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Range of this function

  1. Aug 31, 2010 #1
    [tex]f(x) = \frac{1 - x}{\sqrt{5 + 7x - x^{2}}}[/tex]

    I know that the range of that function is [tex](-\infty, 0) U (0, +\infty)[/tex]

    But how do I get it? I'm having difficulties in isolating x. Can you explain it to me? Thank you. :)
     
  2. jcsd
  3. Sep 1, 2010 #2

    HallsofIvy

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    You don't get that range- it's wrong. It is easy to see that
    [tex]f(1)= \frac{1- 1}{\sqrt{5+ 7- 1}}= \frac{0}{\sqrt{11}}= 0[/tex]
    so 0 certainly is in the range.

    Since you have determined what x can be, what values of f do those values of x give?
     
  4. Sep 1, 2010 #3
    Moreover, for sufficiently large |x|, the denominator is not real.
     
  5. Sep 2, 2010 #4

    HallsofIvy

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    Right. [itex]5+ 7x- x^2= 0[/itex] when [itex]x= (7\pm\sqrt{49+ 20})/2[/itex] so that the domain of the function is only from [itex](7- \sqrt{69})/2[/itex] to [itex]7+\sqrt{69})/2[/itex], or from between -1 and 0 to between 7 and 8. The function goes to infinity as x approaches the end points. The lower bound on the range will occur at the minimum of the function.
     
  6. Sep 2, 2010 #5

    uart

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    Assuming we're talking about real functions then the domain is [itex](7-\sqrt{69})/2 < x < (7+\sqrt{69})/2[/tex] and the range is all real [itex]y[/tex].

    Halls, note that the function goes to +infinity at the lower end of it's domain and to -infinity at the upper end of it's domain and in continuous in between.
     
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