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?
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 $1+x^2\ge 1&gt;0$.  $0$ is not in the range, since the fraction is never zero; however we can get arbitrarily close.

Hence the range is: $$(0,1]$$
Your answer is correct, y should be greater than or equal to 1.