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  1. #1
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    functions ...

    f(x) = √(2x + 12)

    find inverse.

    what i was taught was to write f(x) = y

    so y = √(2x + 12)

    then switch x and y

    so x = √(2y + 12)

    then rearrange for y

    so i get

    y = (1/2)√(x - 12)



    could anyone confirm this is right or wrong?
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by djmccabie View Post
    f(x) = √(2x + 12)

    find inverse.

    what i was taught was to write f(x) = y

    so y = √(2x + 12)
    You're right up to here for sure.

    then switch x and y

    so x = √(2y + 12)

    then rearrange for y

    so i get

    y = (1/2)√(x - 12)



    could anyone confirm this is right or wrong?
    What I was taught after putting f(x) = y is make x the subject, written in terms of y:

    y = \sqrt{2x^2+12}

    2x^2+12 = y^2

    <br />
2x^2 = y^2-12

    <br />
x^2 = \frac{1}{2}(y^2-12)

    x = \pm \sqrt{\frac{1}{2}(y^2-12)} = \pm \frac{\sqrt2}{2}{\sqrt{y^2-12}}

    Finally replace x with f^{-1}(x) and y with x:

    f^{-1}(x) = \pm \frac{\sqrt2}{2}{\sqrt{x^2-12}}

    Note that the plus/minus sign means it's not one to one and so not a function unless the domain is restricted
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  3. #3
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    mr fantastic's Avatar
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    Quote Originally Posted by djmccabie View Post
    f(x) = √(2x + 12)

    find inverse.

    what i was taught was to write f(x) = y

    so y = √(2x + 12)

    then switch x and y

    so x = √(2y + 12)

    then rearrange for y

    so i get

    y = (1/2)√(x - 12)



    could anyone confirm this is right or wrong?
    It should be y = \pm \sqrt{\frac{x^2 - 12}{2}}.
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  4. #4
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    Quote Originally Posted by e^(i*pi) View Post
    You're right up to here for sure.



    What I was taught after putting f(x) = y is make x the subject, written in terms of y:

    y = \sqrt{2x^2+12}

    2x^2+12 = y^2

    <br />
2x^2 = y^2-12

    <br />
x^2 = \frac{1}{2}(y^2-12)

    x = \pm \sqrt{\frac{1}{2}(y^2-12)} = \pm \frac{\sqrt2}{2}{\sqrt{y^2-12}}

    Finally replace x with f^{-1}(x) and y with x:

    f^{-1}(x) = \pm \frac{\sqrt2}{2}{\sqrt{x^2-12}}

    Note that the plus/minus sign means it's not one to one and so not a function unless the domain is restricted

    This is actually the same method except you switch x and y at the end
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