Theory of Intelligence • First-Principles AI & STEM Education

Express $f(x)$ in vertex form:

\begin{aligned} f(x) &= (x^2 + 6x + 9) - 9 + 4 \\ &= (x + 3)^2 - 5 \end{aligned}

Given the domain restriction $x < -3$ :

x + 3 < 0 \implies (x + 3)^2 > 0 \implies (x + 3)^2 - 5 > -5

Thus, the range of $f$ is strictly greater than $-5$ :

\text{Ran}(f) = (-5, \infty)

Set $y = (x + 3)^2 - 5$ and isolate $(x + 3)^2$ :

\begin{aligned} y + 5 &= (x + 3)^2 \\ x + 3 &= \pm\sqrt{y + 5} \\ x &= -3 \pm\sqrt{y + 5} \end{aligned}

Since the restricted domain requires $x < -3$ , we have $x + 3 < 0$ .

The positive branch $-3 + \sqrt{y+5} \ge -3$ would contradict the domain condition.
Therefore, we must select the negative branch:

x = -3 - \sqrt{y + 5}

Interchanging variables:

f^{-1}(x) = -3 - \sqrt{x + 5}

The domain and range swap cleanly:

\begin{aligned} \text{Dom}(f^{-1}) &= \text{Ran}(f) = (-5, \infty) \\ \text{Ran}(f^{-1}) &= \text{Dom}(f) = (-\infty, -3) \end{aligned}