In this problem, we're going to take the F of the two endpoints, so F of -2 and F of three. So if I stick a -2, we get the square root of the opposite of -2 ^2 + 9, sqrt -4 + 9 or sqrt 5 and we put in three, we get the opposite of 3 ^2 + 9 or the opposite of -9 + 9 or 0. Now we have to take the first derivative to find if there are any extreme points. So the first derivative here is going to be, it's a square root. So we have 1 / 2 square roots negative X ^2 + 9 and that gets multiplied by the derivative of the inside and the derivative of the inside is -2 X. So simplifying this, the twos are going to cancel and we get a negative X over square root negative X ^2 + 9. We set the first derivative equal to 0 to figure out where the horizontal tangents are. So we get 0 equaling negative X or X equaling zero. We also have to pay attention to when our derivative is undefined. While our derivative would be undefined at sqrt -X ^2 + 9 equaling 0, if we square each side we get negative X ^2 + 9 equaling 0, X ^2 equal 9. So X is going to equal positive or negative square roots of Nope, positive, negative or three. The -3 is not within our range and the three was already one of our endpoints. So now all we do is we find our F of 0 sticking it back into the original square root. The opposite of 0 ^2 + 9 is going to give a square in a nine or three. So if we thought about this with our derivative 0 here -2 here and three here on a number line, if we think to the in between zero and three, maybe at one, looking at this original derivative, we would get a negative value -1 over square root of opposite of 1 ^2 + 9. That zero came from that numerator. It's got a multiplicity of 1. So we know on the other side it's going to be positive. So as we're doing this problem, if it's positive, we know that the original function is increasing. Then we know when it turns negative, the original function's decreasing. So the zero's got to be an absolute maximum at X = 0 for the height of three, because this height 3 is bigger than any of the others. Zero is smaller than any of the others, so the three is going to be my absolute minimum at X equal 340, and this one is going to be a local minimum at -2 four square roots of five.