 Remember that in a polynomial function, the variables can only be raised to whole number powers. So what about non-pollinomial functions? We still want to find the domain. We still want to find the set of all possible input values. But how can we do that for non-pollinomial functions? Here's a useful guide to life. It's sometimes easier to figure out what is forbidden than what is allowed. Hey, get out of the cookie jar. You'll spoil your dinner. Sorry about that. In general, a denominator can never be zero. And also, the radicand of a square root can never be negative. For example, let's find the domain of f of x equals 1 divided by 3x plus 8. So this is not a polynomial because our x is the denominator of a fraction. And because this is a fraction, it's important to remember that the denominator of a fraction can never be zero. And so first we'll find the forbidden values. Since the denominator can't be zero, we require the denominator not be equal to zero, or since our denominator is 3x plus 8, we require 3x plus 8 not be zero. Well, this is an inequality, and so we can solve an inequality by ignoring it and finding the critical values. So let's solve the equation 3x plus 8 equals zero, which gives us our critical value x equals minus 8 thirds. But, like a good mathematician or a good human being, we don't ignore the inequality forever. We do have to address it. Since the critical value solves the equality, 3x plus 8 equals zero, but we want to solve 3x plus 8 not equal to zero, the critical value is not included in the solution set. So let's graph the critical value on our number line, and since it's not included, we'll put an open circle at x equals minus 8 thirds. The critical value splits the number line into two parts, and we test a point in each part. To the right, we see that x equals zero is included in the right interval, so let's test that point. If x equals zero, then our inequality, 3x plus 8 not equal to zero, is true, so we include the region to the right. To the left, we'll test x equals minus, so how about 10,000? If x equals minus 10,000, 3x plus 8 not equal to zero is true, so we include the region to the left. And so our domain is all this stuff on the left, together with all this stuff on the right, and in interval notation, we can write this as, how about a square root function? Let's find the domain of g of x equals square root 12 minus 1 third x. So, remember the radicand of a square root can never be negative. So we require the radicand be greater than or equal to zero. And again, we'll solve by ignoring the inequality. Let's get rid of this fraction first. We'll multiply both sides by the denominator 3, simplify, and solve. And remember, x equals 36 solves the equality, 12 minus 1 third x equals zero, and the inequality allows for equality, so x equals 36 is part of the solution. So we'll graph that with a closed circle of x equals 36, and our number line has been split into two parts, the part to the right and the part to the left. So, we'll pick a few test points. How about, well, I don't know which ones you picked, but I'm going to test x equals 1 million and x equals zero. At x equals 1 million, our inequality is both. So we do not include the section to the right of the critical value. At x equals zero, our inequality is true. So we do include the region to the left of the critical value. And so our solution will be, how about a horrible frightening function? Well, actually, this isn't too bad. As long as you keep in mind, the type of expression is the last thing that you do. So this is a quotient. The last thing we do in this function is to divide the thing, 8 plus 3x, by the other thing, square root 12 minus 2x. And remember, the denominator of a fraction can never be zero. And so we require square root of 12 minus 2x not be equal to zero. But wait, there's more. This is a square root. And since we have a radical, remember the radicand of a square root can never be negative. And so we require 12 minus 2x be greater than or equal to zero. Now, we have two inequalities, both of which must be satisfied. But note that if we require 12 minus 2x be strictly greater than zero, then the radicand will never be negative and the denominator will never be zero. And so we can combine these two requirements into a single requirement. 12 minus 2x must be strictly greater than zero. So we'll solve this. We'll ignore the inequality and solve the equation, getting the critical value x equals 6. And because x equals 6 solves the equation, but we don't allow equality, x equals 6 is not part of the solution, and so we put an open circle at x equals 6. And we'll test x equals zero and x equals 40 billion. And if we test these values, we find if x equals zero, our inequality is true, and so we want to shade the region to the left. If x equals 40 billion, our inequality is false, so we don't want to shade the region to the right. And finally, we can express our answer in interval notation.