 Hi and 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 x2 minus 3, if x is less than equal to 2, fx is equal to x square plus 1, if x is greater than 2. First of all let us understand that function f is discontinuous and x is equal to a is, left hand side limit of the function is not equal to right hand side limit of the function at x is equal to a. This is the key idea to solve the given question. Let us now start the solution. We are given fx is equal to x2 minus 3, if x is less than equal to 2 and fx is equal to x square plus 1, if x is greater than 2. fx is equal to x cube minus 3, if x is less than 2. This is a polynomial function and we know polynomial function is continuous at every real number. This implies function f is continuous at every real number less than 2. We are given fx is equal to x square plus 1, if x is greater than 2. This is again a polynomial function and we know polynomial function is continuous at every real number. This implies function f is continuous at every real number greater than 2. The continuity of the function at x is equal to 2. Clearly we can see function is defined at x is equal to 2. So we can write at x is equal to 2. Function f is defined. Let us now find out right hand side limit of the function at x is equal to 2. That is limit of x tending to 2 plus fx is equal to limit of x tending to 2 plus x square plus 1. This is further equal to 2 square plus 1 equal to 5. So we get right hand side limit of the function at x is equal to 2 as 5. Let us now find out left hand side limit. That is limit of x tending to 2 minus fx is equal to limit of x tending to 2 minus x cube minus 3 which is further equal to 2 cube minus 3 which is equal to 5. We get limit of x tending to 2 minus fx as 5. Or we can say left hand side limit of the function at x is equal to 2 is 5. Let us now find out value of the function at x is equal to 2. You know f2 is equal to 2 cube minus 3 which is further equal to 8 minus 3 which is equal to 5. So we get value of the function at x is equal to 2 as 5. Now clearly we can see right hand side limit, left hand side limit and value of the function at x is equal to 2 are equal. So we get limit of x tending to 2 plus fx is equal to limit of x tending to 2 minus fx is equal to f2 is equal to 5. This implies function f is continuous at x is equal to 2. So function f is continuous at all the real values of x greater than 2 less than 2 and equal to 2. So we can say function f is continuous at all real numbers. You can write function f is continuous. All real numbers there is no point of discontinuity. So this is our required answer. This completes the session. Hope you understood the session. Goodbye.