Give an example of a continues function from an open and bounded interval $(a,b)rightarrow I $ where...
0
$begingroup$
Give an example of a continues function from an open and bounded interval $(a,b)rightarrow I $ where $f(I)=ℝ$
I'm having trouble coming up with a function that is has a bounded co-domain, yet an infinite domain for any values of a and b.
functions continuity
asked Dec 3 '18 at 17:43
CruZCruZ
547
$endgroup$
add a comment |
0
$begingroup$
Give an example of a continues function from an open and bounded interval $(a,b)rightarrow I $ where $f(I)=ℝ$
I'm having trouble coming up with a function that is has a bounded co-domain, yet an infinite domain for any values of a and b.
functions continuity
asked Dec 3 '18 at 17:43
CruZCruZ
547
$endgroup$
add a comment |
0
0
0
$begingroup$
Give an example of a continues function from an open and bounded interval $(a,b)rightarrow I $ where $f(I)=ℝ$
I'm having trouble coming up with a function that is has a bounded co-domain, yet an infinite domain for any values of a and b.
functions continuity
asked Dec 3 '18 at 17:43
CruZCruZ
547
$endgroup$
Give an example of a continues function from an open and bounded interval $(a,b)rightarrow I $ where $f(I)=ℝ$
I'm having trouble coming up with a function that is has a bounded co-domain, yet an infinite domain for any values of a and b.
functions continuity
functions continuity
asked Dec 3 '18 at 17:43
CruZCruZ
547
asked Dec 3 '18 at 17:43
CruZCruZ
547
asked Dec 3 '18 at 17:43
CruZCruZ
547
asked Dec 3 '18 at 17:43
CruZCruZ
547
asked Dec 3 '18 at 17:43
CruZCruZ
547
547
add a comment |
add a comment |
1 Answer
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Consider $tan(x)$ restricted to the interval $(-pi/2,pi/2)$.
answered Dec 3 '18 at 17:48
MelodyMelody
81012
$endgroup$
$begingroup$
Yes, however as far as I understand the exercise, it needs to be ANY interval e.g. (0,1)... Is there any continues function where this is possible?
$endgroup$
– CruZ
Dec 3 '18 at 17:54
2
$begingroup$
Then shift and stretch the one given.
$endgroup$
– Randall
Dec 3 '18 at 18:01
1
$begingroup$
Also, I don't think it reads that way. It reads as if the domain is fixed to be $(a,b)$ at the outset. (Assuming that $(a,b) to I$ means $(a,b)=I$.)
$endgroup$
– Randall
Dec 3 '18 at 18:02
$begingroup$
Ah okay, that makes sense. Thanks a lot! :)
$endgroup$
– CruZ
Dec 3 '18 at 18:06
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
$begingroup$
Consider $tan(x)$ restricted to the interval $(-pi/2,pi/2)$.
answered Dec 3 '18 at 17:48
MelodyMelody
81012
$endgroup$
$begingroup$
Yes, however as far as I understand the exercise, it needs to be ANY interval e.g. (0,1)... Is there any continues function where this is possible?
$endgroup$
– CruZ
Dec 3 '18 at 17:54
2
$begingroup$
Then shift and stretch the one given.
$endgroup$
– Randall
Dec 3 '18 at 18:01
1
$begingroup$
Also, I don't think it reads that way. It reads as if the domain is fixed to be $(a,b)$ at the outset. (Assuming that $(a,b) to I$ means $(a,b)=I$.)
$endgroup$
– Randall
Dec 3 '18 at 18:02
$begingroup$
Ah okay, that makes sense. Thanks a lot! :)
$endgroup$
– CruZ
Dec 3 '18 at 18:06
add a comment |
2
$begingroup$
Consider $tan(x)$ restricted to the interval $(-pi/2,pi/2)$.
answered Dec 3 '18 at 17:48
MelodyMelody
81012
$endgroup$
$begingroup$
Yes, however as far as I understand the exercise, it needs to be ANY interval e.g. (0,1)... Is there any continues function where this is possible?
$endgroup$
– CruZ
Dec 3 '18 at 17:54
2
$begingroup$
Then shift and stretch the one given.
$endgroup$
– Randall
Dec 3 '18 at 18:01
1
$begingroup$
Also, I don't think it reads that way. It reads as if the domain is fixed to be $(a,b)$ at the outset. (Assuming that $(a,b) to I$ means $(a,b)=I$.)
$endgroup$
– Randall
Dec 3 '18 at 18:02
$begingroup$
Ah okay, that makes sense. Thanks a lot! :)
$endgroup$
– CruZ
Dec 3 '18 at 18:06
add a comment |
2
2
2
$begingroup$
Consider $tan(x)$ restricted to the interval $(-pi/2,pi/2)$.
answered Dec 3 '18 at 17:48
MelodyMelody
81012
$endgroup$
Consider $tan(x)$ restricted to the interval $(-pi/2,pi/2)$.
answered Dec 3 '18 at 17:48
MelodyMelody
81012
answered Dec 3 '18 at 17:48
MelodyMelody
81012
answered Dec 3 '18 at 17:48
MelodyMelody
81012
answered Dec 3 '18 at 17:48
MelodyMelody
81012
81012
$begingroup$
Yes, however as far as I understand the exercise, it needs to be ANY interval e.g. (0,1)... Is there any continues function where this is possible?
$endgroup$
– CruZ
Dec 3 '18 at 17:54
2
$begingroup$
Then shift and stretch the one given.
$endgroup$
– Randall
Dec 3 '18 at 18:01
1
$begingroup$
Also, I don't think it reads that way. It reads as if the domain is fixed to be $(a,b)$ at the outset. (Assuming that $(a,b) to I$ means $(a,b)=I$.)
$endgroup$
– Randall
Dec 3 '18 at 18:02
$begingroup$
Ah okay, that makes sense. Thanks a lot! :)
$endgroup$
– CruZ
Dec 3 '18 at 18:06
add a comment |
$begingroup$
Yes, however as far as I understand the exercise, it needs to be ANY interval e.g. (0,1)... Is there any continues function where this is possible?
$endgroup$
– CruZ
Dec 3 '18 at 17:54
2
$begingroup$
Then shift and stretch the one given.
$endgroup$
– Randall
Dec 3 '18 at 18:01
1
$begingroup$
Also, I don't think it reads that way. It reads as if the domain is fixed to be $(a,b)$ at the outset. (Assuming that $(a,b) to I$ means $(a,b)=I$.)
$endgroup$
– Randall
Dec 3 '18 at 18:02
$begingroup$
Ah okay, that makes sense. Thanks a lot! :)
$endgroup$
– CruZ
Dec 3 '18 at 18:06
$begingroup$
Yes, however as far as I understand the exercise, it needs to be ANY interval e.g. (0,1)... Is there any continues function where this is possible?
$endgroup$
– CruZ
Dec 3 '18 at 17:54
$begingroup$
Yes, however as far as I understand the exercise, it needs to be ANY interval e.g. (0,1)... Is there any continues function where this is possible?
$endgroup$
– CruZ
Dec 3 '18 at 17:54
2
2
$begingroup$
Then shift and stretch the one given.
$endgroup$
– Randall
Dec 3 '18 at 18:01
$begingroup$
Then shift and stretch the one given.
$endgroup$
– Randall
Dec 3 '18 at 18:01
1
1
$begingroup$
Also, I don't think it reads that way. It reads as if the domain is fixed to be $(a,b)$ at the outset. (Assuming that $(a,b) to I$ means $(a,b)=I$.)
$endgroup$
– Randall
Dec 3 '18 at 18:02
$begingroup$
Also, I don't think it reads that way. It reads as if the domain is fixed to be $(a,b)$ at the outset. (Assuming that $(a,b) to I$ means $(a,b)=I$.)
$endgroup$
– Randall
Dec 3 '18 at 18:02
$begingroup$
Ah okay, that makes sense. Thanks a lot! :)
$endgroup$
– CruZ
Dec 3 '18 at 18:06
$begingroup$
Ah okay, that makes sense. Thanks a lot! :)
$endgroup$
– CruZ
Dec 3 '18 at 18:06
add a comment |
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