Let us quickly train our mind with more problems
on composition of functions. Consider this function f from the set of all
real numbers to real numbers defined as f(x)=x2 and g from R to Z, real numbers to integers
defined as g(x)=modulus of x. Now what is f composition g? f composition g(x) is f(g(x)). This is how composition is defined as. Now what is g(x)? It is modulus of x and f(x) is x2 and hence
f(g(x)) happens to be f of modulus of x. And this is going to be my term now and hence
f of modulus of x will be modulus of x the whole square. Correct? Yes. Now I will end this problem by asking a nice
question. Is f composition g same as g composition f? I leave it to you to try out. You may want to compute g composition f and
try proving that both of them are not the same. The last question. Consider this
function f from integers to integers defined as f(x)=x2 + 1 and the function g defined
from again integers to integers as g(x)=3x. Let us compute g composition f and f composition
g, both of them. g composition f is given to be g composition
f(x), which is g(f(x)), which is equal to g of what is f(x)? It is x2+1. Now if this is going to be my term in the
domain of g, then what is g of this term? Let’s say this is y. So what is g(y) going to be? It is going to be 3y. So y is x2+1 and hence it is 3(x2+1). What is f composition g? f composition g(x) is equal to f(g(x). g(x) is 3x. Now this is my y here. What is f(y)? f(y) is x2+1. Okay. Let me say f(z) because I have already used
y. So f(z) is z2+1 and what is my z here. It is 3x actually. I can substitute 3x as z and hence f composition
g(x) is (3x)2 + 1. This is my answer and hence what did we observe? g composition f is 3(x2+1) and f composition
g is (3x)2 + 1.