1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Graph problem

  1. Feb 5, 2008 #1
    1. The problem statement, all variables and given/known data
    http://img220.imageshack.us/img220/2813/graphmo4.jpg [Broken]​

    2. Relevant equations
    Not necessary

    3. The attempt at a solution
    The answers that I have up there are correct, but I was wondering why graph 2 did not look something like this:
    http://img218.imageshack.us/img218/256/graph2rx8.jpg [Broken]
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Feb 5, 2008 #2


    User Avatar
    Homework Helper

    Let y=x^2 so x=sqrt(y)

    so [itex]f^{-1}(x)=\sqrt{x}[/itex] and [itex]f^{-1}(f(x))=\sqrt{x^2}[/itex]
  4. Feb 6, 2008 #3
    In fact, the inverse of [itex]y=x^2[/itex] is [itex]y=\sqrt{x}, x \geq 0[/itex] and [itex]y=-\sqrt{x}, x < 0[/itex]. It doesn't matter in the end, however, because the definition of an inverse says that [itex]f^-1(f(x)) = x[/itex]
  5. Feb 6, 2008 #4


    User Avatar
    Science Advisor

    Actually, that's a seriously bad question! Since y= x2 is not one-to-one, it does not have an inverse! temaire, the graph you showed would be the "inverse function" to x2 but, as you can see from the graph, there are two values of x for each positive y and so it is not a function.,
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook