Find the domain of f ( x ) = 1 x 2 − 5 x + 6 f\left(x\right)=\frac{1}{x^2-5x+6} f ( x ) = x 2 − 5 x + 6 1 . Express your answer using interval notation.

Hey, everyone. So let's see if we can solve this problem. In this problem, we are given the function f of x is equal to 1 over x squared minus 5x plus 6, and we're asked to find the domain of this function and express our answer using interval notation. Now when finding the domain of a function written as an equation, you need to ask yourself what values for x are going to break our function. And in this situation, I can see that we have a fraction. For fractions, you cannot have the denominator of your fraction equal to 0. So what we could do is say this entire denominator, x squared minus 5x plus 6, cannot be equal to 0. Now, what I can do from here is solve this like it's just an equation where we solve for x. And notice that we have a polynomial where this is actually a quadratic. We've actually discussed strategies for solving quadratics in previous videos. For this one, one of the quick strategies we can use is to see what two numbers are going to multiply to give us positive 6 and add to give us negative 5. We can see that we have a one in front of the x squared, so this is really all we need to figure out. Now the two numbers that I can think of, well, I could try 23. We know that these multiply to give us positive 6, but what do they add to give us? Well, they add to give us positive 5, and we're looking for negative 5. So this is not the correct pair. So what I can do instead is try, let's say, negative 2 and negative 3. Well, if I try this, negative 2 times negative 3, the negatives cancel, and it does give us positive 6. So that looks good. And negative 2 plus negative 3 is negative 5. So these two numbers actually do work. These check out. So this is the pair we're looking for. So that means that in this case, x squared minus 5x plus 6 is going to factor into x minus 2 times x minus 3, and then this whole thing cannot be equal to 0. Now I can set these up as 2 separate equations with the x minus 2 and x minus 3, because I can say that x minus 2 cannot be equal to 0, and x minus 3 cannot be equal to 0. I can go ahead and add 2 on both sides of this top equation, that'll get the 2's to cancel up here, giving me that x cannot equal 2. Now I can do the same thing for this, this x minus 3 down here I can add 3 to both sides of the equation that'll get the threes to cancel on the left, giving me that x cannot equal positive 3. So we end up with the x cannot equal to 2, and x cannot be equal to 3. So these are the two restrictions for x. So when writing our domain out, we can say that the domain for our function is that x can go from negative infinity to positive 2. This is going to be one of the situations where we cannot include 2, because we're not allowed to have it equal to 2. We also can go from 2 to 3, so we could have, like, 2.5 or 2.7 or 2.9, but we cannot equal 3 or 2 because these are the two restrictions, and we can go from 3 to infinity, not including 3. So this right here would be our domain for the function that we have. So this is how you can find the function or excuse me. This is how you can find the domain of the function we have up here. Hopefully, you found this video helpful, and let's move on.