Problem

Use the guidelines of this section to sketch the …

04:52

Question

Answered step-by-step

Problem 31 Hard Difficulty

Use the guidelines of this section to sketch the curve.

$ y = \sqrt[3]{x^2 - 1} $

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 here we know the domain is exes, all riel. There are no ass in totes because it's not a rational function. So no, ask him to its And we know that the symmetry is even because when we do f of X is equal to f of negative X. So this is even and for our intercepts, we have, um if we said why equal to zero, we have zero equals the cube root of X squared minus one. And so take, um, this side to the power of three, this side to the power of three. And we're left with zero equals X squared minus one. So we can see that X is equal to plus or minus one. And similarly, if we plug in zero for X, we're going to see Mom. Why equals the cube root of zero minus one? So we see that this is 010 negative. One is an intercept here. So zero negative one is an intercept as well as from this one. We see it's plus or minus. Come on, plus or minus 10 for other intercept intercepts. And now we have to find why prime? Because the first riveted test allows us to see increasing and decreasing intervals. So if this is why, then why prime is 1/3 X squared, minus one to the power of negative to over three times in the inside. So to X, why prime equals to over three x times X squared minus one to the power of negative to over three. And if we set this equal to zero to find critical points X is equal to zero and X is also equal to plus or minus one. So it's true. The first derivative test. It's a crime and put her critical points. We have zero negative ones and positive one. So we test values in the first derivative. Anything smaller than negative one negative. Anything in between. Negative one and zero negative between zero and one positive and anything beyond one is positive. So here, we see, is going from negative to positive. So this is Amen. So we're done in the first derivative test. So now we have to look a con cavity. So now we have our first derivative. We have to find second derivative Furqan cavity. So why double prime is equal to negative two x squared plus six favor nine times X squared minus one to the power of five over three. And if we set that equal to zero, we see that X is approximately plus or minus 1.7. It's about it's a squared of three. So let's make a line for the second derivative test Testing for con cavity Negative 1.7 in positive, 1.7. So here we see anything smaller than 1.7 is going to be negative. Anything. And we're testing in the second derivative, remember? And then anything between negative 1.7 and 1.7 is positive than anything beyond 1.7 is negative. So here we see. This is positive. So this is Khan gave up and here it's concrete down. And here it's concrete down. And also we see a sign change from positive to negative. Negative 1.7 and from positive to negative, Um, at 1.7. So this is an inflection point. And so is this inflection point at one place a negative 1.7 in 1.7. So this tells us that con cavity for Agra. Now we can grab it. So here we have zero negative one. So zero negative one. This isn't intercept positive one and negative one. Her also intercepts, and it's even. So we're gonna see symmetry across the Y axis. And now we see that it's there's an inflection point at around negative 1.7. So around here and around 1.7, so around here, So it's gonna change con cavity. So over here it's gonna be con cave down and around 1.7, it's gonna switch to Khan gave up, it's gonna hit the axis, it's gonna be concave down. And then it's gonna go back up here, and it's going to go from 1.7 and beyond. Its gonna be back to con cave down. So about 1.7 and the sources which backed it down. And this is the graph for the function

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