Problem

Sketch the graph by hand using asymptotes and int…

08:08

Question

Answered step-by-step

Problem 13 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{(x + 4)(x - 3)^2}{x^4(x - 1)} $

Video Answer

Solved by verified expert

Textbook Answer

Official textbook answer

Video by Bobby Barnes

Numerade Educator

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Related Courses

Calculus 1 / AB

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

Section 6

Graphing with Calculus and Calculators

Discussion

You must be signed in to discuss.

Top Calculus 2 / BC Educators

Watch More Solved Questions in Chapter 4

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Video Transcript

we want to sketch the graft by hand, using incentives and intercepts. Oh F of X is equal to X plus four times X minus three. Squared over. Excellent forth times x, Nice one. Then we want to use our sketches, a guide to producing graphs that will display the major features of the Kurt and going to use that graft to estimate the maximum had minimal Bowden's. So let's go ahead and first just figure out world are we're all of our vertical and horizontal plastic tubes were going to be So let me just go ahead and draw little graph here. So 1234 1234 Now remember, we get vertical asking totes when our denominator is equal to zero. So that tells us X is equal to zero or ex physical one. And something else we're going to want to know is the multiplicity of these So multiplicity. So this here is for an ex is just going to be one now to find our next let's pluck those first So X is it to zero and X is equal to one Now let's go ahead and find our horizontal and so in this case, remember, What we do is we look for our largest power in the new Miranda nominator, which in this case, is gonna be X to the fifth. And I'm gonna multiply the top bottom by excellently on DDE. Doing that will give us that. Next plus four x minus three squared over won over extracted times extra force X minus one over X. Now the x of the fourth annex that it just becomes one of her ex. And I should be the rest of that into the denominator to just get one over one sex. And in the numerator, I can distribute two of them inside of X minus three. And doing that will give X minus three over one one over three over X squared and the other three I can distribute, Expose for and doing that would give one over X squared plus four over X and of God that little bit clearer before over x cubed plus one over X squared some now as X approaches infinity Well, all the one of her ex terms. We're going to go to zero so I can rewrite that. A zero plus zero one Myers zero squared over one minus zero, which goes to zero. So that tells us our horizontal ask himto is that why is equal to zero? Let's go ahead. And that is all right, easy to deserve. And then, lastly, for intercepts, you go ahead and set the numerator equal deserve at least far Exeter's ups and doing that well, tell us that X should be negative for or X is equal to debris. And again, we should just go ahead and save the multiplicity of these are over four or negative towards one and for three, All right, now what we want to do is to decide if we're going to start above or below, um, firm on one side of our class in tow, either going to negative or positive. So I'm just gonna go ahead and look at X is equal to zero. So let's look at the limit as X approaches zero from the right of X plus four X minus three squared over X to the fourth X minus one. Now zero our experts, for if I approach this from the right, Bull X is going to be a positive number. If I add up to something positive that'll be positive now if I approach fix my story from the right, it's going to be negative. Three. But if I square that it becomes positive, my approach next to the 4th 1 right, Will I be raising a positive number two or so? That's gonna be positive and then X minus one. That's going to be negative. So I have three positives, one negative. So this tells me you should go too negative in Bennett. So to the right of excessive zero, I'm going to go to negative 30. And now this is all I really need to know. And I'll just go ahead and also plot my interception quickly. So three and negative for so between X is equal to 01 I know I have no intercepts, so it's going to curve up like this and then have to curve back down at some point and then, since exited, the one has a multiplicity of one for its horizontal or vertical acid. That means it's gonna have to start from the opposite side like this and then is going to approach our X intercept, which is three. And since X is equal to two, was a multiplicity, too We know is goingto balance off. Yes, I was gonna have to increase for a little bit and then asked Certain decreasing again. So we will approach our why her horizontal ass. Until that, why is it zero? So that takes care of everything on the right side of our graph now on the left side. So X equals zero has a multiplicity of four. So since it's even, they should start on the same side, or they should be on the same side of our critical aspect. So this I will also go to negative infinity, and then it's just gonna go up until we hit our X intercept. And then I'm going to pass through the X intercept and then at some point, has been after peak so I can start decreasing and approach my Why access again? So, looking at this, the thing they tells we want to do is to determine what these maximum and minimums are going to be. The little green men below exes instead of dots are going to be so we know we're gonna have a local backs slightly to the left of our intercept at negative or somewhere between zero and one. We're going to have a maximum local. Maxine at X is equal to three. We know we're gonna have a local men. And somewhere after X is equal to three, we will have a local max so we can go ahead and use this now to actually produce a graft. And so I went ahead and did that. All right. And this is the nice little graph I got. So all I did to get this waas I kept zooming out until I actually saw this part here between zero and one. And then I just kept on going on the X and Y direction until I actually found values where it looked like it starts to decrease again. So I went ahead and estimated these points and we have that at 0.8 to 1 minus 28146 We have a local max to the left of our intercept of negative four. We have negative by 0.57 and 0.18 We know for sure that we have a minimum at 03 so we don't run. You testified that one and then to the right of that, we have a maximum at 5.249 and 0.1

Get More Help with this Textbook

James Stewart

Calculus: Early Transcendentals

View More Answers From This Book

Find Another Textbook

Study Groups

Study with other students and unlock Numerade solutions for free.

Problem