) = 3ax(squared) + 2bx
    f " (x) = 3ax + 2b


    3. The attempt at a solution

    f ' (x) = 3ax(squared) + 2bx
    -12 = 3a2(squared) + 2b(2)
    -12 = 12a + 4b
    b = -3a - 3

    -12 = 3a(2)(sqrd) + 2(-3a -3)
    a = -1

    b = 0

    -11 = ax(cubed) + bx(sqrd) + c
    c = -3
     
  2. jcsd
  3. Mar 29, 2007 #2
    That is correct.

    Another way to do it, and probably the one "expected" from you, is to use the fact that at the point of inflection, f''(x)=0.
     
  4. Mar 29, 2007 #3
    not exactly sure how to do that..

    6ax + 2b = 0?
     
  5. Mar 29, 2007 #4
    Yes, that's right. Therefore b = -6a. Solve for a.
     
  6. Mar 29, 2007 #5
    so b = 6 not 0?
     
  7. Mar 29, 2007 #6
    No, no!!
    b = -6a is one relation, and is b = -3a - 3 is another. Now solve for a. Sorry for the confusion.
     
  8. Mar 29, 2007 #7
    ur solution in OP is wrong . Check out step 5.
     
  9. Mar 29, 2007 #8
    which do you consider step 5?
     
  10. Mar 29, 2007 #9
    -12 = 3a(2)(sqrd) + 2(-3a -3)(?)
    a 2 is missing in (?)
     
  11. Mar 29, 2007 #10
    ok so a = 0 now
     
  12. Mar 29, 2007 #11
    Sorry, my bad. As Sourabh has pointed out, your solution in the first post is in fact incorrect. a is not 0. What you would get from "step 5" is 0=0. That is because you got the relation bet. a and b from that very equation.

    As I said earlier, use b = -6a and b = -3a -3 to solve for a.
     
  13. Mar 29, 2007 #12
    -12=3ax(sqrd)+2(-6a)x
    a=-1
    b=6
    c=-27
     
    Last edited: Mar 29, 2007
  14. Mar 29, 2007 #13
    Check again.

    Once you get values for a, b and c, you can check if they are right by putting x=2 in the original equation and looking at the value you end up with.
     
  15. Mar 29, 2007 #14
    ok im confused now.. ive found new values fro a(1),b(6) and c(5) but x doesnt = 2 when i enter it into the original equation.
     
    Last edited: Mar 29, 2007
  16. Mar 29, 2007 #15
    i think there's again a calculation mistake in #12.
    -12 = 3a2(sqrd) + 2(-6a)2
    -12 = -12a
    a=1.
     
  17. Mar 29, 2007 #16
    ok i got that one are the other two right? b=6 and c=5?
     
  18. Mar 29, 2007 #17
    b = -6 (b = -6a, remember?)
     
  19. Mar 29, 2007 #18
    no. b=-6a=-6

    -11 = 1*2(cubed) - 6*2(sqrd) +c
     
  20. Mar 29, 2007 #19
    a=1, b=-6, c =5
     
  21. Mar 29, 2007 #20
    is this correct?
     
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