We can explicitly find the inverse. If [tex]f^{-1}(x)[/tex] is the inverse of [tex]f(x)[/tex], then
[tex]f\left(f^{-1}(x)\right) = \dfrac{f^{-1}(x)+2}{f^{-1}(x)+6} = x[/tex]
Solve for the inverse :
[tex]\dfrac{f^{-1}(x) + 2}{f^{-1}(x) + 6} = x[/tex]
[tex]\dfrac{f^{-1}(x) + 6 - 4}{f^{-1}(x) + 6} = x[/tex]
[tex]1 - \dfrac4{f^{-1}(x) + 6} = x[/tex]
[tex]1 - x = \dfrac4{f^{-1}(x) + 6}[/tex]
[tex]f^{-1}(x) + 6 = \dfrac4{1-x}[/tex]
[tex]\implies f^{-1}(x) = \dfrac4{1-x} - 6[/tex]
Then when x = -6, we have
[tex]f^{-1}(-6) = \dfrac4{1-(-6)} - 6 = \dfrac47-6 = \boxed{-\dfrac{38}7}[/tex]
Alternatively, we can first solve for x such that [tex]f(x) = -6[/tex]. Then taking the inverse of both sides, [tex]x = f^{-1}(-6)[/tex]. (The difference in this method is that we don't compute the inverse for all x.)
We have
[tex]\dfrac{x+2}{x+6} = -6[/tex]
[tex]x + 2 = -6 (x + 6)[/tex]
[tex]x + 2 = -6x - 36[/tex]
[tex]7x = -38[/tex]
[tex]\implies x = \boxed{-\dfrac{38}7}[/tex]
