Problem

Use the guidelines of this section to sketch the …

07:53

Question

Answered step-by-step

Problem 45 Easy Difficulty

Use the guidelines of this section to sketch the curve.

$ y = \dfrac{1}{x} + \ln x $

Video Answer

Solved by verified expert

Textbook Answer

Official textbook answer

Video by Jamie M

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 5

Summary of Curve Sketching

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 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

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Problem 60

Problem 61

Problem 62

Problem 63

Problem 64

Problem 65

Problem 66

Problem 67

Problem 68

Problem 69

Problem 70

Problem 71

Problem 72

Problem 73

Problem 74

Problem 75

Problem 76

Video Transcript

So for the domain, we know that it's gonna be, um X is greater than zero because it has to be first of all, ex cannot be zero because of first part. And the second part tells us that it has to be positive. So x is greater than zero intercepts. There are none symmetry, there is none. And for Assam totes, if we look for a horizontal, ask himto, for example, we have to set limits. Um, as X approaches infinity or limit as X approaches. Negative infinity. We see that there's no horizontal. However, there is a vertical ass until at X equals zero. Because we set our denominator for one of her ex. We said the denominator to zero. So x zero is our, ah, vertical ass in two. And now we look for increasing decreasing intervals. So you find the first derivative y prime, and that is one of her ex minus one over X squared. And if we set our first derivative to zero, we confined critical points and we see that X is one is a critical point. So if we do a first derivative test, so this is for F prime and we substitute, um, and research for inside of our derivative. We're gonna plug values that are smaller than one. What we have to remember to stay in our domain. So if it's smaller than one, it's going to be negative. If it's greater than one, it's gonna be positive. So that means that it's decreasing until it gets to one and increasing after that. So if it's going from negative to positive, so decreasing to increasing its gonna be men a local minimum value. And if you want to know the why value of our critical point we go F one. So we're putting it back into her original equation to find one. So 11 is our minimum local minimum value for our graph. And now we can find, um, we can look at kong cavity for this graph by finding the second derivative. So why double prime equals negative, and this negative is on the outside of the entire fraction. So negative X minus two over X cubed. So if we set that equal to zero, we see that X is equal to two and now our second derivative test, so double prime, and this is too and we have to make sure to test values in our domain. So, for example, if it says 0.5 for over here so smaller than two. But in still inside of her domain, we're going to get positive. So Khan came up, and beyond that, it's gonna be Khan came down so negative. Okay? And now we can graph with the information that we got. So first thing I'm gonna put on the graph is the, um, vertical ascent. Tow X equals zero, and we can see that, um, we have a minimum value at 11 So let's say this is one. This is one. So that's our local minimum value. And, um, we see that we haven't Oh, yeah. This is an inflection point because it's switching signs from Khan gave up to conquer it down. So we can say that this is an inflection point at two. All right, so, um, say this is about to So this is a rough sketch of our graph. It's gonna be decreasing until it hits our minimum value so it cannot touch the ass in tow. So decreasing until hits the minimum value, then it's gonna increase from then on. So it's so we have con cave up. That's correct. And then it's gonna be calling came down after it gets to our inflection permit. So it's Kong cave up, but still increasing until it's our inflection point that it's gonna be con cave down but still increasing, and that's a rough sketch.

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