Question

Are Composite Functions Commutative?

Answer 1

Step1. Defining composite function.

A function whose value are determine by two given function by applying one function to an independent variable and then applying second function to the results whose domain consist of all values of the independent variable for which for which the result yielded by the first function lies in the domain of the second.

for eg:

consider two function

$f\left(x\right)=x^2$ &

$g\left(x\right)=2x+5$

Then composition of the function is given by. $fog$ & $gof$ also written as$f\left(g\right(x\left)\right)$ & $g\left(f\right(x\left)\right)$.

Step2. Verifying commutative property

Now, $f\left(g\right(x\left)\right)$ mean putting the value of $g\left(x\right)$in $f\left(x\right)$

Therefore,

$$\begin{gathered} \Rightarrow f\left(x\right)=x^2 \\ \Rightarrow f\left(g\right(x\left)\right)={\left[g\right(x\left)\right]}^2 \\ ={(2x+5)}^2[\because g(x)=(2x+5\left)\right] \\ =4x^2+20x+25 \end{gathered}$$

Similarly,

$$\begin{gathered} \Rightarrow g\left(x\right)=2x+5 \\ \Rightarrow g\left(f\right(x\left)\right)=2f\left(x\right)+5 \\ =2x^2+5[\because f(x)=x^2] \end{gathered}$$

From the above we conclude that $\mathit{f}\mathbf{\left(}\mathit{g}\mathbf{\right(}\mathit{x}\mathbf{\left)}\mathbf{\right)}\mathbf{\ne }\mathit{g}\mathbf{\left(}\mathit{f}\mathbf{\right(}\mathit{x}\mathbf{\left)}\mathbf{\right)}$

Hence, commutative property not applicable on composite function.