 Hello and welcome to the session. In this session we discussed the following question that says, find the intervals in which the function f given by fx equal to x cube plus 1 upon x cube x is not equal to 0 is increasing and decreasing. Let's move on to the solution now. We are given the function fx equal to x cube plus 1 upon x cube x is not equal to 0. We have to find the intervals in which this function is increasing and decreasing. So first of all we will find out f dash x this is equal to 3x square minus 3 upon x to the power of 4 where x is not equal to 0. Now we find the intervals in which this function is increasing. The function fx is increasing if and only if we have f dash x is greater than equal to 0. That is 3x square minus 3 upon x to the power of 4 is greater than equal to 0 where x is not equal to 0. So this means 3 into x to the power of 6 minus 3 greater than equal to 0 where x is not equal to 0 or you can say x to the power of 6 minus 1 is greater than equal to 0. x is not equal to 0 or x square the whole cube minus 1 greater than equal to 0 x is not equal to 0 or x square the whole cube is greater than equal to 1 where x is not equal to 0. Or you can say further x square is greater than equal to 1 where x is not equal to 0. This means we have x square minus 1 is greater than equal to 0 where x is not equal to 0. Now we factorize x square minus 1 so we have x plus 1 whole multiplied by x minus 1 is greater than equal to 0 where x is not equal to 0. So either both these factors would be greater than equal to 0 or both would be less than equal to 0 so that the product is greater than equal to 0. So this means we have x plus 1 greater than equal to 0 and x minus 1 greater than equal to 0 or x plus 1 less than equal to 0 and x minus 1 less than equal to 0 where again x is not equal to 0. So this means we have x is greater than equal to minus 1 and x is greater than equal to 1 or x is less than equal to minus 1 and x is less than equal to 1 again x is not equal to 0. Now the common portion to x greater than equal to minus 1 and x greater than equal to 1 is x greater than equal to 1 so here we have x greater than equal to 1 or the common portion to x less than equal to minus 1 and x less than equal to 1 is x less than equal to minus 1. here again x is not equal to 0. So the interval for x greater than equal to 1 is x belongs to the interval closed at 1 open at infinity or x belongs to now the interval for x less than equal to minus 1 would be the interval open at minus infinity and closed at minus 1. So that is we get x belongs to the interval open at minus infinity closed at minus 1 union interval closed at 1 open at infinity. Thus the function fx is increasing in the interval open at minus infinity closed at minus 1 union close at 1 open at infinity. So this is the interval in which the function fx is increasing. Now let's find out the interval in which the function fx is decreasing. We know that a function fx is decreasing if and only if f dash x is less than equal to 0. This means 3x square minus 3 upon x to the power 4 is less than equal to 0 where again x is not equal to 0. So this would mean 3 into x to the power 6 minus 3 is less than equal to 0 where x is not equal to 0. So further we get x to the power 6 minus 1 is less than equal to 0 where x is not equal to 0 or square the whole cube minus 1 is less than equal to 0 where x is not equal to 0. This means x square the whole cube is less than equal to 1 where x is not equal to 0 or x square is less than equal to 1 where x is not equal to 0. Further x square minus 1 is less than equal to 0 where x is not equal to 0. This means x plus 1 whole multiplied by x minus 1 is less than equal to 0 where x is not equal to 0. So this means we have x belongs to the interval closed at minus 1 open at 0 or x belongs to the interval open at 0 closed at 1 since x is not equal to 0. Thus we get x belongs to the interval closed at minus 1 open at 0 union interval open at 0 closed at 1 thus sx is decreasing in the interval closed at minus 1 open at 0 union the interval open at 0 closed at 1. So this is our final answer. This completes the session. Hope you have understood the solution of this question.