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Math Help - Functions.

  1. #1
    Super Member Showcase_22's Avatar
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    Functions.

    Let g: \Re \rightarrow \Re be defined by g(x)=\frac{x}{1+x^2}. Find a necessary and sufficient condition on x_1 and x_2 for g(x_1)=g(x_2). What is the largest interval I you can find on which g is injective? Find g(I) (ie. find the set {g(x): x \in I}).

    Find an explicit formula for the inverse to g:I \rightarrow g(I)
    g(x_1)=g(x_2)

    \frac{x_1}{1+(x_1)^2}=\frac{x_2}{1+(x_2)^2}

    \frac{1+(x_2)^2}{1+(x_1)^2}=\frac{x_2}{x_1}

    x_2 \neq 1+(x_2)^2 by the quadratic formula. The same applies for x_1.

    Since (x_2)^2 >0, \ 1+(x_2)^2 \neq 0 . Hence the necessary and sufficient condition is x_1=x_2.

    Since this is a map from the reals to the reals, I=\Re.

    Can someone check those?

    The inverse is a bit harder to find.

    x=\frac{y}{1+y^2}

    x(1+y^2)=y

    and from here all the ways i've tried have lead to y's on both sides of the equation. If you substitute x={1}{k} into g(x) you end up with the same function (in terms of k). I haven't been able to use this to my advantage though
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  2. #2
    MHF Contributor

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    Quote Originally Posted by Showcase_22 View Post
    g(x_1)=g(x_2)

    \frac{x_1}{1+(x_1)^2}=\frac{x_2}{1+(x_2)^2}

    \frac{1+(x_2)^2}{1+(x_1)^2}=\frac{x_2}{x_1}

    x_2 \neq 1+(x_2)^2 by the quadratic formula. The same applies for x_1.

    Since (x_2)^2 >0, \ 1+(x_2)^2 \neq 0 . Hence the necessary and sufficient condition is x_1=x_2.

    Since this is a map from the reals to the reals, I=\Re.

    Can someone check those?
    Yes, x_2^2\neq x_2^2+ 1 and 1+ x_2^2\ne 0 but how does it follow that x_1= x_2?

    I would do this in a slightly different way: \frac{x_1}{x_1^2+ 1}= \frac{x_2}{x_2^2+ 1} gives x_1(x_2^2+ 1)= x_2(x_1^2+ 1; x_1x_2^2+ x_1= x_2x_1^2+ x_2 so x_1x_2^2- x_2x_1^2= x_2- x_1 or (x_2-x_1)(x_1x_2)= x_2- x_1 so either x_2= x_1 or x_1x_2= 0. Since x_1= 0 or x_2= 0, the equation is not satisfied, we must have x_1= x_2.

    The inverse is a bit harder to find.

    x=\frac{y}{1+y^2}

    x(1+y^2)=y

    and from here all the ways i've tried have lead to y's on both sides of the equation. If you substitute x={1}{k} into g(x) you end up with the same function (in terms of k). I haven't been able to use this to my advantage though
    That's a quadratic equation in y: xy^2- y+ x= 0 so by the quadratic formula, y= \frac{1\pm\sqrt{1- 4x^2}}{2x}. It is that " \pm" that means it does NOT have, strictly speaking, an inverse.
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