 Hello and welcome to the session. I am Deepika here. Let's discuss the question. What are the intervals in which the calling function is strictly increasing or decreasing? x plus 1q into x minus 3 whole cube. Let's start the solution. Now we know that a function is strictly increasing on an open interval where its derivative is positive and the function is strictly decreasing on an open interval where its derivative is negative. Now our given function is fx is equal to x plus 1 whole cube into x minus 3 whole cube. Now after x is equal to, here we will apply the product truth. x plus 1q into 3 into x minus 3 square plus x minus 3 whole cube into 3 into x plus 1 whole square and this is equal to 3 into x plus 1 whole cube into x minus 3 whole square plus 3 into x minus 3 whole cube into x plus 1 whole square and this is equal to, let us say 3 into x plus 1 whole square into x minus 3 whole square common. We get 1 plus x and this is equal to 3 into x plus 1 whole square into x minus 3 whole square into 2x minus 2 and this is equal to, let us take 2 common from here, 6 into x plus 1 whole square into x minus 3 whole square into x minus 1. Now equal to 0 implies either x plus 1 whole square is equal to 0 or x minus 3 whole square is equal to 0 or x minus 1 is equal to 0. This implies either x plus 1 is equal to 0 or x is equal to 0 or x is equal to 1. This implies either x is equal to minus 1 or x is equal to 3 or x is equal to 1. Hence x is equal to minus 1, x is equal to 1 and x is equal to 3 divides the real line into 4 disjoint intervals which are minus infinity to minus 1, minus 1 to 1, 1 to 3, 3 to infinity. Now our f dash x is 6 into x plus 1 whole square into x minus 3 whole square into x minus 1. Now for less than minus 1 dash x is less than 0 therefore f is decreasing in an open interval minus infinity to minus 1. Let us take the next open interval which is minus 1 to 1. We will see whether our given function is strictly increasing or decreasing for minus 1 less than x, x less than 1 again f dash x is less than 0. Therefore decreasing in an open interval minus 1 to 1. Now for 1 less than x, x less than 3 or f dash x is greater than 0 therefore f is strictly increasing in an open interval 1 to 3. Now for x greater than 3 dash x is greater than 0 therefore f is strictly increasing in an open interval 3 to infinity. Hence the answer to our question is function is strictly increasing in an open interval 1 to 3 and 3 to infinity and strictly decreasing in minus infinity to minus 1 and an open interval minus 1 to 1. So this is our answer. I hope the solution is clear to you. Bye and take care.