 Hi and welcome to the session. Let us discuss the following question. Question is show that function f from closed interval minus 1 1 to r where r is the set of all real numbers given by fx is equal to x upon x plus 2 is 1 1. Find the inverse of the function f from closed interval minus 1 1 to range f. First of all let us understand the key idea to solve the given question. If we have given a function f from x to y such that it is invertible then there exists a function g from y to x such that g o f is equal to identity function on set x and f o g is equal to identity function on set y. g is the inverse of function f. It is denoted by f inverse. This is the key idea to solve the given question. Let us now start the solution. We have given function f from closed interval minus 1 1 to r where r is the set of all real numbers. It is given by fx is equal to x upon x plus 2. Now let us consider any two arbitrary values x 1 x 2 belonging to interval minus 1 1 which is closed at both the ends. fx 1 must be equal to fx 2. This implies x 1 upon x 1 plus 2 is equal to x 2 upon x 2 plus 2. This implies x 1 multiplied by x 2 plus 2 is equal to x 2 multiplied by x 1 plus 2. This implies x 1 x 2 plus 2 x 1 is equal to x 1 x 2 plus 2 x 2. Now x 1 and x 2 will get cancelled. x 1 x 2 cancel each other and we get 2 x 1 equal to 2 x 2. Now this implies dividing both sides by 2 we get x 1 is equal to x 2. Clearly we can see for x 1 x 2 belonging to interval minus 1 1 which is closed at both the ends. fx 1 is equal to fx 2 implies x 1 is equal to x 2. So the given function is a 1 1 function. So we can write f is a 1 1 function. Now we have to find the inverse of function f from closed interval minus 1 1 to range f and by the arbitrary element of range f is in the interval minus 1 1 x upon x 1 plus 2. So we can write y is equal to x upon x plus 2. This implies y multiplied by is equal to x. This implies y x plus 2 y is equal to x. This implies 2 y is equal to x minus y x. This implies x is equal to 2 y upon 1 minus. This function g is equal to x plus 2 y to close interval minus 1 1 is not equal to value of x lies in the closed interval minus 1 1. Find out g of fx. g of fx is equal to g of fx. Now we have fx is equal to x upon x plus 2. So we can write g of x upon x plus 2. g of x upon x plus 2 is equal to 2 multiplied by x upon x plus 2 upon x plus 2 is equal to x upon x 1 minus x plus 2. Simplifying we get 2x upon plus 2 minus x plus x minus x will get cancelled and we get 2x upon 2 g of fx equal to which is further equal to identity function on the interval minus 1 1 which is closed at both the ends. Now let us now find out f of g y is equal to f of g y. Now we know g y is equal to 2 y upon 1 minus y. We can write f of 2 y upon 1 minus y. This is further equal to 2 y upon 1 minus y plus 2 minus 2 y plus 2 y minus 2 y cancels each other. We will get 2 y upon 2. 2 and 2 will get cancelled and we get y. So f of g y is equal to identity function on range x, identity function on the closed interval minus 1 1 and f of g y is equal to identity function on range y. This implies that f is invertible and g is the inverse of f. So we can write this implies f is invertible and function g is the inverse of function f. So our final answer is f inverse is given y equal to 2 y upon 1 minus y where y is not equal to 1. This completes the session. Hope you understood the session. Goodbye.