Horizontal asymptotes of a function

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Homework Help Overview

The discussion revolves around the function f(x)=(2+4e^x)/(2+e^x), focusing on its increasing nature and the determination of horizontal asymptotes on both sides. Participants explore the implications of limits as x approaches infinity and negative infinity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the method of finding horizontal asymptotes using limits and question the approach of "plugging" in infinity. There is an exploration of the function's behavior for both positive and negative values of x to establish its increasing nature.

Discussion Status

Several participants have offered guidance on how to approach the problem, including suggestions to find the derivative for part a) and to manipulate limits for part b). There is an acknowledgment of different interpretations regarding the asymptotic behavior of the function.

Contextual Notes

Some participants note the confusion around using infinity in calculations and the need for a rigorous approach to limits, while others highlight the importance of understanding the function's behavior across its entire domain.

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Homework Statement


Consider the function f(x)=(2+4e^x)/(2+e^x). a) Show that f(x) is increasing for all x. b) Find the horizontal asymptote on the left and right side.


Homework Equations


Use of the lim x->oo to find HA's



The Attempt at a Solution


Seems like an easy question, but it's got me slighty confused. For a) I tried plugging in oo (refer to NOTE at the bottom) for x and solving it that way, but this comes out undefined. Would this be the correct answer or do I need to use another method to prove that f(x) is increasing for all x? For b) I don't know where to start, as I have no idea how to show it has two asymptotes. I was able to find one of the asymptotes (y=4) by factoring e^x out, cancelling it off and then placing in oo, but I don't know how to get the other HA of y=1. Any help would be greatly appreciated, thanks in advance.

NOTE: I used oo to represent infinity since latex wasn't working.
 
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You don't "plug" in infinity, you allow the function to go to an arbitrarily large number. As for the left side you do the same but negative

As for a) when a function increases for all 'x', that means no matter what number you plug in the resulting number is greater than the value of the function at x-1.

Think about:
What happens when you plug in negative numbers? what about positive numbers? Does the function still increase?
 
Think about the following quotients when N gets very large, say a billion or a quintillion (1018). What number is the quotient close to being (Ie., if you were using an 8-digit calculator, what would it display)?

1) N/N

2) (N + 1)/N (Your calculator doesn't have enough decimal digits to display both the leading 1 and the trailing 1 for 1/1018. What will it round off to?)

3) (N + 2)/N

4) (N + N)/N

5) N/(N + 1)

6) N/(N + N)

7) (N + 1)/(N + 2)

8) (2N + 2)/(N + 2)

9) (4N + 2)/(N + 2)

The last is pretty much what they mean when they say "let x approach infinity" in your problem. When they say this, simply let N be extremely large compared to the largest constant in the problem. You can make "extremely large" rigorous by going through the epsilon-delta method. Analysis lives on approximation (made rigorous by epsilon-delta arguments).
 
Last edited:
For a), to demonstrate that f(x) is an increasing function for all x, find f'(x) and look for any maximums or minimums; since f(x) should be increasing if you are asked to demonstrate this, there should be no maximums or minimums. Then take f'(c) for some c in the domain of f(x) and f'(x). If f'(c)>0, the function is increasing on all x.

To find the horizontal asymptotes, take lim_x>inf f(x) and lim_x>-inf f(x). Do not simply "plug" infinity into f(x); manipulate the limit in order to get an expression that does not result in an indeterminate form.
 
Thanks for all that help guys, I finally solved it and now it makes sense :smile:
 

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