precalc
posted by ANON .
f(x)=2+4xx^2 when x is less than 2
thus f1(x)=2sqrt(x+6)
i need to show that f(f1(x))=x, but i keep getting negative x...any help

by definition ...
f(f^1 (x)) = x , but anyway ...
f( f^1(x) )
= f(2  √(6x) )
= 2 + 4(2  √(6x) )  (2  √(6x) )^2
= 2 + 8  4√(6x)  (4  4√(+x) + 6x)
= 10  4√(6x)  4 + 4√(6x)  6 + x
= x