💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!
Get the answer to your homework problem.
Try Numerade Free for 7 Days
Like
Report
Use the guidelines of this section to sketch the curve.
$ y = \frac{x}{\sqrt{x^2 - 1}} $
see solution
Calculus 1 / AB
Calculus 2 / BC
Chapter 4
Applications of Differentiation
Section 5
Summary of Curve Sketching
Derivatives
Differentiation
Volume
Campbell University
Harvey Mudd College
Baylor University
University of Nottingham
Lectures
01:11
In mathematics, integratio…
06:55
In grammar, determiners ar…
06:31
Use the guidelines of this…
20:32
15:54
14:38
16:12
14:39
16:02
16:20
04:52
16:30
So we know the graph is gonna look something like this because it's a rational function. But, um, first, let's find out the ass in totes so that we can know exactly where the graph is gonna be. First of all, the domain. Um, since there's a square on the bottom, we cannot. Well, since there's a square, yeah, we cannot have a negative under the radical. And also we cannot have the denominator to equal zero. So that means for the domain, it could be anything besides x zero so X can not equals here for the domain and ex cannot equal plus or minus one, but it can't exist anywhere else. Intercepts there are none. So substituting zero for extra Why we see that there are no intercepts. Symmetry. It's odd because F of negative X is equal to negative f of X. So it hasn't always symmetry ass until it's so first was finding vertical ascent oats. So were the denominator equal zero. We already said that was plus or minus one. So let's test out the limit as X approaches, plus or minus one. So let's do one first. Um, and if this equals infinity or negative infinity. We know it's a vertical lassitude. So substituting one, we see it's one over zero, so that's infinity. And if we do the same thing for a negative one, we just know it's gonna be negative. Infinity. So for vertical, ask himto we can say that X is plus or minus one. That's where Assam totes are, which is why the domain cannot be plus or minus one. All right, enough reverted for ah, horizontal as in two. The horizontal ask himto is where the limit approaches infinity or negative infinity. So the limit as X approaches infinity of our function X over swear ou of X squared minus one. So if we factor out in the denominator, we get the limit as X approaches Infinity of X over the square of X squared one minus one over X squared. And if we take that out of the square root. If you take this out of the square root, we get an X and X and X cancel. So we're left with the limit as X approaches. Infinity of one over the square root one minus one over X squared, won over. Affinity is zero. So one minus zero is one. So our limit is one see, and the limits as X approaches infinity or a negative infinity, we're gonna follow very similar steps. The only difference is that when we pull out because it's approaching negative infinity here. When we pull out the X squared out of the radical, we're just gonna make it negative X. So we're gonna be left with, um when they cancel out a negative on the bottom, that's the only difference. So the limit of negative square root of one minus one over X squared is negative one. So we have a horizontal ass in two. That why equals one and why equals negative one. And now we can graft with the information that we have. All right, so we're gonna put in the Assam totes and red, So here we know I should have been angry. So here we know X plus or minus one, one, minus one. That's our vertical asking totes. And we also know that why is one when why is one and negative one? We have horizontal ass in two, and that's where our function can lie. That's how you grow
In mathematics, integration is one of the two main operations in calculus, w…
In grammar, determiners are a class of words that are used in front of nouns…
$ y = \sqrt[…
$ y = \frac{…
$ y = \sqrt{…
$ y = x\sqrt…
$ y = 1 + \f…
