Step-by-step explanation:
We have f(x) = (x+6)² for x ≥ -6
That is
y = (x+6)²
[tex]x+6=\sqrt{y}\\\\x=\sqrt{y}-6\\\\f^{-1}(x)=\sqrt{x}-6[/tex]
We know that negative of square root value is not defined.
So x should be greater than or equal to zero.
So
[tex]f^{-1}(x)=\sqrt{x}-6,x\geq 0[/tex]
Option C is the correct answer.