$f (x) = \{\begin{cases} {(x + 1)}^{2} - 2, x \leq 1 \\ 3, x > 1 \end{cases}$

Domain $\{x \in ℝ\}$ , Range $\{y \in ℝ, y \geq - 2\}$

Solve for

f (x) = x

First solve for the left branch.

${(x + 1)}^{2} - 2 = x$

$x^{2} + x - 1 = 0$

$x = - \frac{1}{2} \pm \frac{\sqrt{5}}{2}$

Now solve for the right branch.

$x = 3$

So our solutions are

$x = 3, x = - \frac{1}{2} + \frac{\sqrt{5}}{2}, x = - \frac{1}{2} - \frac{\sqrt{5}}{2}$