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Domain of Inverse?

  1. Jan 15, 2012 #1
    1. The problem statement, all variables and given/known data

    The function f:x→ 4-x2 for the domain x≤0. Find the inverse of is denoted by f-1 and state the domain and range of f-1.

    2. Relevant equations
    Set equation to 0 and solve for x to find inverse. The D and R is going to be switched for the inverse..?


    3. The attempt at a solution
    I think I have found the inverse and range for this problem:
    f-1(x) = -√4-x

    and has a range of y≤0 (which was given in problem)

    not sure of range. Are my solutions correct?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 15, 2012 #2

    SammyS

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    You should really put 4-x in parentheses (or use LaTeX):
    f -1(x) = -√(4-x)
    The Range of f -1(x) is the same as the Domain of f(x). You are correct about this.

    Do you know the Range of f(x) ?
     
  4. Jan 16, 2012 #3
    No, That is my problem. I tried but I keep getting wrong answer. 4-x^2=0
    x=+/-2
     
  5. Jan 16, 2012 #4

    SammyS

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    How would you describe the graph of y = 4-x2 ?
     
  6. Jan 16, 2012 #5
    Its a parabola. It opens down, with the max point on 4.
     
  7. Jan 16, 2012 #6

    Mark44

    Staff: Mentor

    But you have a restricted domain, with x <= 0.
     
  8. Jan 16, 2012 #7
    I believe the domain and range of the inverse is
    x>4 and y≤0
     
    Last edited: Jan 16, 2012
  9. Jan 16, 2012 #8

    Mark44

    Staff: Mentor

    The domain of f is {x | x <= 0}
    The range of f is what? You know what the graph of f looks like, so you should be able to answer that by looking at the graph.

    Then look at what SammyS said in post #2 about the domain and range of f-1.
     
  10. Jan 16, 2012 #9
    Okay thanks! I know that it is going to be oppsite of whatever the original function was.
    Now that I see the graph, it would be x≤4.
     
  11. Jan 16, 2012 #10

    Mark44

    Staff: Mentor

    That's better.
     
  12. Jan 16, 2012 #11
    Thanks so much for your help!! I always have trouble with the domain and range.
     
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