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, \[a*b=b*a\], where \[*\] 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, \[f(g(x))=2({{x}^{2}})=2{{x}^{2}}\]
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, \[a*b=b*a\], where \[*\] 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,
\[f(g(x))=2({{x}^{2}})=2{{x}^{2}}\]
Taking the RHS now, we will get,
\[g(f(x))={{(2x)}^{2}}=4{{x}^{2}}\]
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.