 Hello and welcome to the session. Let's work out the following question. It says that i, b, a, d interval disjoint from minus 1, 1 proves that the function f given by fx is equal to x plus 1 by x is strictly increasing on i. Let's now move on to the solution. The given function is x plus 1 by x. We have to show that this is an increasing function and we know that to prove that a function is increasing, we need to prove that its first order derivative is greater than 0 in the open interval. So, we first need to find the first order derivative of this function. So, f dash x is equal to 1 minus 1 upon x squared derivative of x is 1 and derivative of 1 upon x is minus 1 upon x squared. Now, we have to show that f dash x is greater than 0 in the interval i, which is disjoint from minus 1 and 1. So, for x belonging to i and x does not belong to the open interval minus 1 and 1, x squared will be greater than 1, right? Since x does not belong to the open interval minus 1 and 1, x squared will be greater than 1. You can take any example. Let's say if you take minus 2. So, square would be 4, which is greater than 1. So, x squared would be greater than 1 and in that case, x squared minus 1 will be greater than 0 and from here, we can say f dash x is x squared minus 1 upon x squared and of course, x squared is greater than 0 in the perfect square. Therefore, f dash x, which is x squared minus 1 upon x squared is greater than 0. For x belonging to the interval i and x is not an element of the open interval minus 1, 1. Hence, f is an increasing function. In fact, it is a strictly increasing function. So, this completes the question on the session. Why for now, I think you have a good day.