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Help with hard question

  1. Jun 11, 2007 #1
    hey here is my questions:
    Find all y satisfying the inequality z<y^2-6/y-1

    problem is i do not know where to being, considering there is 2 unknowns. And the top doesnt break down which could simplify the question. Could someone help me out or help me start this question off?

    thank you.
     
  2. jcsd
  3. Jun 11, 2007 #2

    malawi_glenn

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    and is y a real or complex number? or integer?..

    if y is a real number:

    set f(y) = y^2 - 6/y - 1 - z
    and search for intervals of y, that satisfy f(x) > 0

    i.e "draw" the graph of f(y)

    N.B : The interval may contain some factors of z.. ;)
     
    Last edited: Jun 11, 2007
  4. Jun 11, 2007 #3
    how can we solve anything, or find the intervals with 2 unknowns? .. we dont have 2 solutios 2 unknowns, so im kinda lost, i dont think i can graph it either with all 2 unknowns.. still need a bit more guidance sry, im really lost with this one.
     
  5. Jun 11, 2007 #4

    malawi_glenn

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    as I said, but not in exact words, you are supposed to give the intervall of y, expressed i z...

    also you must say what z and y is for numbers.. real, integer or complex?!
     
  6. Jun 11, 2007 #5
    the question was just given to me like this .. im going to assume it is for real numbers.
     
  7. Jun 11, 2007 #6
    and i forget, how would i solve for the interval? and would it have 2?
     
  8. Jun 11, 2007 #7

    malawi_glenn

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    you can always push the "edit" button if you want to add.

    well if we assume real number, we can differentiate the function f(y)

    then do the usual things, to the limits at the singularities (+/- infintiy and y=0), and find max/min and so on...

    I assuming that this is a course in one dimensional analysis, you should know how to do limits, do derivatives, find and interprent the df(y) = 0. finding max- and min of functions and so on.
     
    Last edited: Jun 11, 2007
  9. Jun 11, 2007 #8

    HallsofIvy

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    What you can do with 2 variables is graph it, in a "y-z plane". The graph of the equation z= y^2-6/y-1 shows where the two sides are equal. On one side of that graph, z< y^2-6/y-1, on the other, z> y^2-6/y-1.
     
  10. Jun 11, 2007 #9
    for3 you use limit, so what does it do when x approaches + - infinity.

    so, you would get it down to= y^2 -6/y -1 -z

    then divide each by y^2

    = 1- 6/y^3 - 1/y^2 - z/y^2

    so when x approaches + - infiinity the limit is 1.


    hmm never thought of using the y-z plane
     
  11. Jun 11, 2007 #10

    malawi_glenn

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    no you can not divide as you like..

    you must have the same function..

    the limit when y goes to +/- inf is +inf scince y^2 goes to + inf and all others goes to 0 or is constant.
     
  12. Jun 11, 2007 #11
    so what you are saying is to just graph it, then it will show we 2 equations where Z is < and > then y^2 - 6 / y-1 .. then what do i do with these to solve the equation?
     
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