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Math Help - Left and Right inverse

  1. #1
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    Left and Right inverse

    Dear everyone.
    I am really sorry.
    Could you help me out?

    1. Consider the function f : R -> R defined by
    f(x) = x^2 + 2 if x > 0 and x + 1 if x <= 0.
    Find a left inverse of f.


    2. Consider the function f : R -> R defined by
    f(x) = x^2 - 2 if x > 0, x - 1 if x <= 0.
    Find a right inverse of f.

    Can I ask if a left and right inverse means just swap x with y?
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  2. #2
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    Re: Left and Right inverse

    No. I'm afraid that the terminology "left inverse" and "right inverse" being used here are being used in the wrong context. My guess is that because a function can only have an inverse if it is one-to-one on its domain, it is asking you to break up the function into the (maximal) parts where the function is one-to-one (one will point left and the other will point right), and then find their corresponding inverse functions.
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  3. #3
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    Re: Left and Right inverse

    A left inverse is a function g such that g(f(x)) = x for all x in \mathbb{R}, and a right inverse is a function h such that f(h(x)) = x for all x in \mathbb{R}. An inverse is both a right inverse and a left inverse.

    In problem 1, the range of f is [-\infty,1] \cup [2,\infty]; you can define the function piecewise - something for [-\infty,1] and something else for [2,\infty]. The function needs to undo what f has done, and it doesn't matter what g does on (1,2) since g will never see an input in that interval.

    For problem 2, you need to think the other way - you need to figure out what to feed f so that it will give you back your original input. The range of f is \mathbb{R} since the two pieces of f overlap on the interval [-2,-1]. On that interval, you can choose what you want h to do - either output something that will trigger the first piece and give you back x or output something that will trigger the second piece and give you back x. Above and below [-2,-1] you don't have a choice - only one piece will work.

    - Hollywood
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    Re: Left and Right inverse

    I deeply appreciate it.
    With all due respect, may I ask the answers as well?

    As for the right inverse, would it be:

    h(x) = x+1 , when x <= 0

    and h(x) = +- root(x+2), when h(x) is in R?

    May I ask if I am doing this in a correct way?
    Last edited by yanirose; January 26th 2014 at 11:13 PM.
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  5. #5
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    Re: Left and Right inverse

    Quote Originally Posted by hollywood View Post
    A left inverse is a function g such that g(f(x)) = x for all x in \mathbb{R}, and a right inverse is a function h such that f(h(x)) = x for all x in \mathbb{R}. An inverse is both a right inverse and a left inverse.

    In problem 1, the range of f is [-\infty,1] \cup [2,\infty]; you can define the function piecewise - something for [-\infty,1] and something else for [2,\infty]. The function needs to undo what f has done, and it doesn't matter what g does on (1,2) since g will never see an input in that interval.

    For problem 2, you need to think the other way - you need to figure out what to feed f so that it will give you back your original input. The range of f is \mathbb{R} since the two pieces of f overlap on the interval [-2,-1]. On that interval, you can choose what you want h to do - either output something that will trigger the first piece and give you back x or output something that will trigger the second piece and give you back x. Above and below [-2,-1] you don't have a choice - only one piece will work.

    - Hollywood
    I deeply appreciate it.
    With all due respect, may I ask the answers as well?

    As for the right inverse, would it be:

    h(x) = x+1 , when x <= 0

    and h(x) = +- root(x+2), when h(x) is in R?

    May I ask if I am doing this in a correct way?
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  6. #6
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    Re: Left and Right inverse

    No...that's not right.

    Ok, if h(x) = x+1 (we'll work on the domain later), and f(x) = x-1, then:

    f(h(x)) = (x+1)-1 = x + (1-1) = x + 0 = x.

    Now, we can only use f(x) = x - 1, when x ≤ 0. Since we are taking f(h(x)), we actually want h(x) ≤ 0, so setting:

    h(x) = x + 1 ≤ 0, we find x ≤ -1.

    In other words, if we set h(x) = x + 1 for all x ≤ -1, then h(x) ≤ 0, so we can "hit" h(x) with the part of f where f(x) = x - 1.

    So what do we do when x > -1?

    We need a function h so that h(x) > 0, when x > -1. It can be LOTS bigger than 0, it doesn't have to be something like:

    (positive something) + 1.

    Since for x > 0, a "normal" inverse for f(x) = x2 - 2 is: k(x) = √(x + 2)

    it seems natural to use h(x) = √(x + 2) for x > -1 (this makes sense, h(x) is defined for all x ≥ -2).

    Then: f(h(x)) = (√(x + 2))2 - 2 = |x + 2| - 2 = x + 2 - 2 = x + 0 = x (we know that |x + 2| = x + 2, since x + 2 > -1 + 2 = 1 > 0).

    We could however, define as well:

    h(x) = x + 1, for x ≤ -2
    h(x) = √(x + 2) for x > 2

    In fact, we can pick ANY point a in (-2,-1] and set:

    h(x) = x + 1, for x ≤ a
    h(x) = √(x + 2) for x > a (this is what hollywood was getting at).

    Even more complicated choices are possible, but that gets into some complicated set-theory, so we'll stick with simple intervals.

    Note that our choice of "a" gives us a DIFFERENT h each time, meaning the right inverse of f is not UNIQUE (we have lots of possible ones to choose from).

    Your proposal, choosing a = 0, does not work, though. To see this, suppose x = -1/2.

    For that x, you have h(x) = x + 1 (since -1/2 ≤ 0).

    now h(-1/2) = -1/2 + 1 = 1/2 > 0.

    so we must use the form of f: f(x) = x2 - 2, and so:

    f(1/2) = 1/4 - 2 = -7/4, so that f(h(-1/2)) = -7/4 ≠ -1/2.

    One more point:

    h(x) = √(x + 2) is NOT A FUNCTION. Functions must be "single-valued": for each input, only one output must be defined.
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