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Finding Critical Points

  1. Nov 28, 2011 #1
    1. The problem statement, all variables and given/known data
    I need to find the critical points of this functionhttp://www4a.wolframalpha.com/Calculate/MSP/MSP84719i6a1iabg11598d000049hhhie62iif1c81?MSPStoreType=image/gif&s=16&w=140&h=46 [Broken]

    3. The attempt at a solution
    I'm having trouble solving the first derivative, which is http://www4a.wolframalpha.com/Calculate/MSP/MSP12919i6a079bg6ed96c00003ed60iee4d866bb3?MSPStoreType=image/gif&s=39&w=493&h=48 [Broken]. Well more specifically solving the 5th degree polynomial. It needs to be solved without using a graphing calculator.
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 28, 2011 #2


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    Please write out the function/polynomial.

    Use Latex or text with [ /sup] tags as appropriate. Remove the space between [ /sup when using the superscript tag.
  4. Nov 28, 2011 #3
    Oh heres the derivative i need to solve -ex(3x5+x4-34x3+55x2+169x-216)/(x2-9)2
  5. Nov 28, 2011 #4

    Ray Vickson

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    You should show the function itself as well. How do we know your derivative is correct?

  6. Nov 28, 2011 #5
    Good point. Sorry im new to this.

    The function is (3x3+4x2-7x+17)/(ex(x2-9))
  7. Nov 28, 2011 #6


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    That derivative is correct.

    Unfortunately, the polynomial in the numerator does not appear to factor easily. The denominator is factorable.
  8. Nov 28, 2011 #7
    Ohhhhh so i now have (x+3)(x-3)(x+3)(x-3) in the denominator.. :/ sorry i dont see how thats suppose to help.
  9. Nov 28, 2011 #8


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    Unless it cancels things in the numerator (and you can check that by seeing if x= 3 or x= 3 will make the numerator equal to 0), factoring the denominator is irrelevant. A fraction, a/b, is equal to 0 only if a= 0. You don't need to worry about the denominator. It is also true that [itex]e^x[/itex] is never 0 so it all reduces to solving [itex]3x^5+ x^4- 34x^3+ 55x^2+ 169x- 216= 0[/itex].
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