" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
g(x) = 3 - 2x --> g⁻¹(x) = -1/2 (x-3)
a) (gog⁻¹)(x) = g(g⁻¹) = 3 - 2 {-1/2 (x-3)}
gog⁻¹(x)= 3 + x - 3 = x
b) g⁻¹og (x)= g⁻¹(g) = -1/2 ( 3-2x -3)
g⁻¹og (x) = -1/2 (-2x) = x
c) (fof⁻¹ og) (x) =
(fof⁻¹)(x) = f( f⁻¹) = 10 ( 1/10 (x+6))- 6 = x + 6 - 6 = x
fof⁻¹ (g(x)) = {fof⁻¹ (g(x))} = g(x)= 3-2x
d) (gof⁻¹og⁻¹)(x)= ...
gof⁻¹ = g (f⁻¹) = 3 - 2 (1/10 (x+6)) = 3 - 1/5 (x+6)
gof⁻¹og⁻¹ (gof⁻¹)(g(x)) = 3 - 1/5 (3- 2x +6)
gof⁻¹og⁻¹(x)= 3 - 3/5 + 2/5 x - 6/5
gof⁻¹og⁻¹(x) = 1/5 (15-3 + 2x -6) = 1/5 (2x + 6)