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Math Help - Finding the inverse of f(x)

  1. #1
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    Finding the inverse of f(x)

    The given function f is one-to-one. Find f-1.
    f(x)=\sqrt{x-7}

    I can't seem to do the correct algebra to get the inverse. If someone could guide me a little, that would be wonderful. Thank you.
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  2. #2
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    Let y = f^{-1}(x) and change the x's and f(x)'s:
    x = \sqrt{y-7}

    Simply solve for y.
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  3. #3
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    Thanks, I understand that part. Afterwards, however, do I move the squareroot to the x? I tried...
    x^2=(y-7)
    x=(y+7)^2

    So is that the inverse (incomplete answer)? I don't believe that's right. i'm just not getting something in the algebra. ugh..
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  4. #4
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    You're just solving for y in terms of x:
    \begin{array}{rcl} x & = & \sqrt{y - 7} \\ x^2 & = & y - 7 \\ & \vdots & \end{array}

    There's only one step left to do ...
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  5. #5
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    y+x^2=-7
    y=-x^2-7 ??
    Then turn it into...
    f^-1(x)=-x^2-7??

    Is that it? Things are always much simpler than they seem, lol. Thank you!
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  6. #6
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    Quote Originally Posted by tiar View Post
    y+x^2=-7
    y=-x^2-7 ??
    Then turn it into...
    f^-1(x)=-x^2-7??

    Is that it? Things are always much simpler than they seem, lol. Thank you!
    Close it is

    x^2 + 7 = y
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  7. #7
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    You copied incorrectly.

    \begin{array}{rcl} x^2 & = & y - 7 \\ x^2 + 7 & = & y \qquad \text{Added 7 to both sides} \\ y & = & x^2 + 7 \end{array}

    However, note that the range of f(x) was f(x) \geq 0. Since we interchanged x's and y's, the domain of f^{-1}(x) is x \geq 0
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  8. #8
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    Thanks both of you. It was illogical of me to move the "y" over when it could have been isolated right away. At least I see that now.

    I have a question in regards to range. You are saying it is f(x) \geq 0 because in the original equation, f(x)=\sqrt{x-7}, we cannot have a negative under the square root thus it must be greater than zero. How do you know that this is your range and not domain? The domain would be f(x)= \geq7, correct? Wait, did I just answer my own question? I just need clarification.
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  9. #9
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    When finding inverses just switch the domain and range.

    So if f(x) domain is A and f(x) range is B then f^{-1}(x) domain is B and f^{-1}(x) range A

    Does this make sense?
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  10. #10
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    Yes, that makes perfect sense, thanks.
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