Range of this function

  • Thread starter yik-boh
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  • #1
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[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. :)
 

Answers and Replies

  • #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?
 
  • #3
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Moreover, for sufficiently large |x|, the denominator is not real.
 
  • #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.
 
  • #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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