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Math Help - Please help- one-to-one functions

  1. #1
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    Please help- one-to-one functions

    Could someone please explain to me how you can tell if a function is one-to-one? I don't understand...
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by runner07 View Post
    Could someone please explain to me how you can tell if a function is one-to-one? I don't understand...
    do you know what it means to be one-to-one? what is you're definition of one-to-one functions.


    graphically: we can show that a graph is a one-to-one relation by the horizontal line test. drawing any horizontal line will cut the graph ONLY once


    analytically: we can show a graph f(x) is one-to-one by showing that if f(x_1) = f(x_2) then x_1 = x_2

    or equivalently, if we can find any two none equal elements in the domain, x_1 \neq x_2 then we would find that f(x_1) \neq f(x_2)
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  3. #3
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    i understand that one-to-one is using the horizontal line test but i cant always draw a graph to check. its the analytical part that confuses me.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by runner07 View Post
    i understand that one-to-one is using the horizontal line test but i cant always draw a graph to check. its the analytical part that confuses me.
    example:

    the function y = x + 1 is one-to-one, since if we have:

    x_1 + 1 = x_2 + 1 (which is f(x_1) = f(x_2))

    we can subtract 1 from both sides to get:

    x_1 + 1 - 1 = x_2 + 1 - 1

    \Rightarrow x_1 = x_2

    thus the function is one-to-one




    counter-example:

    the function y = x^2 is not one-to-one

    since if we have f(x_1) = f(x_2) it is not necessarily true that x_1 = x_2

    look: x_1^2 = x_2^2
    square rooting both sides we obtain:

    x_1 = \pm x_2 in general

    a specific example would be, f(1) = f(-1) since 1^2 = (-1)^2 = 1 (you could just state this example by the way to prove it's not one-to-one, no explanation needed)

    so there are two elements in the domain that map to a single element in the range. thus we have a many-to-one relation, not one-to-one
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  5. #5
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    when you take the y=x^2 and you do this..

    x1^2 = x2^2

    why is it only the x2 that becomes +- when you square root it? why doesnt x1 become +- too?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by runner07 View Post
    when you take the y=x^2 and you do this..

    x1^2 = x2^2

    why is it only the x2 that becomes +- when you square root it? why doesnt x1 become +- too?
    that's why i said "in general."

    let's say i did do that:

    \pm x_1 = \pm x_2

    how many possible combination of signs can we have? well 4 of course

    1. x_1 = x_2

    2. x_1 = - x_2

    3. -x_1 = -x_2

    4. -x_1 = x_2

    Of course, we realize that 1 and 3 are saying the same thing, and 2 and 4 are saying the same thing. so i can summarize these 4 possibilities into two possibilities

    1. x_1 = x_2 .............(this is equation 1 above, which is the same as equation 3)

    2. x_1 = -x_2 ...........(this is equation 2 above, which is the same as equation 4)

    Now I can summarize these two equations into one:

    namely, x_1 = \pm x_2


    but like i said, this kind of reasoning is not necessary, saying "since f(1) = f(-1), we have that f(x) is NOT one-to-one" will suffice
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