A continuous function from (0,1) to [0,1]
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So today my professor gave a self-check to us to write a continuous function defined on (0,1) whose range is [0,1]. And I thought of a function which I am not sure if it is considered to be continuous or not.
f(x) is a piece-wise function defined as:
begin{cases}
0, &quadtext{if x < 1/10}\
5/4(x-1/10), &quadtext{if 1/10 <= x <= 9/10}\
1, &quadtext{if x > 9/10 }\
end{cases}
Thank you for your help and/or hints in prior.
real-analysis functions
asked Jan 8 at 16:13
AllorjaAllorja
789
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add a comment |
$begingroup$
So today my professor gave a self-check to us to write a continuous function defined on (0,1) whose range is [0,1]. And I thought of a function which I am not sure if it is considered to be continuous or not.
f(x) is a piece-wise function defined as:
begin{cases}
0, &quadtext{if x < 1/10}\
5/4(x-1/10), &quadtext{if 1/10 <= x <= 9/10}\
1, &quadtext{if x > 9/10 }\
end{cases}
Thank you for your help and/or hints in prior.
real-analysis functions
asked Jan 8 at 16:13
AllorjaAllorja
789
$endgroup$
add a comment |
$begingroup$
So today my professor gave a self-check to us to write a continuous function defined on (0,1) whose range is [0,1]. And I thought of a function which I am not sure if it is considered to be continuous or not.
f(x) is a piece-wise function defined as:
begin{cases}
0, &quadtext{if x < 1/10}\
5/4(x-1/10), &quadtext{if 1/10 <= x <= 9/10}\
1, &quadtext{if x > 9/10 }\
end{cases}
Thank you for your help and/or hints in prior.
real-analysis functions
asked Jan 8 at 16:13
AllorjaAllorja
789
$endgroup$
So today my professor gave a self-check to us to write a continuous function defined on (0,1) whose range is [0,1]. And I thought of a function which I am not sure if it is considered to be continuous or not.
f(x) is a piece-wise function defined as:
begin{cases}
0, &quadtext{if x < 1/10}\
5/4(x-1/10), &quadtext{if 1/10 <= x <= 9/10}\
1, &quadtext{if x > 9/10 }\
end{cases}
Thank you for your help and/or hints in prior.
real-analysis functions
real-analysis functions
asked Jan 8 at 16:13
AllorjaAllorja
789
asked Jan 8 at 16:13
AllorjaAllorja
789
edited Jan 8 at 16:43
Allorja
asked Jan 8 at 16:13
AllorjaAllorja
789
asked Jan 8 at 16:13
AllorjaAllorja
789
asked Jan 8 at 16:13
AllorjaAllorja
789
789
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
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Your function is not continuous (or even defined in all places you need it). You can mend your idea by letting
$$f(x)=max{0,min{1,tfrac 54(x-tfrac1{10})}}.$$
Remark: It is also possible to find a cubic polynomial function $f(x)=ax^3+bx^2+cx+d$ that solves the problem.
answered Jan 8 at 16:18
Hagen von EitzenHagen von Eitzen
280k23272504
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add a comment |
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Your idea is good. But your presentation isn't entirely correct. You need to use $leq$ instead of $<$ in a few strategic places, so that the function is actually defined on all of $(0,1)$. And you need to look at that middle expression (i.e. $x$) to make sure it actually says what you want it to say.
Alternately, if you don't like defining by cases, you could use something like $sin^2(10x)$.
answered Jan 8 at 16:16
ArthurArthur
116k7116198
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$begingroup$
With the edit that I just made, my function is continuous right?
$endgroup$
– Allorja
Jan 8 at 16:44
1
$begingroup$
Yes, that ought to do it.
$endgroup$
– Arthur
Jan 8 at 17:06
add a comment |
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Your function can be plotted using Desmos. You can fix this by making sure the middle segment actually connects to the two horizontal lines at the $x=frac{1}{10}$ and $x=frac{9}{10}$.
answered Jan 8 at 16:25
DubsDubs
55826
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add a comment |
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3 Answers
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3 Answers
3
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
Your function is not continuous (or even defined in all places you need it). You can mend your idea by letting
$$f(x)=max{0,min{1,tfrac 54(x-tfrac1{10})}}.$$
Remark: It is also possible to find a cubic polynomial function $f(x)=ax^3+bx^2+cx+d$ that solves the problem.
answered Jan 8 at 16:18
Hagen von EitzenHagen von Eitzen
280k23272504
$endgroup$
add a comment |
$begingroup$
Your function is not continuous (or even defined in all places you need it). You can mend your idea by letting
$$f(x)=max{0,min{1,tfrac 54(x-tfrac1{10})}}.$$
Remark: It is also possible to find a cubic polynomial function $f(x)=ax^3+bx^2+cx+d$ that solves the problem.
answered Jan 8 at 16:18
Hagen von EitzenHagen von Eitzen
280k23272504
$endgroup$
add a comment |
$begingroup$
Your function is not continuous (or even defined in all places you need it). You can mend your idea by letting
$$f(x)=max{0,min{1,tfrac 54(x-tfrac1{10})}}.$$
Remark: It is also possible to find a cubic polynomial function $f(x)=ax^3+bx^2+cx+d$ that solves the problem.
answered Jan 8 at 16:18
Hagen von EitzenHagen von Eitzen
280k23272504
$endgroup$
Your function is not continuous (or even defined in all places you need it). You can mend your idea by letting
$$f(x)=max{0,min{1,tfrac 54(x-tfrac1{10})}}.$$
Remark: It is also possible to find a cubic polynomial function $f(x)=ax^3+bx^2+cx+d$ that solves the problem.
answered Jan 8 at 16:18
Hagen von EitzenHagen von Eitzen
280k23272504
answered Jan 8 at 16:18
Hagen von EitzenHagen von Eitzen
280k23272504
answered Jan 8 at 16:18
Hagen von EitzenHagen von Eitzen
280k23272504
answered Jan 8 at 16:18
Hagen von EitzenHagen von Eitzen
280k23272504
280k23272504
add a comment |
add a comment |
$begingroup$
Your idea is good. But your presentation isn't entirely correct. You need to use $leq$ instead of $<$ in a few strategic places, so that the function is actually defined on all of $(0,1)$. And you need to look at that middle expression (i.e. $x$) to make sure it actually says what you want it to say.
Alternately, if you don't like defining by cases, you could use something like $sin^2(10x)$.
answered Jan 8 at 16:16
ArthurArthur
116k7116198
$endgroup$
$begingroup$
With the edit that I just made, my function is continuous right?
$endgroup$
– Allorja
Jan 8 at 16:44
1
$begingroup$
Yes, that ought to do it.
$endgroup$
– Arthur
Jan 8 at 17:06
add a comment |
$begingroup$
Your idea is good. But your presentation isn't entirely correct. You need to use $leq$ instead of $<$ in a few strategic places, so that the function is actually defined on all of $(0,1)$. And you need to look at that middle expression (i.e. $x$) to make sure it actually says what you want it to say.
Alternately, if you don't like defining by cases, you could use something like $sin^2(10x)$.
answered Jan 8 at 16:16
ArthurArthur
116k7116198
$endgroup$
$begingroup$
With the edit that I just made, my function is continuous right?
$endgroup$
– Allorja
Jan 8 at 16:44
1
$begingroup$
Yes, that ought to do it.
$endgroup$
– Arthur
Jan 8 at 17:06
add a comment |
$begingroup$
Your idea is good. But your presentation isn't entirely correct. You need to use $leq$ instead of $<$ in a few strategic places, so that the function is actually defined on all of $(0,1)$. And you need to look at that middle expression (i.e. $x$) to make sure it actually says what you want it to say.
Alternately, if you don't like defining by cases, you could use something like $sin^2(10x)$.
answered Jan 8 at 16:16
ArthurArthur
116k7116198
$endgroup$
Your idea is good. But your presentation isn't entirely correct. You need to use $leq$ instead of $<$ in a few strategic places, so that the function is actually defined on all of $(0,1)$. And you need to look at that middle expression (i.e. $x$) to make sure it actually says what you want it to say.
Alternately, if you don't like defining by cases, you could use something like $sin^2(10x)$.
answered Jan 8 at 16:16
ArthurArthur
116k7116198
answered Jan 8 at 16:16
ArthurArthur
116k7116198
answered Jan 8 at 16:16
ArthurArthur
116k7116198
answered Jan 8 at 16:16
ArthurArthur
116k7116198
116k7116198
$begingroup$
With the edit that I just made, my function is continuous right?
$endgroup$
– Allorja
Jan 8 at 16:44
1
$begingroup$
Yes, that ought to do it.
$endgroup$
– Arthur
Jan 8 at 17:06
add a comment |
$begingroup$
With the edit that I just made, my function is continuous right?
$endgroup$
– Allorja
Jan 8 at 16:44
1
$begingroup$
Yes, that ought to do it.
$endgroup$
– Arthur
Jan 8 at 17:06
$begingroup$
With the edit that I just made, my function is continuous right?
$endgroup$
– Allorja
Jan 8 at 16:44
$begingroup$
With the edit that I just made, my function is continuous right?
$endgroup$
– Allorja
Jan 8 at 16:44
1
1
$begingroup$
Yes, that ought to do it.
$endgroup$
– Arthur
Jan 8 at 17:06
$begingroup$
Yes, that ought to do it.
$endgroup$
– Arthur
Jan 8 at 17:06
add a comment |
$begingroup$
Your function can be plotted using Desmos. You can fix this by making sure the middle segment actually connects to the two horizontal lines at the $x=frac{1}{10}$ and $x=frac{9}{10}$.
answered Jan 8 at 16:25
DubsDubs
55826
$endgroup$
add a comment |
$begingroup$
Your function can be plotted using Desmos. You can fix this by making sure the middle segment actually connects to the two horizontal lines at the $x=frac{1}{10}$ and $x=frac{9}{10}$.
answered Jan 8 at 16:25
DubsDubs
55826
$endgroup$
add a comment |
$begingroup$
Your function can be plotted using Desmos. You can fix this by making sure the middle segment actually connects to the two horizontal lines at the $x=frac{1}{10}$ and $x=frac{9}{10}$.
answered Jan 8 at 16:25
DubsDubs
55826
$endgroup$
Your function can be plotted using Desmos. You can fix this by making sure the middle segment actually connects to the two horizontal lines at the $x=frac{1}{10}$ and $x=frac{9}{10}$.
answered Jan 8 at 16:25
DubsDubs
55826
answered Jan 8 at 16:25
DubsDubs
55826
answered Jan 8 at 16:25
DubsDubs
55826
answered Jan 8 at 16:25
DubsDubs
55826
55826
add a comment |
add a comment |
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