supremum and infinum of exponential function
-1
$begingroup$
Show that the $sup (frac{xe^x}{e^x-1})=1$ where $0<x<1$.
How do I get the sup of this function? Please show me step by step or give me some advice.
real-analysis
asked Dec 13 '18 at 22:31
manman
33
$endgroup$
add a comment |
-1
$begingroup$
Show that the $sup (frac{xe^x}{e^x-1})=1$ where $0<x<1$.
How do I get the sup of this function? Please show me step by step or give me some advice.
real-analysis
asked Dec 13 '18 at 22:31
manman
33
$endgroup$
$begingroup$
I believe $x<0$?
$endgroup$
– Shubham Johri
Dec 13 '18 at 22:58
add a comment |
-1
-1
-1
$begingroup$
Show that the $sup (frac{xe^x}{e^x-1})=1$ where $0<x<1$.
How do I get the sup of this function? Please show me step by step or give me some advice.
real-analysis
asked Dec 13 '18 at 22:31
manman
33
$endgroup$
Show that the $sup (frac{xe^x}{e^x-1})=1$ where $0<x<1$.
How do I get the sup of this function? Please show me step by step or give me some advice.
real-analysis
real-analysis
asked Dec 13 '18 at 22:31
manman
33
asked Dec 13 '18 at 22:31
manman
33
edited Dec 14 '18 at 4:06
man
asked Dec 13 '18 at 22:31
manman
33
asked Dec 13 '18 at 22:31
manman
33
asked Dec 13 '18 at 22:31
manman
33
33
$begingroup$
I believe $x<0$?
$endgroup$
– Shubham Johri
Dec 13 '18 at 22:58
add a comment |
$begingroup$
I believe $x<0$?
$endgroup$
– Shubham Johri
Dec 13 '18 at 22:58
$begingroup$
I believe $x<0$?
$endgroup$
– Shubham Johri
Dec 13 '18 at 22:58
$begingroup$
I believe $x<0$?
$endgroup$
– Shubham Johri
Dec 13 '18 at 22:58
add a comment |
1 Answer
1
active
oldest
votes
0
$begingroup$
Where is the function defined?
Because if it is defined in all R, it has no supreme. Note that
$lim_{xrightarrow infty}=infty$.
answered Dec 13 '18 at 22:39
R. N. MarleyR. N. Marley
788
$endgroup$
$begingroup$
The function is defined for 0<x<1
$endgroup$
– man
Dec 14 '18 at 3:51
$begingroup$
So in that case the supreme is not 1. Note that the function is increasing in (0,1) (compute the first derivative) therefore $$sup (xe^x/(e^x-1))=e/(e-1)$$ for 0<x<1
$endgroup$
– R. N. Marley
Dec 14 '18 at 11:05
add a comment |
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1 Answer
1
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oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
0
$begingroup$
Where is the function defined?
Because if it is defined in all R, it has no supreme. Note that
$lim_{xrightarrow infty}=infty$.
answered Dec 13 '18 at 22:39
R. N. MarleyR. N. Marley
788
$endgroup$
$begingroup$
The function is defined for 0<x<1
$endgroup$
– man
Dec 14 '18 at 3:51
$begingroup$
So in that case the supreme is not 1. Note that the function is increasing in (0,1) (compute the first derivative) therefore $$sup (xe^x/(e^x-1))=e/(e-1)$$ for 0<x<1
$endgroup$
– R. N. Marley
Dec 14 '18 at 11:05
add a comment |
0
$begingroup$
Where is the function defined?
Because if it is defined in all R, it has no supreme. Note that
$lim_{xrightarrow infty}=infty$.
answered Dec 13 '18 at 22:39
R. N. MarleyR. N. Marley
788
$endgroup$
$begingroup$
The function is defined for 0<x<1
$endgroup$
– man
Dec 14 '18 at 3:51
$begingroup$
So in that case the supreme is not 1. Note that the function is increasing in (0,1) (compute the first derivative) therefore $$sup (xe^x/(e^x-1))=e/(e-1)$$ for 0<x<1
$endgroup$
– R. N. Marley
Dec 14 '18 at 11:05
add a comment |
0
0
0
$begingroup$
Where is the function defined?
Because if it is defined in all R, it has no supreme. Note that
$lim_{xrightarrow infty}=infty$.
answered Dec 13 '18 at 22:39
R. N. MarleyR. N. Marley
788
$endgroup$
Where is the function defined?
Because if it is defined in all R, it has no supreme. Note that
$lim_{xrightarrow infty}=infty$.
answered Dec 13 '18 at 22:39
R. N. MarleyR. N. Marley
788
answered Dec 13 '18 at 22:39
R. N. MarleyR. N. Marley
788
answered Dec 13 '18 at 22:39
R. N. MarleyR. N. Marley
788
answered Dec 13 '18 at 22:39
R. N. MarleyR. N. Marley
788
788
$begingroup$
The function is defined for 0<x<1
$endgroup$
– man
Dec 14 '18 at 3:51
$begingroup$
So in that case the supreme is not 1. Note that the function is increasing in (0,1) (compute the first derivative) therefore $$sup (xe^x/(e^x-1))=e/(e-1)$$ for 0<x<1
$endgroup$
– R. N. Marley
Dec 14 '18 at 11:05
add a comment |
$begingroup$
The function is defined for 0<x<1
$endgroup$
– man
Dec 14 '18 at 3:51
$begingroup$
So in that case the supreme is not 1. Note that the function is increasing in (0,1) (compute the first derivative) therefore $$sup (xe^x/(e^x-1))=e/(e-1)$$ for 0<x<1
$endgroup$
– R. N. Marley
Dec 14 '18 at 11:05
$begingroup$
The function is defined for 0<x<1
$endgroup$
– man
Dec 14 '18 at 3:51
$begingroup$
The function is defined for 0<x<1
$endgroup$
– man
Dec 14 '18 at 3:51
$begingroup$
So in that case the supreme is not 1. Note that the function is increasing in (0,1) (compute the first derivative) therefore $$sup (xe^x/(e^x-1))=e/(e-1)$$ for 0<x<1
$endgroup$
– R. N. Marley
Dec 14 '18 at 11:05
$begingroup$
So in that case the supreme is not 1. Note that the function is increasing in (0,1) (compute the first derivative) therefore $$sup (xe^x/(e^x-1))=e/(e-1)$$ for 0<x<1
$endgroup$
– R. N. Marley
Dec 14 '18 at 11:05
add a comment |
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$begingroup$
I believe $x<0$?
$endgroup$
– Shubham Johri
Dec 13 '18 at 22:58