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Function ?

  1. Oct 28, 2004 #1
    how do I solve this one?
    y=-2x^3-9x^2-60x
     
  2. jcsd
  3. Oct 28, 2004 #2
    You're going to have to supply more information about the problem. The actual question would be good.
     
  4. Oct 28, 2004 #3
    function inquiry question?

    this it function inquiry question?
     
  5. Oct 28, 2004 #4
    I don't think that is gonna help.
     
  6. Oct 28, 2004 #5

    T@P

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    you want to solve for the roots of this function?

    in that case, your function is
    [tex] 0 = -2x^3 -9x^2 -60x[/tex] solving for x
    in that case, factor out x from the right hand side,
    you get [tex] x \cdot (-2x^2 -9x -60) = 0 [/tex]
    then you can use the quadratic formula to solve for x.
    x would equalt to [tex] \frac{-b \pm \sqrt{b^2 -4ac}}{2a}[/tex]
    which by pluging and chuging gives you two answer.
    The final answer would then be x = 0, [tex]r_{1}[/tex], [tex]r_{2}[/tex]

    in the case that you are attempting to simply draw it with the help of calculus, i do not see where the problem lies. Simply take the first derivative, set it to 0, and continue normally. perhaps a little more information on what your desired answer is...?
     
  7. Oct 28, 2004 #6
    Perhaps, i am missing something here, but what's all the fuss about???

    Just write : -x(2x^2+9x+60) = 0. So 0 is one solution. Then you have to solve 2x^2+9x+60 = 0. Since D = 81-2*4*60 < 0 You have no real solutions. can you incorporated compelx numbers ? If so, just continue as in the D > 0 case. if not, there are no solutions here...


    marlon
     
  8. Oct 28, 2004 #7

    T@P

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    good pont marlon, there is only one solution. it is x = 0. sorry i didnt even bother to check what D was.
     
  9. Oct 28, 2004 #8
    Most of us are not mindreaders, and don't know what people are really asking when they say "solve" and then post a function.
     
  10. Oct 28, 2004 #9

    Sorry, but do you know another way to solve a function ???

    i think not...

    marlon, the mind-reader
     
  11. Oct 28, 2004 #10
    I don't know of any way to "solve a function", just like I don't know how to "solve a banana".
     
  12. Oct 28, 2004 #11
    mmmmm

    sweden, ja???

    are you into ABBA ???

    marlon
     
  13. Oct 28, 2004 #12
    Nope. How many waffles have you eaten today...?
     
  14. Oct 28, 2004 #13
    5

    marlon
     
  15. Oct 29, 2004 #14

    dav2008

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    Gold Member

    y=x

    solve that function
     
  16. Oct 30, 2004 #15

    hahaha, please ask a more difficult question...


    answer : x = 0

    marlon
     
  17. Oct 30, 2004 #16
    Solving means finding the solutions guys...

    wow, what a revealing theory...

    marlon
     
  18. Oct 30, 2004 #17

    dav2008

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    Gold Member

    That answer is only true if y is 0.
     
  19. Oct 30, 2004 #18
    yes indeed it is...

    marlon
     
  20. Oct 30, 2004 #19
    if y were to be 5 the the question should be : solve x - 5. hence the answer is 5...


    beware that you do not violate the injectivity part of the definition of a function. With one x value, there can be at least one y-value, otherwise we do not have a function. Eg. x = 6 is NOT a function. y = x - 6 sure is...
    Ofcourse with one y-value there can be several x-values like in y = x²


    marlon
     
  21. Oct 30, 2004 #20

    HallsofIvy

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    "Solving means finding the solutions guys...

    wow, what a revealing theory...

    marlon"

    You still haven't understood what everyone is saying! Yes, "solve" means find the solutions- but to what problem?

    The orginal post just said "How do I solve this one? y= 3x<sup>3</sup> -9x<sup>2</sup>- 60x"

    That's not a "problem" that's just a statement. You then ASSUMED that the problem was "find all values of x that make y= 0" but that certainly is NOT the only possible problem that could be associated with a function.
     
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