Finding the image of $f(x)=\frac{1}{1+x^{2}}$ $f(x)=\frac{1}{1+x^{2}}$ and $x\geq0$
To find the image:
$y=f(x)$
$y=\frac{1}{1+x^{2}}$
$x=y^{-1}$
$x=\sqrt{y^{-1}-1}$
$y\geq1$
Then the image of $f(x)=\frac{1}{1+x^{2}}$ is $y\geq1$
Is this correct?  If not, what information is missing?
 A: For $\sqrt{y^{-1}-1}$ to be defined, $y^{-1}\ge 1$, i.e. $\frac{1}{y}\ge 1$ or $y\le 1$.  The range is indeed bounded above by 1; it is bounded below by zero since $1+x^2\ge 1>0$.  $0$ is not in the range, since the fraction is never zero; however we can get arbitrarily close.
Hence the range is: $$(0,1]$$
A: Hint:
When $x$ is getting small and small.i.e, $x$ be so close to $0$, then $f(x)$ is getting close and closer to $1$. Make some examples for yourself:
$$x=0.1\to  f(x)\sim 0.9900990$$
$$x=0.01\to f(x)\sim 0.9999000$$
$$x=0.001\to  f(x)\sim 0.9999990$$
$$x=0.0001\to f(x)\sim 0.99999999$$ This can be interpreted that while $x\to 0, ~~x\ge0$ then we have $f(x)\to 1$. Now do the same for the cases in which $x$ is getting large and larger. In these case, obviously, the function is getting small and smaller such that for very big value for $x$ it can be considered a number close to $0$. What these results mean?
Considering the function as a sequence, we can have the below plot. 

A: A new way to get the right answer, Hint:$y=\frac{1}{1+x^{2}}$, So $yx^2+y-1=0$ and solve the equation using the quadratic formula, taking $x$ the variable, and then calculate the domain of the new function you just made.
P.S: $x$ does not need to be positive necessarily. Because for $x=\pm k$ we have $x^2=k^2$.
A: Given $A$, $B$ sets and a function $f:A \to B$, the image of $A$ through $f$ is defined as $f(A)=\{ b \in B \, | \, \exists \, a \in A \text{ s.t. }f(a)=b\} \subseteq B$.
Now let's pick your case: $1/(1+x^2) \le 1 \ \forall \, x \in \mathbb{R}$... so: do you still think that your answer is correct?
A: Your answer is correct, y should be greater than or equal to 1.
