Match Functions to Horizontal Asymptotes

  • Thread starter Slimsta
  • Start date
  • Tags
    Horizontal
In summary, you were trying to solve for f(x) using the limits of arctan(x4 - x2) and cot-1(x). However, you are stuck and need help. You also mentioned another problem that you are struggling with.
  • #1
Slimsta
190
0

Homework Statement


"Match functions and their horizontal asymptotes."


Homework Equations


lim f(x)
x-> + infinity

lim f(x)
x-> - infinity


The Attempt at a Solution


http://img198.imageshack.us/img198/7678/45328107.jpg

im just stuck at this... I am sure that everything is right!
but its wrong :(

i was thinking, arctan(x^4-x^2) = tan-1(x^4-x^2)
i saw on a different post, that someone said, graph x^4-x^2 then take the limit and do tan(L) and that's the answer.. but it doesn't seem to work.

the other one that may be wrong (but I am sure its not) is cot-1(x).. i graphed it as 1/tan-1(x) and got 0.6 and -0.6

please help me!
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
So your problem was with arctan(x4 - x2)?

What is the limit of arctan(x) as x goes to infinity? Does x4 - x2 also go to infinity as x goes to infinity?
 
  • #3
i don't know... I am assuming so.
thats my question, what wrong with my matching in the picture?
 
  • #4
F is correct. Was there another answer you thought it might be? Do you know why B would not be correct?

Also, what did you mean where you said "but it doesn't seem to work."?
 
  • #5
well the system uses all of the answer i input and shows me right/wrong overall.. it doesn't show me which one is wrong.. so i have to go through each one and check it.
i have check each one 20+ times and also graphed it on a graphing calculator and nothing seems to solve my problem because i keep on getting the same thing..
yeah and B wouldn't be correct because it is only + infinity..
and when plugging in -infinity, x^2 and x^4 make it positive anyways..

how about the cot-1(x) is this one correct?
 
  • #6
cot-1x ≠ 1/tan-1x


[tex]y = \cot^{-1}x[/tex]
[tex]\cot y = x[/tex]

[tex]\frac{1}{\cot y} = \frac{1}{x}[/tex]
[tex]\tan y = \frac{1}{x}[/tex]
[tex]y = \tan^{-1}\left(\frac{1}{x}\right)[/tex]
[tex]\cot^{-1}x = \tan^{-1}\left(\frac{1}{x}\right)[/tex]

You should be able to work with this last one now.
 
  • #7
ohhh really?
so that would mean that the answer is 0!
i thought it's 1/tan-1(x)

so the answer would be G=0?
 
  • #8
Yes, looks like that's the only one wrong.
 
  • #9
the system shows that i already had an answer like this.. which means that something else is wrong.

for (6x+1) / (sqrt(3x^2+1))
i get hor.asymptote as pi and -pi..
which means its J.. right?

edit:
so my answers are now in this order:
C
F
B
G
J
A

and looks like everything is right but still the system says its wrong...
maybe that one with J is wrong?
 
Last edited:
  • #10
That one is J (why [itex]\pi[/itex] and -[itex]\pi[/itex]?). Do you know what it's horizontal asymptote is?
 
  • #11
yeah.. horizontal asymptote is the y value that's a line which the graph doesn't pass (the limit of it.. gets really close to it.. and goes to infinity but still doesn't touch it)

if that one is right... then i just don't know.
the first one is the only thing left and its %99.999 right

edit:
the last one, when graphing it, i get 2. when solving it i get 0.. maybe i solved that one wrong?
 
  • #12
Slimsta said:
if that one is right... then i just don't know.
the first one is the only thing left and its %99.999 right

Which one are you talking about?
 
  • #13
the first matching fraction.. 6x^3+2x+1 / 2x^3-10x^2+13x+100
but i don't think that's the problem..

im more concerned about the last one
f(x)=2-x + sqrt(x^2+1)

im getting 0 when doing it algebraically, but graphing in on a calculator give me a limit of 2
 
  • #14
Slimsta said:
the first matching fraction.. 6x^3+2x+1 / 2x^3-10x^2+13x+100
but i don't think that's the problem..

That one has a limit of 2, so you're right.

im more concerned about the last one
f(x)=2-x + sqrt(x^2+1)

im getting 0 when doing it algebraically, but graphing in on a calculator give me a limit of 2

Some, like this one, need some creative rewriting so you can find its limit.
 
  • #15
Bohrok said:
That one has a limit of 2, so you're right.



Some, like this one, need some creative rewriting so you can find its limit.


you meant 3 for the first one right?

for f(x)=2-x + sqrt(x^2+1),
what i did is multiplying both top and bottom by -2+x + sqrt(x^2+1)
which led to:
(4x-3) / (sqrt(x^2+1) +x -2)
from there, i divide both top and bottom by x^2..

(4x-3)/x^2 / (sqrt(x^2+1) +x -2)/x^2

and i got (4/x - 3/x ) / (sqrt(1+ 1/x^2) + x -2)
if x-> infinity..

the top will be immediately 0 and the bottom is 1 which mean 0/1 = 0...
right?
 
  • #16
3, you're right. :tongue:

You can rewrite it a little easier so you don't make any mistakes, which I think you did

[tex]\lim_{x\rightarrow\infty} (2 + \sqrt{x^2 + 1} - x) = 2 + \lim_{x\rightarrow\infty}(\sqrt{x^2 + 1} - x)\left(\frac{\sqrt{x^2 + 1} + x}{\sqrt{x^2 + 1} + x}\right)[/tex]
 
  • #17
oh man... that's what i did wrong?!

now i get (1/0) / 2, which means, doesn't exist

which means it would be "none of the above" for the final answer, am i correct?
 
  • #18
No, it does exist as you saw on your calculator.
 
  • #19
[tex]2 + \lim_{x\rightarrow\infty}(\sqrt{x^2 + 1} - x)\left(\frac{\sqrt{x^2 + 1} + x}{\sqrt{x^2 + 1} + x}\right) = 2 + \lim_{x\rightarrow\infty}\frac{x^2 + 1 - x^2}{\sqrt{x^2(1 + \frac{1}{x^2})} + x} = 2 + \lim_{x\rightarrow\infty}\frac{1}{x\sqrt{1 + \frac{1}{x^2}} + x}[/tex]

Can you take it from there?
 
  • #20
Bohrok said:
No, it does exist as you saw on your calculator.

i got,
(x^2 + 1 - x^2) / (sqrt(x^2 +1) +x) =
1/ (sqrt(x^2 +1) +x)

now i divided both top and bottom by x

1/x / (sqrt(x^2 +1) +x) / x

and got:
1/x / (sqrt(1+ 1/x^2) +1)

>> 1/x^2 = 0 so (sqrt(1+ 0) +1) = 2

top: 1/x = 1/infintiy = 0
0/2 = 0

2+0 = 2
oh man.. so that still 2

then what is wrong??
we went over every single one and still its wrong..
 
  • #21
That one is J (again, do you know what the exact limit of that one is?).

Do you know how many you got wrong?
 
  • #22
Bohrok said:
That one is J (again, do you know what the exact limit of that one is?).

Do you know how many you got wrong?
6/sqrt(3) = 3.46.. so its close to pi...
i just multiplied top and bottom by sqrt(3x^2 + 1) and solved it..

and no i don't know how many i got wrong.
the system shows wrong overall.. whichever is wrong, that my job to find i guess..

so i have no clue what can be wrong..
we went over every single one in details and still it's wrong. :(
 
  • #23
I'd say all your answers are correct
C
F
B
G
J
A

The system checking your answers has to be wrong,
 
  • #24
Bohrok said:
I'd say all your answers are correct
C
F
B
G
J
A

The system checking your answers has to be wrong,

omg.. I am in stress now! it's due tomorrow and i have no idea what to do :(
 
  • #25
Someone may have made a simple mistake, such as making the answer to the second one B. All I can suggest is trying different combinations of answers that are very similar to the answers you know are correct.
 
  • #26
Bohrok said:
Someone may have made a simple mistake, such as making the answer to the second one B. All I can suggest is trying different combinations of answers that are very similar to the answers you know are correct.

i have this feeling that
6x+1 / sqrt(3x^2+1)
is going to be something else, because it's so close to pi.. 3.46... but then there is no -pi.
whatever.. i give up. i have 2 more tries left so i will guess them both and see what happens.

thanks a lot for your time and huge help!
 
  • #27
i got it!
apparently, cot-1(x) is the same as arccot(x) which means,
http://web.viu.ca/wattsv/math122/Overheads/Section35/overhead13.bmp

and the answer was E.. wow... pweee... on the 7/8 tries i got it! woohooo
 
Last edited by a moderator:

1. What is a horizontal asymptote?

A horizontal asymptote is a straight line that a graph approaches but never touches. It represents the limit of a function as the input variable (usually represented as x) approaches either positive or negative infinity.

2. How do you find the horizontal asymptote of a function?

To find the horizontal asymptote of a function, you can use the rule that states if the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y=0. If the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

3. Can a function have more than one horizontal asymptote?

Yes, a function can have more than one horizontal asymptote. This usually happens when the function has a complex or unusual behavior as the input variable approaches infinity. In these cases, there may be multiple horizontal asymptotes that the graph approaches from different directions.

4. What does it mean if a function crosses its horizontal asymptote?

If a function crosses its horizontal asymptote, it means that the limit of the function as the input variable approaches infinity does not exist. This can happen if the function has a discontinuity or if the behavior of the function becomes more and more erratic as the input variable increases.

5. Can a function have both a vertical and a horizontal asymptote?

Yes, a function can have both a vertical and a horizontal asymptote. In these cases, the vertical asymptote represents a value that the input variable cannot equal, while the horizontal asymptote represents the limit of the function as the input variable approaches infinity.

Similar threads

  • Calculus and Beyond Homework Help
Replies
6
Views
113
  • Calculus and Beyond Homework Help
Replies
2
Views
746
  • Calculus and Beyond Homework Help
Replies
7
Views
828
  • Calculus and Beyond Homework Help
Replies
3
Views
722
  • Calculus and Beyond Homework Help
Replies
4
Views
281
  • Calculus and Beyond Homework Help
Replies
2
Views
702
  • Calculus and Beyond Homework Help
Replies
32
Views
2K
  • Calculus and Beyond Homework Help
Replies
10
Views
820
  • Calculus and Beyond Homework Help
Replies
15
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
783
Back
Top