Problem

If $ f(x) = 3x^2 - x^3 $, find $ f'(1) $ and use …

01:50

Question

Answered step-by-step

Problem 26 Medium Difficulty

Sketch the graph of a function $ f $ where the domain is $ (-2, 2) $, $ f'(0) = -2 $, $ \displaystyle \lim_{x \to 2^-} f(x) = \infty $, $ f $ is continuous at all numbers in its domain except $ \pm 1 $, and $ f $ is odd.

Video Answer

Solved by verified expert

Textbook Answer

Official textbook answer

Video by Anjali Kurse

Numerade Educator

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

Watch More Solved Questions in Chapter 2

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

Video Transcript

Okay, so here we have this function that we have to sketch actually when we have a bunch of facts about it. So let's start with the domain. I'm just going to make each of these tick marks equal to one half. So that this graph goes, sorry, go to one half. So that goes like this. Okay, so our graph and notice that there are open parentheses around the domain negative to the positive too. So our graph isn't going to include those values of X if they were brackets, we would include those values, but they're not so we're not going to Okay, so we have the domain. Now we have the fact F prime of zero equals negative two. What does this mean? Well, f crime, F prime refers to the slope of f slope of F. So this means that at the value x equals zero. The slope of F Is going to be equal to -2. So we can't really draw that but we could draw the shape of it negative two looks a negative two, slope looks something like that. Right? That means our graph has to follow that path for F prime of zero to have a slope for f of zero to have a slope of negative two. Okay, our next step, uh the limit as X approaches two from the left, the limit as X approaches to from the left of f of X is positive infinity, positive infinity. So what does that mean? Well, we have our domain sketched in here, the limit as F of X, which might look like whatever here as it approaches to the X. Value to from the left, it's going to head up to positive infinity. So the graph is going to head up that way. Okay And then f is continuous at all numbers in its domain except at plus or -1 and F is odd. Okay, so plus or -1 At closer -1. We have to have some kind of discontinuity. This continuity. Uh huh. And and F is odd. Odd. So what does odd mean odd means symmetrical about the origin, symmetrical about the origin. So think of a graph like execute execute me or tangent that looks similar where it's symmetrical about the origin. So if you flip it up over the X axis it would look like this. And then if you reflected it over across the UAE it would look just like that. So that's why it's symmetrical about the origin. If you cross it across the X. And the Y. They matches up to the other side. Okay, so what does that mean for our graph that means that if this goes up to negative and positive infinity at this side we have to have something going down to negative infinity Because if we use this graph as a model again, we have to be able to flip it over and over. So think over here over here and then over here it would match up with this negative infinity with this positive infinity. Sorry? Okay and then our discontinuities are going to have to match. So how about we just try to sketch something and see if it if it works properly basically. So let's say we have a graph that I'm just going to redraw this, that looks something like this. We and then let's say we have a jump here just for fun and then it has to be able to copy so just like that. Let's see if does this work? Let's see. So if we flip this over so just think about flipping this part over the y axis first, so it'll look like this. What? And then you flip it up so this part comes down, it matches up perfectly. Okay, cool, so I can arrest this part. Yeah, and then can we flip this part? Well, if we flip it over, it'll look like that like that. And then if we flip it up, don't match like that. So it does take some imagination here to figure out even an odd and how to make this graph look just like this. But this works. Let's double check it. So f of negative zero is negative two, yep, that's a slope of negative something right there. It looks like about negative two to me. And then the limit as X approaches to from the left is positive infinity. There we go and it plus or minus one. We have a jump dis continuity. Cool. And then it's odd. So we have symmetry about the origin, and the domain is negative to positive, too. So this works. You could have another variation of this, you could add a cusp here, A -1 and positive one a different kind of discontinuity. But this function and it shows if we flip it across the X and the Y is symmetrical about the origin and fits all of the other conditions about this question. Yeah.

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

If $ f(x) = 3x^2 - x^3 $, find $ f'(1) $ and use …

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

Anna Marie Vagnozzi

Heather Zimmers

Kristen Karbon

Joseph Lentino

Calculus 1 / AB Courses

Recommended Videos

Watch More Solved Questions in Chapter 2

Video Transcript

James Stewart

Calculus: Early Transcendentals

Anna Marie Vagnozzi

Heather Zimmers

Kristen Karbon

Joseph Lentino

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

Calculus: Early Transcendentals

Anna Marie Vagnozzi

Heather Zimmers

Kristen Karbon

Joseph Lentino

James StewartCalculus: Early Transcendentals

Anna Marie Vagnozzi

Heather Zimmers

Kristen Karbon

Joseph Lentino

James Stewart

Calculus: Early Transcendentals