Problem

Use the guidelines of this section to sketch the …

Problem 15 Easy Difficulty

Use the guidelines of this section to sketch the curve.
$$y=\frac{x-1}{x^{2}}$$

Answer

$\left(3, \frac{2}{9}\right)$

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Video Transcript

we want to sketch the curve of the graph, why is equal to x minus one over X squared. So in this chapter we're given this laundry list of steps we should follow to grapple function. So we're going to follow those stuff. Buy stuff. So the first thing they suggest is that we determine our filming and we know for rational functions as long as our denominator is not equal to zero. So we need to make sure that X squared does not equal zero. We will have part of me. We're just going to be negative in any 20 union, zero to infinity. The next thing we are going to look for is our intercepts. Well, we're going to know that we're going to have no Y intercept since our function is not defined at X is equal to zero. So we can go ahead and find our X intercept though. So X intercepts is going to be where we set y equal to zero. Get exercised over expert And we know our denominator can never make this equal to zero. So we just have the number equals zero and zero is equal to x minus one or X is one. So at 10 we will have a exit or something. The next thing they suggest we do is to look for symmetry. Well, special functions aren't very known for being periodic. So we can check to see if this is going to be even or odd by looking at f of negative looks. So plugging at it, we'll get negative x minus one over x squared and we can factor that negative X. Out to get X plus one over X squared. Now this does not equal to ethel X. Nor is it equal to negative fx. So there is no symmetry at least around the origin or across the UAE access. The next thing they want us to do is to look for horizontal supports or horizontal and vertical institutes acid toots. Now we know from Chapter 26 that if we look at the in behavior of irrational function, it should be the same On either side for this one due to the fact that our denominator is larger than our denominator is larger than the power of the numerator. So we know that this function is going to have a horizontal store it. Why is it zero? Now to find our vertical ascent, we know that if we plug in zero or denominator becomes zero. So we need to look for the behavior to the left and right of zero. So this is going to be zero from the right -1 Over zero from the right square. So if I have some number slightly to the right of zero and I subtract one from it that's going to be negative in the numerator. And if I take something slightly to the right of zero and I swear to that it's going to be pasta. So I have a negative number over a positive number. So this is going to go to negative infinity. And we know since that term is a even degrees vertical ascent. Hope that it should have the same in behavior or the same behavior about the vertical ascent. O as we approach it from the right, so you can go through the same thing, Look at zero slightly to the left and we'll come to the same negative over positive. But that's just a quicker way to cocktail. The next thing they want us to find is intervals where the function is increasing or decreasing and we want to find any local maxes or mins. So to do this, we're going to need to figure out what why prime is and let's go ahead and do that on another page. So why is able to X -1 over expert. And now I'm going to just go ahead and divide the expert into each of those to make a little bit easier. Two take this derivative or maybe a little quicker. So I don't have to use questionable. So why prime? Well, one of our X is really X to the negative first and one of our X square is really extra negative second. So we can use the power rule to take care of the beach those. So we're going to end up with negative one over X squared and then going to have minus times negative two over execute. Now those negatives there can cancel out and we can go ahead and combine this into one fraction to give us X plus or negative X negative X plus two over X. Cute. Yeah. Now let's go ahead and find our Critical points. So again, nothing I can plug into denominator will ever make this function from zero. So I just set the numerator equal to zero and that tells me X is equal to two is a possible local man. Our necks now also remember we would want to check where the function is undefined for critical values but since we place that are derivative is undefined is at X equals zero. And that is also not in our domain. We don't need to worry about. All right. So Here we'll have 2 -1 over execute. Now this function is going to be out of increasing When my private strictly larger than zero and already went ahead and solve this and this should be Between zero and 2. And we know the function is going to be decreasing when Y prime is strictly less than zero. And when I did this I got negative 50-0. Union two. So we are told that are possible point that may be a maximum is excited to So To the left of two the function is increasing and to the right of to the function is decreasing. So this tells us we have a local max At X is equal to two. Well, the next thing we want to do is to go ahead and find our con cavity and any inflection points we might have. So that means we're gonna use okay, Y double prime and why double prime is going to be equal to? So again, I'm going to take the derivative using this one first. Uh just because I don't have to use questionable. So we're gonna need to use power rule once again. So we're going to end up with negative two or negative one times negative to let me go ahead and take that. So negative form times negative two. All over X. Can't plus two times negative three Over X to the 4th. Now these negatives here will cancel out with each other so we just get to ever execute and that will become negative six. So if we were to add these together, We should get two x 56 over X to the fourth. And we can go ahead and factor that two out to get two X minus three over X 24. So let's see zero. So we can find a point of inflection. So again, Denominator can ever make this even zero. So we're just gonna set X -3-0 which tells us X is equal to. Mhm. So we have a possible point of collection at X Secret three. So You have two X -3 over X 24. So we can go ahead and figure out where the smokers can't keep up and can't keep down. So it's going to be concave up when Y double prime is strictly larger than new Zealand And this is going to be on the interval 300 charity and we know it's going to be concave down when light of the prime is strictly less than zero, which would just be the rest of our interval excluding zero. So negative infinity. Two, zero union zero to All right. And so remember we're just excluding that zero there sense our function is not defined right? And now we can go ahead and look at X is equal 3-4 of the study. Concave down our timetable. So to the left of XC, three function is going to be competent down and to the right of three function is going to be concave up. So we do have a change an inflection. So we know this year should be a inflexible. No, that was the last thing they suggested we do before we start graphing. So let's go ahead and graph now. So we had a intercept at X0-1. So let's go ahead and put that right here. We had no symmetry, we have some ascetics so why is equal to zero? We have a horizontal velocity and at X is equal to zero. We will have a horizontal, isn't it? So to the right of zero we're gonna approach and 30 negative in the end to the left, zero, we're going to approach negative infinity. So we have a local max at the end of the tube. So let's just go ahead and put max here and we know we're going to have a change in inflection at X is equal to three. No, we go ahead and start sketching. So let's start to the left of X is equal to zero. So we're starting from negative infinity and we're going to go until we hit some important value. But here we don't really have any changing from cavities maximums or anything like that. So they're just going to get closer and closer to our horizontal. Isn't that why is it observed? And now on the right side, well we have our intercept at X. Z to the ones that are going to pass through that. Then we know after we hit that we're going to have to hit a maximum at X. Is equal to. So it's going to come over here and right now a little bit and then at Next is equal to three. I should have a change of con cavity like you back. So this here is at least in my opinion, are good enough sketch something you could go back and do is maybe going and say, well what is this value the maximum? What is the value at the point of inflection. But at least to me, I don't really think it matters that much

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