Problem

At what point on the curve $ y = \sqrt {1 + 2x} $…

01:54

Question

Answered step-by-step

Problem 59 Hard Difficulty

Find all points on the graph of the function $ f(x) = 2 \sin x + \sin^2 x $ at which the tangent line is horizontal.

Video Answer

Solved by verified expert

Textbook Answer

Official textbook answer

Video by Heather Zimmers

Numerade Educator

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

Watch More Solved Questions in Chapter 3

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

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

Problem 77

Problem 78

Problem 79

Problem 80

Problem 81

Problem 82

Problem 83

Problem 84

Problem 85

Problem 86

Problem 87

Problem 88

Problem 89

Problem 90

Problem 91

Problem 92

Problem 93

Problem 94

Problem 95

Problem 96

Problem 97

Problem 98

Problem 99

Problem 100

Video Transcript

Our goal here is to find all points on this function where the tangent line is horizontal and if a tangent line is horizontal than its slope will be zero. That means we're looking for all the points that have a derivative of zero. So let's find the derivative. But first I'm going to rewrite the function as to sign of X plus the sign of X quantity squared, which is what it means if we write science Codex. But it's a little bit more difficult to use the chain rule when it's written the science cortex. Okay, so now for the derivative, the derivative of To Sign Next is to co sign X and for the derivative of sign of X quantity squared will use the chain rule first, the derivative of the outside. We bring down the two and we raise sign next to the first and then the derivative of the inside. The director of the sign is co sign. Okay, We want to know where this equals zero so that we get the tangent lines to be horizontal. So let's set this equal to zero and notice that both terms have a to co sign next that we can factor out. Yes, we have to co sign X times one plus sine X is equal to zero, so we can set each factor equal to zero when we set to coast Line X equal to zero and then divide by two co sign X equals zero X would be pi over two or negative pi over two. And when we set one plus sign of X equal to zero, subtract one from both sides. Sign of X equals negative. One X would be negative pi over two, which is the answer we already got in the previous part. So we have two answers with pi over two and negative pi over, too. But don't forget that those air only the answers when you've gone around the unit circle one time you could go around again infinitely many times, and you could go in the other direction as well. So they're infinitely many answers that give you the same co sign and sign values so we'll write those as hi over to plus two pi n to Pai n represents going around the circle end times and could be one and could be too, and could be three and could be any integer and the other one would be negative. Pi over two plus two Pion. Same idea. Okay, The problem asks for the points on the graph, so that means it wants X coordinates and y coordinates. We have X coordinates. We need to find Why Coordinates. So let's do that next. So let's take our function. F of X equals two sine x plus the sign of X squared and let's find the Y Coordinate for Pi over two. So that would be two times. Sign of Pi over two, plus a sign of pi over two school. Where'd sign of Pi over two is one. So if two times one plus one squared, so we get three. And now let's find f of negative pi over two. Same idea. We have two times a sign of negative pi over two, plus the sign of negative pi over two squared and the sign of negative pi over to his negative one. So if two times negative one plus native one squared and that gives us negative one. So let's go back and write these ordered pairs. So we had high over two plus two pi n for the first set of X coordinates comma three for their Y coordinates. And then we had a negative pi over two plus two pi n representing the other answers and was positive one negative one negative one for the y coordinates Those air all the points where the horizontal where the tangent line is horizontal.

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

At what point on the curve $ y = \sqrt {1 + 2x} $…

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

More Answers

Related Courses

Calculus: Early Transcendentals

Related Topics

Discussion

Top Calculus 1 / AB Educators

Grace He

Caleb Elmore

Samuel Hannah

Joseph Lentino

Calculus 1 / AB Courses

Recommended Videos

Watch More Solved Questions in Chapter 3

Video Transcript

James Stewart

Calculus: Early Transcendentals

Grace He

Caleb Elmore

Samuel Hannah

Joseph Lentino

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

Calculus: Early Transcendentals

Grace He

Caleb Elmore

Samuel Hannah

Joseph Lentino

James StewartCalculus: Early Transcendentals

Grace He

Caleb Elmore

Samuel Hannah

Joseph Lentino

James Stewart

Calculus: Early Transcendentals