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Math Help - Inverse functions and symmetry

  1. #1
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    Inverse functions and symmetry

    I have the line y=ax + b
    I am told that y=-ax + b is an inverse of the first line. Why is this so? I know that inverses are symmetrical about the line y=x. I graphed y=ax+3 and y=-ax+3, and they do not seem symmetrical about the line y=x to me.
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    MHF Contributor Siron's Avatar
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    Re: Inverse functions and symmetry

    Because the statement is not true, you can determine the inverse of y=ax+b by changing y and x and solve back to y=.., like this:
    y=ax+b
    Calculating the inverse:
    x=ay+b \Leftrightarrow y=\frac{x-b}{a}

    For example:
    y=3x+2
    The inverse function has to be
    y=\frac{x-2}{3}

    Now look at the graphs of this functions, they're symmetrical towards the first bissectrice (y=x).
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  3. #3
    Junior Member BERMES39's Avatar
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    Re: Inverse functions and symmetry

    To find the inverse function of y= ax+b, replace x for y and y for x, then solve for y:
    x=ay+b inverse: y= (x-b)/a
    for instance: y = 3x +2 The inverse: y = (x-2)/3. Try the graph, they are symmetrical to the line y=x.
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    Junior Member BERMES39's Avatar
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    Re: Inverse functions and symmetry

    Hey Siron we must have similar minds!
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    MHF Contributor Siron's Avatar
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    Re: Inverse functions and symmetry

    Quote Originally Posted by BERMES39 View Post
    Hey Siron we must have similar minds!
    Yes, I was thinking the same!
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  6. #6
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    Re: Inverse functions and symmetry

    So the two equations I are NOT inverses?
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  7. #7
    MHF Contributor Siron's Avatar
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    Re: Inverse functions and symmetry

    y=-ax+b is indeed not the inverse function of y=ax+b

    Like you said, if you graph these lines then they're not symmetrical to the first bissectrice ( y=x).

    Have you graphed the example I and BERMES39 have given you?
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