Finding the image of an absolute value function I am trying to find the image of the following function:
$$f:\mathbb{R}\rightarrow\mathbb{R}, \\f(x)=|1-x^2|.$$
I know trivially that the image is in ${[0,\infty[}$, but I'm looking for a way to algebraically solve it.
So I have tried taking a $y\in im(f)$
$\Rightarrow$ there must be a $x\in dom(f)$ such that $f(x)=y$
$\Rightarrow |1-x^2|=y$.
And at this point I do not know what to do. Do I separate into cases? What is the next logical algebraic reasoning?
Any help is much appreciated.

Please note that there are usually three ways to find the range of a function. I show how they work by finding the range of the given function through these ways.
First Way
We can find the range of a function by starting from the known functions, of which we know the range, and then trying to transform them into the given function.
As we know, the range of the function $g(x)=x^2$ is $R_g=[0, \infty )$. So we have$$-x^2 \le 0 \quad \Rightarrow \quad 1-x^2 \le 1.$$Now, according to the definition of the absolute value function$$|u|= \begin{cases}u, & \text{if } u \ge 0 \\ -u, & \text{if } u \lt 0 \end{cases}.$$Since $1-x^2\le 1$,$$0\le 1-x^2 \le 1 \qquad \text{ or } \qquad 1-x^2 \lt 0.$$So we have$$\begin{cases} 0 \le 1-x^2 \lt 1 \\ 1-x^2 \lt 0 \end{cases}$$ $$\Rightarrow \quad |1-x^2|= \begin{cases}1-x^2, & \text{if } 0 \le 1-x^2 \le 1 \\ -(1-x^2), & \text{if } 1-x^2 \lt 0 \end{cases}$$ $$\Rightarrow \quad |1-x^2|= \begin{cases}1-x^2 \ge 0, & \text{if } 0 \le 1-x^2 \le 1 \\ -(1-x^2) \gt 0, & \text{if } 1-x^2 \lt 0 \end{cases}.$$Thus, we conclude that the range of the function $f(x)=|1-x^2|$ is$$R_f=[0, \infty ).$$
Second Way
We can find the range of a function by finding the inverse map of the function; the range of the function is the domain of its inverse map.
So, let us find the inverse map of the function $f(x)=|1-x^2|$, by using the definition of the absolute value function mentioned above, as follows.$$y=|1-x^2| \quad \Rightarrow \quad y=\begin{cases} 1-x^2, & \text{if } 1-x^2 \ge 0 \\-(1-x^2), & \text{if } 1-x^2 \lt 0 \end{cases}$$ $$\Rightarrow \quad \begin{cases} y=1-x^2, & \text{if } x^2 \le 1 \\y=-1+x^2, & \text{if } x^2 \gt 1 \end{cases}$$ $$\Rightarrow \quad \begin{cases} x=\pm \sqrt{1-y}, & \text{if } -1\le x \le 1 \\ x=\pm \sqrt{1+y}, & \text{if } x \in (-\infty , -1) \cup (1, \infty ) \end{cases}.$$ The domain of the radical function in the first case is $1-y \ge 0$, that is, $y \le 1$. But, since $x$ is restricted to $(-1, 1)$, $y$ can be never less than $0$. So the domain of the first case is $y \in [0, 1]$.
The domain of the radical function in the second case is $1+y \ge 0$, that is, $y \ge -1$. But, since $x$ is restricted to $(-\infty , -1) \cup (1, \infty )$, $y$ can be never less than or equal to $0$. So the domain of the first case is $y \in (0, \infty )$.
Please note that the domain of a piecewise-defined function equals the union of the domain of the pieces. So the domain of the inverse map is$$[0,1] \cup (0, \infty ) = [0, \infty ).$$Thus, we conclude that the range of the function $f(x)=|1-x^2|$ is$$R_f=[0, \infty ).$$
