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(a) Find the $x$ -intercept(s); (b) Find the vertical asymptotes; (c) Find the horizontal asymptotes. (d) Sketch the graph$$f(x)=\frac{2 x}{\sqrt{x^{2}-1}}$$

(a) $x=0$(b) $x \pm 1$(c) $y=\pm 2$$(d)$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 4

Limits at Infinity, Infinite Limits and Asymptotes

Derivatives

Campbell University

University of Michigan - Ann Arbor

Idaho State University

Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

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(a) Determine the $x$ -int…

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For the graph of $y=f(x)$<…

in this problem, we're going to be finding the X intercepts and ask some totes for the function F of X is equal to two. X divided by the square root of x squared minus one, and then sketching a graph at the end. So for part A here, finding these X intercepts, we know we set our function equal to zero and solve for X. You can see multiplying both sides by the denominator just leaves us with two. X is equal to zero, which is obviously saying that our X intercept occurs when X is equal to zero. We can plot that on our graph over here for part B. Finding these vertical ascent oats, we know that we like to look at our denominator. That's really the important part for the vertical ascent oats. And we just set that denominator equal to zero and solve for X. Again, taking the square root obviously leaves us with X squared minus one is equal to zero, which gives us X squared is equal to positive one. And taking the square root there, that'll leave us with X being equal to plus and minus square root of one. So plotting that now for our vertical assam totes we can give ourselves a nice started line here. And really all we're doing on the graph right now is giving ourselves this basic outline for when we do go ahead and draw the curve and for part C finding these horizontal asientos. Now this occurs or we find it by taking the limit as X approaches infinity of our functions. So two X divided by the square root of X squared minus one. And we want to pay attention to our dominant terms in both the numerator and the denominators of the numerator that's going to be just to access the only term we have and in our denominator we see it's going to be X squared But to make a little easier on ourselves, you can see that we're taking the square root of X squared, which is really just X. So we're gonna be left with taking the limit as X approaches infinity of two. X divided by X. Our exes canceled. And we're left with two but jumping back and remembering that we just took the square root, it's actually going to be a plus or minus two. So we have a horizontal ass into it when y is equal to plus and minus two. So we can sketch that on here as well. And going ahead now and sketching this if you want, you can plug the function into your graphing calculator to get a better idea of when the turns take place. But ultimately you'll see that it does follow what we just sketched and it looks something similar to this

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