Graph function by determining the key features of the curve represented by Y equal 3 square root of three X + 6 cosine X from zero, less than or equal to X less than or equal to 2π. Identify any local and absolute extreme points and inflection points. So the first thing we're going to do is find the first derivative and we get 3 root 3 -, 6 sine X. Setting that first derivative equal to 0 and solving, we can see that X = π thirds and 2π thirds. So if I put that on a number line and figure out where my positive and negative occurs, IE where my increasing and decreasing occurs, I can see that zero to π thirds is increasing π thirds to 2π thirds is decreasing, and 2π thirds to 2π is increasing. Remember, I'm referring to the original function. Then we're going to take the 2nd derivative and we get -6 cosine X, setting it equal to 0. We know that cosine X is zero at π halves and three Pi halves, so I'm going to put that on the number line. Also, I'm going to test those locations to figure out the concavity of the original function. So the original function is going to be concave down from zero to π halves, concave up from π halves to three Pi halves, and concave down again from three Pi halves to 2π. I need to figure out all of my key locations so my two end points on the closed and bounded interval, so Y of 0 and Y at 2π. Then I need to find my key points from my Y prime, so Y of Y of π thirds and Y of 2π thirds. And then I finally need to find my Y values of my Y double prime, so Y of π halves and Y of three Pi halves. So when we do all of that, we can see our values. Now if we know that the original graph is back here, let's use red. We know that the original graph is increasing but concave down. South increasing, concave down. Then we know the original graph is decreasing and still concave down until we hit π halves and at π halves it's still decreasing, but now it's concave up. At 2π thirds it's going to be increasing and concave up. At 3 Pi halves it's still going to be increasing, but now it's going to be concave down South. A rough sketch. So we can see that this has got to be a local min here π thirds is going to be a local Max here. At 2π, thirds is going to be a min again, and at 2π is going to be a Max to find which ones are going to be the absolutes. So those are all locals. To find the absolutes, we're actually going to look at the Y values. So the two Max's, this is a local Max at π thirds and another local Max at 2π. So to find the one that's the absolute, we look at those two Y values and figure out which one's the bigger Y value. So 2π is going to be my absolute Max, whereas π thirds is just going to be a local Max. To find the absolute min, we look at zero and we look at 2π thirds. And if we look at zero and we look at 2π thirds, 6 is less. So this is my absolute min, and this one is a local min. And then the other two points, the Pi halves and three Pi halves. Because concavity changes. These are points of inflection. Now we could graph this much more accurately by actually using the Y values. I just did a very rough sketch there.