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AK
Numerade Educator

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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.

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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.

AK
The University of Alabama
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