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# What is the difference between (fo(goh)(x) and ((fog)oh)(x)?

is associative, that is

##(f@(g@h))(x) = ((f@g)@h)(x)##

There is no difference in the result, though the steps may be expressed differently.

##(f@(g@h))(x) = f(g(h(x))) = ((f@g)@h)(x)##

For example, suppose:

##f(x) = x^2##

##g(x) = 1/x##

##h(x) = x + 1##

Then:

##(g@h)(x) = g(h(x)) = 1/(x+1)##

##(f@(g@h))(x) = f((g@h)(x)) = f(1/(x+1)) = 1/(x+1)^2##

##(f@g)(x) = f(g(x)) = 1/x^2##

##((f@g)@h)(x) = (f@g)(h(x)) = (f@g)(x+1) = 1/(x+1)^2##