Theory of Intelligence • First-Principles AI & STEM Education

Factor out $2$ from the $x$ -terms and complete the square:

\begin{aligned} f(x) &= 2(x^2 - 4x) + 11 \\ &= 2(x^2 - 4x + 4) + 11 - 8 \\ &= 2(x - 2)^2 + 3 \end{aligned}

For $x \ge 2$ , $(x - 2) \ge 0 \implies 2(x - 2)^2 \ge 0$ . The minimum value is $f(2) = 3$ . Hence:

\text{Ran}(f) = [3, \infty)

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

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

Because $x \ge 2$ , the deviation from $2$ must be non-negative ( $x - 2 \ge 0$ ).

The minus branch $2 - \sqrt{\frac{y-3}{2}}$ would yield values $\le 2$ , which lies outside the domain restriction for $x > 2$ .
Thus, we must choose the positive branch:

x = 2 + \sqrt{\frac{y - 3}{2}}

Interchanging variables $x$ and $y$ :

f^{-1}(x) = 2 + \sqrt{\frac{x - 3}{2}}

Stating the swapped intervals:

\begin{aligned} \text{Dom}(f^{-1}) &= \text{Ran}(f) = [3, \infty) \\ \text{Ran}(f^{-1}) &= \text{Dom}(f) = [2, \infty) \end{aligned}