Suppose the rule of the function f is "add one" and the rule of the...

Given:
Suppose the rule of the function f is "add one" and the rule of the function g is "multiply by 4".
Calculation:
To express the function algebraically:
Here, f and g are function of x.
From the given information,
${\displaystyle f\left(x\right)=x+1}$
${\displaystyle g\left(x\right)=4x}$
${\displaystyle (f\circ g)\left(x\right)=f\left(g\left(x\right)\right)}$
${\displaystyle =f\left(4x\right)}$
${\displaystyle (f\circ g)\left(x\right)=4x+1}$, (using ${\displaystyle f\left(x\right)=x+1}$)
${\displaystyle (g\circ f)\left(x\right)=g\left(f\left(x\right)\right)}$
${\displaystyle =g(x+1)}$
${\displaystyle =4(x+1)}$, (using ${\displaystyle g\left(x\right)=4x)}$
${\displaystyle (g\circ f)\left(x\right)=4x+4}$

We have to find the algebraically function f and g where f is "add one" and g is "multiply by 4"
Solution: ${\displaystyle f(\in put)=\in put+1}$
${\displaystyle \Rightarrow f\left(x\right)=x+1}$
and ${\displaystyle g\left(x\right)=4x}$
${\displaystyle (f\circ g)\left(x\right)=f\left(g\left(x\right)\right)=f\left(4x\right)=4x+1}$
${\displaystyle (f\circ g)\left(x\right)=g\left(f\left(x\right)\right)=g(x+1)=4(x+1)}$
${\displaystyle (f\circ g)\left(x\right)=4x+4}$