by definition ...
f(f^-1 (x)) = x , but anyway ...
f( f^-1(x) )
= f(2 - √(6-x) )
= 2 + 4(2 - √(6-x) ) - (2 - √(6-x) )^2
= 2 + 8 - 4√(6-x) - (4 - 4√(+-x) + 6-x)
= 10 - 4√(6-x) - 4 + 4√(6-x) - 6 + x
= x
f(x)=2+4x-x^2 when x is less than 2
thus f-1(x)=2-sqrt(-x+6)
i need to show that f(f-1(x))=x, but i keep getting negative x...any help
1 answer