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as level p1 function πππ?
2 Answers
- az_lenderLv 71 year agoFavorite Answer
In this country we don't write "gf" -- we write either gof or g(f(x)).Β I'm going to assume that your "gf" means g(f(x)) rather than f(g(x)).
(i)Β The requirement that the argument of g should be less than or equal to -1 forces the argument of f to be less than or equal to -2/3.Β What they want is a = -2/3.
(ii)Β f(g(x)) = 3(-1 - x^2) + 14 = 0 =>
-1 - x^2 = -14/3 =>
x^2 = 11/3 =>
x = -11/3 (because x must be <= -1).
(iii)Β g(f(x)) = -1 - (3x+1)^2 <= -50 =>
(3x+1)^2 >= 49 =>
|3x + 1| >= 7 =>
-3x - 1 >= 7 (because x must be negative), and so
-3x >= 8 =>Β
x <= -8/3.
