Graph of a function - asymptotes properties
Is it possible that the graph of a function has vertical asymptote if $D_{f} = mathbb{R}$?
Also, is it possible that the graph of a function intersects its asymptote? (horizontal, slant or vertical)
The answer to both questions seems a straight-forward NO to me, but can someone help prove that? Thanks!
limits graphing-functions
asked Nov 24 at 12:52
weno
817
|
show 1 more comment
Is it possible that the graph of a function has vertical asymptote if $D_{f} = mathbb{R}$?
Also, is it possible that the graph of a function intersects its asymptote? (horizontal, slant or vertical)
The answer to both questions seems a straight-forward NO to me, but can someone help prove that? Thanks!
limits graphing-functions
asked Nov 24 at 12:52
weno
817
Consider the function: f(x)= 1/x if x is not 0, f(0)= 1. That function has a vertical asymptote at x= 0 but is defined for x= 0.
– user247327
Nov 24 at 12:57
How can f(x) = 1/x be defined for x = 0, if x=/=0, hence domain D = R{0}? I'm not sure I understand.
– weno
Nov 24 at 12:58
I did not say anything at all about f(x)= 1/x. Go back and read my post again.
– user247327
Nov 24 at 13:01
I understand now. But since y = 1 for x = 0, how can there be an asymptote then?
– weno
Nov 24 at 13:03
What is your definition of "asymptote"?
– user247327
Nov 24 at 13:04
|
show 1 more comment
Is it possible that the graph of a function has vertical asymptote if $D_{f} = mathbb{R}$?
Also, is it possible that the graph of a function intersects its asymptote? (horizontal, slant or vertical)
The answer to both questions seems a straight-forward NO to me, but can someone help prove that? Thanks!
limits graphing-functions
asked Nov 24 at 12:52
weno
817
Is it possible that the graph of a function has vertical asymptote if $D_{f} = mathbb{R}$?
Also, is it possible that the graph of a function intersects its asymptote? (horizontal, slant or vertical)
The answer to both questions seems a straight-forward NO to me, but can someone help prove that? Thanks!
limits graphing-functions
limits graphing-functions
asked Nov 24 at 12:52
weno
817
asked Nov 24 at 12:52
weno
817
asked Nov 24 at 12:52
weno
817
asked Nov 24 at 12:52
weno
817
asked Nov 24 at 12:52
weno
817
817
Consider the function: f(x)= 1/x if x is not 0, f(0)= 1. That function has a vertical asymptote at x= 0 but is defined for x= 0.
– user247327
Nov 24 at 12:57
How can f(x) = 1/x be defined for x = 0, if x=/=0, hence domain D = R{0}? I'm not sure I understand.
– weno
Nov 24 at 12:58
I did not say anything at all about f(x)= 1/x. Go back and read my post again.
– user247327
Nov 24 at 13:01
I understand now. But since y = 1 for x = 0, how can there be an asymptote then?
– weno
Nov 24 at 13:03
What is your definition of "asymptote"?
– user247327
Nov 24 at 13:04
|
show 1 more comment
Consider the function: f(x)= 1/x if x is not 0, f(0)= 1. That function has a vertical asymptote at x= 0 but is defined for x= 0.
– user247327
Nov 24 at 12:57
How can f(x) = 1/x be defined for x = 0, if x=/=0, hence domain D = R{0}? I'm not sure I understand.
– weno
Nov 24 at 12:58
I did not say anything at all about f(x)= 1/x. Go back and read my post again.
– user247327
Nov 24 at 13:01
I understand now. But since y = 1 for x = 0, how can there be an asymptote then?
– weno
Nov 24 at 13:03
What is your definition of "asymptote"?
– user247327
Nov 24 at 13:04
Consider the function: f(x)= 1/x if x is not 0, f(0)= 1. That function has a vertical asymptote at x= 0 but is defined for x= 0.
– user247327
Nov 24 at 12:57
Consider the function: f(x)= 1/x if x is not 0, f(0)= 1. That function has a vertical asymptote at x= 0 but is defined for x= 0.
– user247327
Nov 24 at 12:57
How can f(x) = 1/x be defined for x = 0, if x=/=0, hence domain D = R{0}? I'm not sure I understand.
– weno
Nov 24 at 12:58
How can f(x) = 1/x be defined for x = 0, if x=/=0, hence domain D = R{0}? I'm not sure I understand.
– weno
Nov 24 at 12:58
I did not say anything at all about f(x)= 1/x. Go back and read my post again.
– user247327
Nov 24 at 13:01
I did not say anything at all about f(x)= 1/x. Go back and read my post again.
– user247327
Nov 24 at 13:01
I understand now. But since y = 1 for x = 0, how can there be an asymptote then?
– weno
Nov 24 at 13:03
I understand now. But since y = 1 for x = 0, how can there be an asymptote then?
– weno
Nov 24 at 13:03
What is your definition of "asymptote"?
– user247327
Nov 24 at 13:04
What is your definition of "asymptote"?
– user247327
Nov 24 at 13:04
|
show 1 more comment
1 Answer
1
active
oldest
votes
If a function is continuous on $mathbb{R},$ then doesn't have a vertical asymptote. If $f$ is not continuous, it can happen what explained in comments.
It is not possible that a graph of a function cuts its vertical asymptote. That would contradict the definition of the function.
For the second part, a quote from Wikipedia:
Asymptote of a curve is a line such that the distance between the curve and the line approaches zero as one or both of the x or y coordinates tends to infinity.
The distance between the courbe and the asymptote (horizontal or oblique/slant) has to tend to $0$ with $xtoinfty,$ intersections are allowed.
As an example, consider $$f(x)=frac x2 + frac{cos x}{sqrt x}.$$ Since $limlimits_{xtoinfty} left(f(x)-frac x2 right)=0,$ the line $y=frac x2$ is an asymptote of the graph of $f$ (see figure). However, they cut, even infinitely times.
answered Nov 25 at 10:12
user376343
2,7882822
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
If a function is continuous on $mathbb{R},$ then doesn't have a vertical asymptote. If $f$ is not continuous, it can happen what explained in comments.
It is not possible that a graph of a function cuts its vertical asymptote. That would contradict the definition of the function.
For the second part, a quote from Wikipedia:
Asymptote of a curve is a line such that the distance between the curve and the line approaches zero as one or both of the x or y coordinates tends to infinity.
The distance between the courbe and the asymptote (horizontal or oblique/slant) has to tend to $0$ with $xtoinfty,$ intersections are allowed.
As an example, consider $$f(x)=frac x2 + frac{cos x}{sqrt x}.$$ Since $limlimits_{xtoinfty} left(f(x)-frac x2 right)=0,$ the line $y=frac x2$ is an asymptote of the graph of $f$ (see figure). However, they cut, even infinitely times.
answered Nov 25 at 10:12
user376343
2,7882822
add a comment |
If a function is continuous on $mathbb{R},$ then doesn't have a vertical asymptote. If $f$ is not continuous, it can happen what explained in comments.
It is not possible that a graph of a function cuts its vertical asymptote. That would contradict the definition of the function.
For the second part, a quote from Wikipedia:
Asymptote of a curve is a line such that the distance between the curve and the line approaches zero as one or both of the x or y coordinates tends to infinity.
The distance between the courbe and the asymptote (horizontal or oblique/slant) has to tend to $0$ with $xtoinfty,$ intersections are allowed.
As an example, consider $$f(x)=frac x2 + frac{cos x}{sqrt x}.$$ Since $limlimits_{xtoinfty} left(f(x)-frac x2 right)=0,$ the line $y=frac x2$ is an asymptote of the graph of $f$ (see figure). However, they cut, even infinitely times.
answered Nov 25 at 10:12
user376343
2,7882822
add a comment |
If a function is continuous on $mathbb{R},$ then doesn't have a vertical asymptote. If $f$ is not continuous, it can happen what explained in comments.
It is not possible that a graph of a function cuts its vertical asymptote. That would contradict the definition of the function.
For the second part, a quote from Wikipedia:
Asymptote of a curve is a line such that the distance between the curve and the line approaches zero as one or both of the x or y coordinates tends to infinity.
The distance between the courbe and the asymptote (horizontal or oblique/slant) has to tend to $0$ with $xtoinfty,$ intersections are allowed.
As an example, consider $$f(x)=frac x2 + frac{cos x}{sqrt x}.$$ Since $limlimits_{xtoinfty} left(f(x)-frac x2 right)=0,$ the line $y=frac x2$ is an asymptote of the graph of $f$ (see figure). However, they cut, even infinitely times.
answered Nov 25 at 10:12
user376343
2,7882822
If a function is continuous on $mathbb{R},$ then doesn't have a vertical asymptote. If $f$ is not continuous, it can happen what explained in comments.
It is not possible that a graph of a function cuts its vertical asymptote. That would contradict the definition of the function.
For the second part, a quote from Wikipedia:
Asymptote of a curve is a line such that the distance between the curve and the line approaches zero as one or both of the x or y coordinates tends to infinity.
The distance between the courbe and the asymptote (horizontal or oblique/slant) has to tend to $0$ with $xtoinfty,$ intersections are allowed.
As an example, consider $$f(x)=frac x2 + frac{cos x}{sqrt x}.$$ Since $limlimits_{xtoinfty} left(f(x)-frac x2 right)=0,$ the line $y=frac x2$ is an asymptote of the graph of $f$ (see figure). However, they cut, even infinitely times.
answered Nov 25 at 10:12
user376343
2,7882822
answered Nov 25 at 10:12
user376343
2,7882822
answered Nov 25 at 10:12
user376343
2,7882822
answered Nov 25 at 10:12
user376343
2,7882822
2,7882822
add a comment |
add a comment |
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Consider the function: f(x)= 1/x if x is not 0, f(0)= 1. That function has a vertical asymptote at x= 0 but is defined for x= 0.
– user247327
Nov 24 at 12:57
How can f(x) = 1/x be defined for x = 0, if x=/=0, hence domain D = R{0}? I'm not sure I understand.
– weno
Nov 24 at 12:58
I did not say anything at all about f(x)= 1/x. Go back and read my post again.
– user247327
Nov 24 at 13:01
I understand now. But since y = 1 for x = 0, how can there be an asymptote then?
– weno
Nov 24 at 13:03
What is your definition of "asymptote"?
– user247327
Nov 24 at 13:04