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Range and inverse of bijective functions

  1. Nov 15, 2009 #1
    1. The problem statement, all variables and given/known data
    For each of the following bijective functions find the range S and the inverse:

    a.) f : x |→ x² - 1 (x ∈ R, x ≥ 0)

    b.) f : x |→ (x + 1)² ((x ∈ R, x ≤ -2)........Not sure how to do this one, help please

    2. The attempt at a solution

    a.) http://hotmath.com/images/gt/lessons/genericalg1/parabola.gif
    the range (S) for the graph is (0,∞)
    Inverse = x² - 1
    x = √(y + 1)
     
  2. jcsd
  3. Nov 16, 2009 #2
    Anybody able to help?
     
  4. Nov 16, 2009 #3

    HallsofIvy

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    2. The attempt at a solution

    a.) http://hotmath.com/images/gt/lessons/genericalg1/parabola.gif
    the range (S) for the graph is (0,∞)
    Inverse = x² - 1
    x = √(y + 1)[/QUOTE]
    a) No, the range is NOT (0,∞). Are you clear on what range is? What is f(0)? Your graph is for y= x2, not y= x2- 1.

    And you haven't completely described the inverse until you have told what its domain is.

    b) x+1 is just x, "shifted over 1". Do you know what the range of x2 is?
     
  5. Nov 16, 2009 #4
    The range is the set of output numbers of a function. f(0) would be -1
    The domain is the set of inputs for a function.

    The range of x2 is 0 or and postive R.
     
  6. Nov 16, 2009 #5
    Someone able to help?
     
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