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University of North Texas



Problem 14 Medium Difficulty

Sketch the graph by hand using asymptotes and intercepts, but not derivatives. Then use your sketch as a guide to producing graphs (with a graphing device) that display the major features of the curve. Use these graphs to estimate the maximum and minimum values.

$ f(x) = \dfrac{(2x + 3)^2(x - 2)^5}{x^3(x - 5)^2} $




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Video Transcript

we want to sketch. The graph of level X is equal to two x plus three squared times X minus two to the fifth all over X Q. Times X minus five Squared by hand on Lee using s imports and intercepts and no derivatives. Then once we get our sketch, we want to use as a guide so we can get a nice graph, uh, of this using some kind of graphic device. And then we want to use that graph to estimate our maximum and our minimum Bowden's. So the first thing we should do is go ahead and identify what are, um, intercepts. And our horizontal or vertical acid trip should be so. Remember X intercepts or where our numerator is 80 So tell me, x zero to negative Three hops and X is equal to are vertical. Awesome tips are where are denominator is equal to zero, so X is equal to three and X is equal to bye. And then we should also find our horizontal Ask him to, and in this case, we know it's going to not have so no ask him to. Due to the fact that our power in the new mayor is seven and our power and our denominator is five. So since, um, numerator power, you got a seven and the denominator power is equal to you. Could go to the whole process of multiplying the top and bottom by one of her extra seventh and then actually evaluating the one that as excuse and fitting in negative 30. But this is just a nice little thing toe, remember? Now the only other thing we need to know is the in behavior on one side of our vertical ascent. Oh, and then weaken sketched a graft from there. So let's look at the limit as extra pictures. Zero from the right of two x plus three square X minus two raised to the fifth all over X cubed. That's my eyes. Five. Sweat Now, since we already know this is going to go toe infinity or negative than the all I really care about is, is it going to be positive or negative as we approach so and our numerator will two x plus three we don't really need to care about, since that's going to always be squared, so it's always going to be at least positive now X minus two. Well, the limit as we approached zero there from the right is going to be negative. And if I raise it something negative to the fifth power, it's still gonna be negative If I approach X cubed from the right of zero. Well, I'd be cubing a small positive number, which would be positive and then x minus five if I purchased from the right of zero is gonna be a negative value, but squaring it makes it positive. So, overall, this should be a negative quantity. So I know to the right of zero are vertical ass into is gonna go to negative infinity. Now, this is all the information I believe we need. You actually start graphing this. So let's go ahead and put a little couple of points down. Uh, so we need to at least go out to five on the right or are vertical accident and at least two Tola or are interested. So let's go ahead. Block those. So we're a negative three house there. And over here at X is equal to five. We have a possum as well as that X is equal to zero now. We know to the right of zero. We're approaching negative infinity and I almost forgot to plot our other intercept here at two. Now, I claim this is all we need to know, because now we're going to approach our X intercept to the right of zero, and we need to decide. Are we gonna bounce off this or go through the point? So the multiplicity add x is equal to two is going to be five. So that's on meaning we're going to pass the ruined, so it's gonna kind of curve up a little and then start increasing again, and then we have no other in points here. So then we need to hit our next vertical ascent. Oh, which would then go to infinity and now X to the five where X is equal to five hasn't even multiplicity. So that means is going to be the same value on either side of orbital class into, So it should start from positive indeed to the right and then I have no intercepts over here, so I know is going to have to decreased down and then increased like that since our end behavior is going to be going to infinity or negative infinity. All right, Now for excessive zero, this has a odd multiplicity. So that means we're going to start on opposite sides of our vertical. Awesome, too. So, like this. And then we're going to go until we hit our ex innocent at negative three halves. And since it's an even multiplicity of two, we're going to bounce off. And then we're going to approach infinity like that. So you might know this that here at X is equal to negative through halfs, we're going to have a minimum value since we balance off there and between 05 it doesn't really look like we're gonna have a max or men. But then, after exceeding five somewhere, it's goingto have to have a minimum value as well. So between our after excessive five, we just need to make sure that we get this top right portion into the graph we're going to make, and then we can find that minimum about you. So I went ahead and already graph dysfunction here. So just like I said, I just went out high enough on my y access so I can actually get this part of the graph and shot. And then I went ahead and just estimated what, at minimum value right there looks to be and what I got was occurs at X is equal to 7.90 and our output should be about 609 points. 174 And we already knew what the value of our other minimum was going to be being at negative three halfs zero. And I just want to have employed at that point on there.