Functions.

Showcase_22 · #1 $Old$ 11-19-2008, 11:44 AM

Quote:

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

HallsofIvy · #2 $Old$ 11-19-2008, 12:29 PM

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 $x_1= x_2$ .

Quote:

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.