 Hello friends, welcome to the session I am working on. Let us examine the applicability of mean value theorem for all the three functions given in the above exercise 2. Our function is fx equal to x square minus 1 for x being the element of closed interval 1, 2. So let us start with the solution. We are given fx equal to x square minus 1 for x being the element of closed interval 1, 2. Now since we see that the given function is a polynomial function, so it is continuous. Therefore, it is continuous in closed interval 1, hence differentiable in open interval 1. Now we will find f1 which is equal to 1 minus 1 equal to 0, f2 equal to 4 minus 1 equal to 3. We see that f1 is not equal to f2. Since we see that all the three conditions of LMVT are satisfied, Lagrange's, Lagrange's mean value theorem holds good. There exists a point c which is the element of open interval 1, 2. Now the dash c equal to f2 minus f1 upon 2 minus 1. This is our first equation. Now we are given fx equal to x square minus 1. Therefore, f dash x equal to 2x f dash c equal to 2c. Now putting this value in equation first equal to f2 minus f1 f2 is equal to 3 minus f1 is 0 upon 2 minus 1 is 1. This implies 2c equal to 3. This implies c equal to 3 by 2 which is the element of open interval 1, 2. Now hence we see that LMVT theorem holds good. So, hope you understood the solution and enjoyed the session. Goodbye and take care.