f(x) = x2 + 4 ; g(x) – 3x – 2; h(x) = x – 5
To prove fo (goh) = (fog) oh
L.H.S. fo (goh)
goh = g[h(x)]
= g(x – 5)
= 3(x – 5) – 2
= 3x – 15 – 2
goh = 3x – 17
fo (goh) = f [goh (x)]
= f(3x – 17)
= (3x – 17)2 + 4
= 9x2 + 289 – 102 x + 4
= 9x2 – 102x + 293 … (1)
R.H.S. = (fog) oh
fog – f[g(x)]
= f(3x - 2)
= (3x – 2)2 + 4
= 9x2 + 4 – 12x + 4
= 9x2 – 12x + 8
(fog) oh = fog [h(x)]
= fog (x – 5)
= 9(x – 5)2 – 12 (x – 5) + 8
= 9(x2 + 25 – 10x) – 12x + 60 + 8
= 9x2 + 225 – 90x – 12x + 60 + 8
= 9x2 – 102x + 293 ….(2)
From (1) and (2) we get fo (goh) = (fog) oh.
Composition of function is associative