Let [tex]f(x) = x^2+4x+1[/tex]. If [tex]f^{-1}(x)[/tex] is the inverse of [tex]f(x)[/tex], then
[tex]f\left(f^{-1}(x)\right) = x[/tex]
We're told that [tex]f(x)[/tex] has a domain of x ≥ -2, so the above can only be valid for [tex]f^{-1}(x)\ge-2[/tex].
Evaluate the left side in this equation and solve for the inverse :
[tex]f^{-1}(x)^2 + 4f^{-1}(x) + 1 = x \\\\ f^{-1}(x)^2 + 4f^{-1}(x) + 4 = x + 3 \\\\ \left(f^{-1}(x) + 2\right)^2 = x + 3 \\\\\sqrt{\left(f^{-1}(x)+2\right)^2} = \sqrt{x+3} \\\\ \left|f^{-1}(x) + 2\right| = \sqrt{x+3}[/tex]
Since [tex]f^{-1}(x)\ge-2[/tex], or [tex]f^{-1}(x)+2\ge0[/tex], by definition of absolute value we have
[tex]\left|f^{-1}(x)+2\right| = f^{-1}(x)+2[/tex]
Then it follows that
[tex]f^{-1}(x) + 2 = \sqrt{x+3} \implies \boxed{f^{-1}(x) = \sqrt{x+3}-2}[/tex]
