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Math Help - Proving inverses

  1. #1
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    Proving inverses

    Work in groups of two or three to

    a. find f^-1 and show that (f of f^-1)(x)= (f^-1 of f)(x) = x
    b. graph f and f^-1 in the same viewing window.

    f(x)=2-3x

    note: the of is in replacement of the that little unfilled circe that's supposed to be between f and f^-1, and f^-1 and f.

    Here's what i did so far and where I got stuck:

    y=2-3x
    x=2-3y
    x-2=3y
    y=(x-2)\3

    I now found my inverse equation by switching the x and y values and solving.

    Now when I'm trying to prove that (f of f^-1)(x)= (f^-1 of f)(x) = x ... you have to put the equations into each other and come out with an answer of x, so...

    y=2-3x
    y=(x-2)/3

    2-3(x-2/3)
    2-3x+6 ... now what happens when I multiply the -3 with the 3 in the denominator, does it come out to be -9 or do they cancel out to a negative? I'm lost from that point on.
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  2. #2
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    Here's what i did so far and where I got stuck:

    y=2-3x
    x=2-3y
    x-2=-3y
    y=(x-2)\-3
    I put the revisions in red. Now when you are trying to show f(f^{-1}(x)) you should get:

    y=2-3x

    y=2-3(\frac{x-2}{-3})

    y=2+(x-2)

    y=x

    You would get the exact same answer if you switched which function you were composing the other of.
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  3. #3
    MHF Contributor

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    Quote Originally Posted by fezz349 View Post
    Work in groups of two or three to

    a. find f^-1 and show that (f of f^-1)(x)= (f^-1 of f)(x) = x
    b. graph f and f^-1 in the same viewing window.

    f(x)=2-3x

    note: the of is in replacement of the that little unfilled circe that's supposed to be between f and f^-1, and f^-1 and f.

    Here's what i did so far and where I got stuck:

    y=2-3x
    x=2-3y
    x-2=3y
    y=(x-2)\3

    I now found my inverse equation by switching the x and y values and solving.

    Now when I'm trying to prove that (f of f^-1)(x)= (f^-1 of f)(x) = x ... you have to put the equations into each other and come out with an answer of x, so...

    y=2-3x
    y=(x-2)/3

    2-3(x-2/3)
    2-3x+6 ... now what happens when I multiply the -3 with the 3 in the denominator, does it come out to be -9 or do they cancel out to a negative? I'm lost from that point on.
    As long as a is not 0, a/a= 1. "/a" indicates you are dividing not multiplying.
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