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Trouble understanding range values

  1. Jan 5, 2013 #1
    For example: f(x) = 2 - √(x+9), x ≥ -9

    I'm asked to find the inverse and the domain of the inverse which is the range of f(x). I've found the inverse but for the domain, I put x ε ℝ however the answer is x≥2

    I can see how they got the answer but why is my answer incorrect. This is how I'm seeing it:

    for example: x = - 8 => f(x) = 2 - ±√(-8+9) leading to f(x) = 1, 3: but they seem to be taking the positive root only. Why?
     
  2. jcsd
  3. Jan 5, 2013 #2

    Ray Vickson

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    By definition, the "square-root" function is always ≥ 0. That is, the two roots of an equation of the form ##u = v^2## are ##v = \pm \sqrt{u}##; ##\sqrt{u}## itself is never < 0 (for non-negative real values of u).

    In your example, the values of ##y = 2 - \sqrt{x+9}## are always ≤ 2, and for any such y the corresponding x is obtained from
    [tex]\sqrt{x+9} = 2-y \: \Longrightarrow x+9 = (2-y)^2,[/tex]
    hence
    [tex] f^{-1}(y) = (2-y)^2 - 9, \: y \leq 2.[/tex]

    Of course, you can re-write this by using the symbol x instead of the symbol y; the domain would then be x ≤ 2 (NOT x ≥ 2).
     
  4. Jan 5, 2013 #3
    I see, thanks.
     
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