Due to the fact that this is a piecewise function, we're going to take the derivative. So Y prime equal -2 X -9. We're going to look at that derivative when we set it equal to 0. So if we set -2 X -9 equal to 0, we're going to get -9 halves. We're going to take the second piece of the original function, the derivative -2 X +3 and we're going to set it equal to 0. So X equal 3 halves. Now we have to figure out at one, is it a smooth continuous curve? Does the limit as X goes from one to the left equal the limit as X goes to one from the right? It could be a defined point, but if it's not a smooth point, it doesn't have a derivative there and hence it be undefined. And that would be an example of like a corner or a cusp. So when we stick in one from the left, we get a -11 and when we stick in one from the right, we get one. So Y prime of 1 does not exist. So our three points that we need to check out are -9 halves, three halves and one. When I put that on the number line, if I look to the right of three halves, I'm going to use the bottom equation. Say we put in five, we're going to get a negative to the right of three halves. Three halves has an odd multiplicity. So if it's negative on one side, it's got to be positive on the other. So between 1:00 and 3:00 halves, it's positive. Now in between -9 halves and one, we have to use the correct derivative, which is that top equation, because we need to be less than or equal to one. If we put in a value, say zero there, we end up with -9 So we know between -9 halves and one, we're going to be negative because of the multiplicity. On the other side of -9 halves, we're going to be positive. So if we look at it that and we think about the original function between negative Infinity and -9 halves, we've got to be increasing from -9 halves to one, we've got to be decreasing. From one to three halves, we've got to be increasing. From three halves to Infinity, we've got to be decreasing. So we need to actually find the original function values at those locations. So F and -9 halves is going to be 93 fourths, making sure to use the top equation F of one. Once again, we're going to use the top equation is -7 and F of three halves using the bottom equation because I needed X greater than one is -27 fourths. So our critical points where the F prime was undefined is that X equal 1 our our X values where the first derivative of 0 is -9 halves and three halves. Our local Max increasing to decreasing is at -9 halves, 93 fourths and three halves -27 fourths, and our local min decreasing to increasing is at one -7.