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Thread: Inverse Fcns: For f(x) = 1/(3x + 2), f^(-1)(x) = (1 - 2x)/(3x), show f(f^(-1)(x)) = x

  1. #1

    Inverse Fcns: For f(x) = 1/(3x + 2), f^(-1)(x) = (1 - 2x)/(3x), show f(f^(-1)(x)) = x

    yo. whats the missing math here.......i cant seem to figure it out tonight.


    [tex]f(f^{-1}(x)) = \frac{1}{3(\frac{1-2x}{3x})+2}=...=x[/tex]


    right? the composition of a function and its inverse is x?


    i just get [tex]\frac{1-2x}{x}+2 = \frac{1}{x}[/tex]



    Edit: nevermind i got it....writing it down on here seemed to help for some reason.
    Last edited by doughishere; 11-23-2017 at 10:53 PM.
    LaTeX Cookbook (How to write equations) - http://www.personal.ceu.hu/tex/cookbook.html
    Highly Recommend -
    1) College Algebra and Problem Solving - https://courses.edx.org/courses/course-v1:ASUx+MAT117x+1T2016/course/
    2) Precalculus - ​https://courses.edx.org/courses/course-v1:ASUx+MAT170x+2T2017/course/



  2. #2
    Junior Member
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    Quote Originally Posted by doughishere View Post
    yo. whats the missing math here.......i cant seem to figure it out tonight.

    [tex]f(f^{-1}(x)) = \frac{1}{3(\frac{1-2x}{3x})+2}=...=x[/tex]

    right? the composition of a function and its inverse is x?

    i just get [tex]\frac{1-2x}{x}+2 = \frac{1}{x}[/tex]

    nevermind i got it....writing it down on here seemed to help for some reason.
    Yes, sometimes explaining your thinking to someone else (or to yourself!) can help you see errors.

    Presumably you found that the inverse of [tex]f(x) = \frac{1}{3x+2}[/tex] is [tex]f^{-1}(x) = \frac{1-2x}{3x}[/tex] and your check initially failed, because you made an algebra error. But now you see that it succeeds, right?

  3. #3
    yup.... when you actually understand this mathematics its amazing. and its so much more fun when you get the right answers or better yet truly understand whats happening.
    Last edited by doughishere; 11-23-2017 at 10:52 PM.
    LaTeX Cookbook (How to write equations) - http://www.personal.ceu.hu/tex/cookbook.html
    Highly Recommend -
    1) College Algebra and Problem Solving - https://courses.edx.org/courses/course-v1:ASUx+MAT117x+1T2016/course/
    2) Precalculus - ​https://courses.edx.org/courses/course-v1:ASUx+MAT170x+2T2017/course/



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