Question

Are composite functions commutative ?

Answer 1

Hint: We are asked if the composite functions are commutative or not. Composite functions are functions which have function within a function. Commutative property involves operations that does not dependent on the order of the functions involved and still give us the same result, that is, $$a\ast b=b\ast a$$, where $$\ast$$ is any operation. We will apply commutative law in the composite function and check if the order has any effect on the commutativity.

Complete step-by-step solution:
According to the given question, we have to check whether the composite functions are commutative or not. We will begin by knowing what a composite function is.
A composite function is a function which is obtained when a function is substituted into another function.
For example – let us take two functions namely, $$f,g$$.
So, we have, $$f(x)=2x$$
 And $$g(x)=x^2$$
So, the composite function, $$f(g(x))=2(x^2)=2x^2$$
Commutative property involves carrying out an operation in which even if the order of the entities involved in the operation is changed, the answer still remains the same.
That is, $$a\ast b=b\ast a$$, where $$\ast$$ is any operation
For example – commutative law of addition: Adding two numbers in any order will result in the same answer. That is, $$2+3=3+2=5$$.
So, now, we have to use commutativity in a composite function.
Let there be two functions namely, $$f$$ and $$g$$.
So, the function $$f(x)=2x$$ and the function $$g(x)=x^2$$
To check for commutativity, we will have to see if,
$$f(g(x))=g(f(x))$$
Taking the LHS first, we have,
$$f(g(x))=2(x^2)=2x^2$$
Taking the RHS now, we will get,
$$g(f(x))={(2x)}^2=4x^2$$
Clearly, $$LHS\ne RHS$$
That means, composite functions do not follow commutative law.
Therefore, composite functions are not commutative.

Note: While computing the composite functions, the substitution should be done step wise to prevent mistakes. $$f(g(x))$$ is a composite function and it means that $$f$$ is the base function and the function $$g$$ is substituted in place of function f’s independent variable.