 Hi, and welcome to the session. This is Kassel according question. The question, say, is find the intervals in which the function f given by fx equals to x cubed plus 1 by x cubed, where x is not equal to 0, is increasing. Second part is increasing. Now, we can get the solution. We are given that fx is equal to x cubed plus 1 by x cubed, where x is not equal to 0. We have to find the intervals in which this function is increasing and decreasing. For this, we have to first find f dash x. So let us first differentiate both sides of this with respect to x. On differentiating, we get f dash x has 3x squared minus 3x squared divided by x to the power 6. This is equal to 3x squared minus 3 by x to the power 4. Now, fx increases, dash x is greater than 0. That is, x squared minus 3 by x to the power 4 is greater than 0. This implies 3 into x squared minus 1 by x to the power 4 is greater than 0. This implies x to the power 6 minus 1 is greater than 0. This implies x cubed minus 1 into x cubed plus 1 is greater than 0. We know that if product of a and b is greater than 0, that this implies that a is greater than 0 and b is also greater than 0. Or a is less than 0 and b is less than 0. So x cubed minus 1 into x cubed plus 1 greater than 0 implies x cubed minus 1 is greater than 0 and x cubed plus 1 is also greater than 0. Or 2 minus 1 is less than 0 and x cubed plus 1 is also less than 0. Now, this implies minus 1, greater than minus 1 implies x less than 1 and x less than minus 1 implies x is less than minus 1. So function decreases. Now, fx decreases is less than 0 plus 1 by x to the power 4 is less than 0. This implies x to the power 6 minus 1 is less than minus 1 into x minus 1 is less than 0, is greater than 0, minus 1 is less than 0, is less than 0, is less than 1, minus 1. And this implies x lies between minus 1 and 1, then 0 is less than 0. Then, now this cannot be possible because we cannot find value of x which is greater than 1 and also which is less than minus 1. So we can say that greater than 1 minus 1 is, minus 1 is less than x is less than 1. See that given function f is increasing when x is less than minus 1 and x is greater than 1 and it is decreasing when x lies between minus 1 and 1. This is our required answer. This competes with the station by intake care.