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Thread: Functions.

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

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

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

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

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

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

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

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

    Can someone check those?

    The inverse is a bit harder to find.

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

    $\displaystyle x(1+y^2)=y$

    and from here all the ways i've tried have lead to y's on both sides of the equation. If you substitute $\displaystyle x={1}{k}$ 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
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    Quote Originally Posted by Showcase_22 View Post
    $\displaystyle g(x_1)=g(x_2)$

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

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

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

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

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

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

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

    The inverse is a bit harder to find.

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

    $\displaystyle x(1+y^2)=y$

    and from here all the ways i've tried have lead to y's on both sides of the equation. If you substitute $\displaystyle x={1}{k}$ 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: $\displaystyle xy^2- y+ x= 0$ so by the quadratic formula, $\displaystyle y= \frac{1\pm\sqrt{1- 4x^2}}{2x}$. It is that "$\displaystyle \pm$" that means it does NOT have, strictly speaking, an inverse.
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