 I am welcome to the session. My name is Shashi and I am going to help you with the following question. Question says, find all points of discontinuity of f, where f is defined by fx is equal to modulus of x upon x, if x is not equal to 0, fx is equal to 0 if x is equal to 0. First recall, let us understand that function f is discontinuous at x is equal to a, if left hand side limit of the function is not equal to right hand side limit of the function at x is equal to a, or we can say limit of x tending to a plus fx is not equal to limit of x tending to a minus fx. This is the key idea to solve the given question, how start the solution? We are given fx is equal to modulus of x upon x, if x is not equal to 0, fx is equal to 0, if x is equal to 0. Now, first recall, let us consider fx is equal to modulus of x upon x, modulus function we know modulus function is continuous at all real numbers and this is a polynomial function, polynomial function is also continuous at all real numbers, so fx is equal to modulus of x upon x is continuous at all real numbers except for x equal to 0, as we can see for x is equal to 0, fx is not defined, so we can write function f is continuous real numbers x is equal to 0, only we can see function is defined at x is equal to 0, so we can write fx is equal to 0, function f is defined, now let us find out left hand side limit of the function, that is limit of x tending to 0 plus fx which is equal to limit of x tending to 0 plus modulus of x upon x, this is further equal to limit of x tending to 0 plus x upon x, now x and x will get cancelled and we get limit of x tending to 0 plus 1 which is equal to 1, here we have taken modulus of x equal to x as we know for x greater than 0 modulus of x is equal to x, so here for right hand side limit modulus of x is equal to x, so we get right hand side limit of the function at x is equal to 0 as 1, so we can write limit of x tending to 0 plus fx is equal to 1, now let us find out left hand side limit of the function at x is equal to 0, so we can write limit of x tending to 0 minus fx is equal to limit of x tending to 0 minus modulus of x upon x, now this is equal to limit of x tending to 0 minus minus x upon x below for x less than 0 modulus of x is equal to minus x, now this is equal to limit of x tending to 0 minus minus 1 x and x will get cancelled and we get minus 1, now this is equal to minus 1, so left hand side limit of the function at x is equal to 0 is equal to minus 1, we know value of the function at x is equal to 0 is 0, so therefore limit of x tending to 0 minus fx is equal to minus x tending to 0 plus fx is not equal to limit of x tending to 0 minus fx is not equal to f0, we know right hand side limit is equal to 1, left hand side limit is equal to minus 1 and value of the function at x is equal to 0 is 0, so all 3 values are unequal, so we get limit of x tending to 0 plus fx is not equal to limit of x tending to 0 minus fx is not equal to f0, so this implies function at x discontinues at x is equal to 0, so this is our required answer, this completes the session, hope you understood the session, good bye.