Problem

Draw a graph of $ f $ and use it to make a rough …

01:47

Question

Answered step-by-step

Problem 57 Medium Difficulty

Draw a graph of $ f $ and use it to make a rough sketch of the antiderivative that passes through the origin

$ f(x) = \dfrac{\sin x}{1 + x^2} $, $ \quad -2\pi \leqslant x \leqslant 2\pi $

Video Answer

Solved by verified expert

Textbook Answer

Official textbook answer

Video by Carson Merrill

Numerade Educator

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

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

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

Video Transcript

So here we're looking to draw the graph and make a rough sketch of the anti derivative as it passes through the origin. So given the sin of X over one plus X square, I'm negative two pi to pie. Or looking at is this region right here. So what we're gonna do is um recognize that our anti derivative, it's going to have zero slope. So it's going to be um like this, the area under the curve is very little and then at this point the area under the curve is negative. So that means that we're going to have a decreasing slope and then the area bullock curve is positive right here. So it's going to be increasing at that point. Um And then it's going to reach a maximum point at zero right here and then um at the zero point because that's the maximum this is going to come down and go toward zero. So what we're going to have is something like this where it comes up like a little hill and then it goes down. Um So an example, this would be like 1/1 class X squared. Something like this would be an example, however, it might dip a little bit lower. Um So it might be something like like this, but maybe we could expect one um be a little bit shifted up more alright, five. So it's something like this would be a little bit more accurate um in representing the anti derivative of this 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