 Hello and welcome to this session. Let us understand the following problem today. Let f be a function from r to r be the signal function defined as fx is equal to 1, where x is greater than equal to 0, fx is equal to 0 and minus 1 is fx is greater than 0. And g be a function from r to r be the greater than d star function given by g of x is equal to x, where x is the greater than d star function less than or equal to x and thus f of g and g of f coincide in closed interval 0, 1 which is open at 0 and close from 1. Now, let us find the solution given to us function f from r to r is the signal function defined as f of x is equal to 1 if x is greater than 0, 0 if x is equal to 0 and minus 1 is x is greater than 0 and g is a function from r to r is defined as g of x is equal to greater than t star of x, where greater than t star or equal to 0, 1 which is open from 0 and close at 1 we check whether f of g of x is equal to g of f of x is greater than 0 that is x belongs to interval 0, 1 which is open at 0 and close at 1 we have fx is equal to 1 and g function from r to r defined by g of x is equal to greater than t star of x. Now, g of f of x is equal to g of f of x which is equal to g of 1 which is equal to 1 and f of g of x is equal to f of g of x which is equal to f of greater than t star function not defined f of g is not equal to g of f therefore f of g and g of f does not go inside interval 0, 1 which is open at 0 and close at 1 as the required answer is no I hope you understood the problem why and have a nice day.