True
In order to prove composition of functions is associative we need to show
[fo(goh)](x) = [(fog)oh](x)
Let us suppose, f(x) = x, g(x) = 2x, h(x) = x + 2
Now,
[fo(goh)](x) = f(g(h(x))) = f(g(x+2)) [∵ h(x) = x + 2 ]
= f(2(x+2)) = f(2x+4)
= 2x+4 (i)
[(fog)oh](x) = (fog)oh(x) = (fog)(h(x))
= (fog)(x+2) = f(g(x+2))
= f(2(x+2)) = f(2x+4)
= 2x+4 (ii)
From (i) and (ii), we observe that
[fo(goh)](x) = [(fog)oh](x)
Thus, composition of functions is associative.
