Prove the range of the following two functions.
1) f(x)=$(\frac{1-x^2}{1+x^2})$ for each x $ \in$ R.
2) g(x)=$\sqrt\frac{1-x}{1+x}$ for each x $ \in$ (-1,1].
Since f(x) is not one to one f inverse can't be computed.In this sort of situations how can I prove what the range is ?
In g(x) since this is one to one I computed the inverse which I get as g$^{-1}$(x)=$\pm$$\sqrt\frac{1-x^2}{1+x^2}$. Here should I take the negative square root also or should it only be the positive one?
When I consider the domain of this inverse function since $\frac{1-x^2}{1+x^2}$>=0.
I get 1>=x$^2$
.
Thus x $\in$ [-1,1]. Hence the rsnge of g is [-1,1]
But I think this is wrong because in g , x $ \in$ (-1,1].When x=1 g(x)=0 and when x is close to -1, g(x) goes to infinity.Thus the range of g should be [0, infinity).
What have I done wrong here ? Please help to find the range of these two functions.
Best Answer
Hints: for $\;a\in\Bbb R\;$
$$\frac{1-x^2}{1+x^2}=a\iff(1+a)x^2=1-a\iff\ldots$$
Can you see what the condition(s) must be to solve the above for $\;x\;$ ?
Likewise with the other function:
$$\sqrt{\frac{1-x}{1+x}}=a\iff 1-x=a^2+a^2x \iff (a^2+1)x=1-a^2\iff\ldots$$
But in this case you also must take into account that the function is defined only when
$$\frac{1-x}{1+x}>0\;,\;x\neq-1\iff \frac{x-1}{x+1}<0\;,\;x\neq -1\iff -1<x<1\;\ldots$$