 Hello and welcome to the session. In this session we are going to discuss the following question and the question says that, find the domain of the following functions. First, f of x is equal to 6th root of minus 9x plus 33 the whole. Second, f of x is equal to cube root of 3x plus 5 the whole. Let us start with the solution of the given question. In the first part of the question, we are given a function f of x is equal to 6th root of minus 9x plus 33 the whole. We have to find the domain of this function. To find the domain of this function, in the first step we determine its index and readyCant. So here index will be equal to 6 which is even and readyCant will be equal to minus of 9x plus 33 since index is an even number. So we must restrict the domain to make readyCant that is the expression inside the root greater than equal to 0. So we set minus 9x plus 33 greater than or equal to 0. Then in the next step, we shall solve this inequality. Now subtracting 33 from both sides we get minus 9x plus 33 minus 33 is greater than equal to 0 minus 33. Which implies that minus 9x is greater than equal to minus 33. Now dividing both sides by minus 9 we get x is less than equal to 33 upon 9. We should note that the inequality reverses when we divide by a negative quantity which further implies that x is less than equal to 11 upon 3. Then in the last step we shall write the answer in interval or set notation. So we have obtained x is less than equal to 11 upon 3. So in set notation domain of the function f of x will be equal to set of all x such that x is less than equal to 11 upon 3 where x belongs to the set of real numbers. And in interval notation domain of the function f of x will be equal to semi open interval from minus infinity to 11 upon 3. Now this is the required answer to the first part of the question. Now we move on to the second part of the question. Here we need to find the domain of the function f of x is equal to cube root of 3x plus 5 the whole. Now here we see that index will be equal to 3 which is odd and redicant will be equal to 3x plus 5. Now we know that if index is an odd number then domain is set of all real numbers as here index is equal to 3 which is an odd number. So in set notation domain of the function f of x will be equal to the set of all x such that x belongs to the set of real numbers. And in interval notation domain of the function f of x will be equal to open interval from minus infinity to infinity. Now this is the required answer to the second part of the question. This completes our session. Hope you enjoyed this session.