How can I prove $\text{Im } f = (0, 1]$ for $f(x) = 1/(1 + x^2)$?

Question

I want to show that for the function $f : [0, \infty) \rightarrow \mathbb{R}$ defined by $f(x) = 1/(1 + x^2)$, we have $\text{Im} f = (0, 1]$ for $x \geq 0$.

I think that I can do this by introducing a new set $S = \{x \mid 0 < x \leq 1\},$ and then showing that the image of $f$ and $S$ are both subsets of each other (and thus, they are equal).

Choose an arbitrary $y \in \text{Im} f$. Then, we have $y = f(a)$ for some $a \geq 0$. Since we have $0 \leq a < \infty$, from properties of inequalities, it follows that $1 < 1 + a^2 < \infty$. Hence, $y \in S$, and therefore, $\text{Im} f \subseteq S$.

Now let $y \in S$. I need help showing $y \in \text{Im} f$.

In fact, I'm not even sure about the first part of my proof, so any help would be appreciated.

Answer 1

For the first part of your proof, I think you meant to write $1 \le 1 + a^2 < \infty$.

Given $y \in S$, consider $f(x)$ where $x = \sqrt{\frac{1}{y} - 1}$. (This last expression is $f^{-1}(y)$.)

Answer 2

You may show it directly as follows:

• Obviously you have $f(x) > 0$ for all $x \in [0,\infty)$. $$\Rightarrow \text{Im} f \subset (0, \infty)$$
• Now, check for what $y>0$ the equation $y=\frac{1}{1+x^2}$ has a solution in $[x,\infty)$: $$y= \frac{1}{1+x^2} \Leftrightarrow x^2 =\frac{1}{y}-1 \stackrel{!}{\geq} 0$$ $$\stackrel{y>0, x\in [0,\infty)}{\Leftrightarrow} \boxed{0<y\leq 1}$$