Mixing
Why is the expression: $frac{1}{x} geq e^{-x} $ is different then the expression: $e^{x} geq x$?
$begingroup$
I was trying to study the function $f(x)=1-xe^{-x}$, and for studying the sign I came to the expression $1-xe^{-x} geq 0$ which can be expressed as $frac{1}{x} geq e^{-x}$ or as $e^{x} geq x$ . But these two inequalities return two different solutions because the first is true only when $xgt0$ while the second is true $forall x in mathbb{R}$, which is the correct solution. Why this happen?
Why is the expression: $frac{1}{x} geq e^{-x} $ is different then the expression: $e^{x} geq x$ ?
calculus inequality
edited Jan 28 at 12:19
Jneven
951322
asked Jan 28 at 12:10
Edoardo GrassiEdoardo Grassi
32
$endgroup$
add a comment |
$begingroup$
I was trying to study the function $f(x)=1-xe^{-x}$, and for studying the sign I came to the expression $1-xe^{-x} geq 0$ which can be expressed as $frac{1}{x} geq e^{-x}$ or as $e^{x} geq x$ . But these two inequalities return two different solutions because the first is true only when $xgt0$ while the second is true $forall x in mathbb{R}$, which is the correct solution. Why this happen?
Why is the expression: $frac{1}{x} geq e^{-x} $ is different then the expression: $e^{x} geq x$ ?
calculus inequality
edited Jan 28 at 12:19
Jneven
951322
asked Jan 28 at 12:10
Edoardo GrassiEdoardo Grassi
32
$endgroup$
1
$begingroup$
If you divide an inequality by a negative number the inequality gets reversed.
$endgroup$
– Kavi Rama Murthy
Jan 28 at 12:15
$begingroup$
How do you express the expression $1-xe^{-x}ge 0$ as $frac{1}{x}ge e^{-x}$ without changing the field of definition for $x$?
$endgroup$
– Ivon
Jan 28 at 12:16
$begingroup$
To go from $e^x geq x$ to an inequality about $1/x$ and $e^{-x}$ involves multiplying $e^{-x}/x$ on both sides which is negative or positive depending on $x$ and respectively affects the inequality
$endgroup$
– Naweed G. Seldon
Jan 28 at 12:17
$begingroup$
Seem my answer to another question at this link.
$endgroup$
– TonyK
Jan 28 at 12:21
add a comment |
$begingroup$
I was trying to study the function $f(x)=1-xe^{-x}$, and for studying the sign I came to the expression $1-xe^{-x} geq 0$ which can be expressed as $frac{1}{x} geq e^{-x}$ or as $e^{x} geq x$ . But these two inequalities return two different solutions because the first is true only when $xgt0$ while the second is true $forall x in mathbb{R}$, which is the correct solution. Why this happen?
Why is the expression: $frac{1}{x} geq e^{-x} $ is different then the expression: $e^{x} geq x$ ?
calculus inequality
edited Jan 28 at 12:19
Jneven
951322
asked Jan 28 at 12:10
Edoardo GrassiEdoardo Grassi
32
$endgroup$
I was trying to study the function $f(x)=1-xe^{-x}$, and for studying the sign I came to the expression $1-xe^{-x} geq 0$ which can be expressed as $frac{1}{x} geq e^{-x}$ or as $e^{x} geq x$ . But these two inequalities return two different solutions because the first is true only when $xgt0$ while the second is true $forall x in mathbb{R}$, which is the correct solution. Why this happen?
Why is the expression: $frac{1}{x} geq e^{-x} $ is different then the expression: $e^{x} geq x$ ?
calculus inequality
calculus inequality
edited Jan 28 at 12:19
Jneven
951322
asked Jan 28 at 12:10
Edoardo GrassiEdoardo Grassi
32
edited Jan 28 at 12:19
Jneven
951322
asked Jan 28 at 12:10
Edoardo GrassiEdoardo Grassi
32
edited Jan 28 at 12:19
Jneven
951322
edited Jan 28 at 12:19
Jneven
951322
edited Jan 28 at 12:19
Jneven
951322
951322
asked Jan 28 at 12:10
Edoardo GrassiEdoardo Grassi
32
asked Jan 28 at 12:10
Edoardo GrassiEdoardo Grassi
32
asked Jan 28 at 12:10
Edoardo GrassiEdoardo Grassi
32
32
1
$begingroup$
If you divide an inequality by a negative number the inequality gets reversed.
$endgroup$
– Kavi Rama Murthy
Jan 28 at 12:15
$begingroup$
How do you express the expression $1-xe^{-x}ge 0$ as $frac{1}{x}ge e^{-x}$ without changing the field of definition for $x$?
$endgroup$
– Ivon
Jan 28 at 12:16
$begingroup$
To go from $e^x geq x$ to an inequality about $1/x$ and $e^{-x}$ involves multiplying $e^{-x}/x$ on both sides which is negative or positive depending on $x$ and respectively affects the inequality
$endgroup$
– Naweed G. Seldon
Jan 28 at 12:17
$begingroup$
Seem my answer to another question at this link.
$endgroup$
– TonyK
Jan 28 at 12:21
add a comment |
1
$begingroup$
If you divide an inequality by a negative number the inequality gets reversed.
$endgroup$
– Kavi Rama Murthy
Jan 28 at 12:15
$begingroup$
How do you express the expression $1-xe^{-x}ge 0$ as $frac{1}{x}ge e^{-x}$ without changing the field of definition for $x$?
$endgroup$
– Ivon
Jan 28 at 12:16
$begingroup$
To go from $e^x geq x$ to an inequality about $1/x$ and $e^{-x}$ involves multiplying $e^{-x}/x$ on both sides which is negative or positive depending on $x$ and respectively affects the inequality
$endgroup$
– Naweed G. Seldon
Jan 28 at 12:17
$begingroup$
Seem my answer to another question at this link.
$endgroup$
– TonyK
Jan 28 at 12:21
1
1
$begingroup$
If you divide an inequality by a negative number the inequality gets reversed.
$endgroup$
– Kavi Rama Murthy
Jan 28 at 12:15
$begingroup$
If you divide an inequality by a negative number the inequality gets reversed.
$endgroup$
– Kavi Rama Murthy
Jan 28 at 12:15
$begingroup$
How do you express the expression $1-xe^{-x}ge 0$ as $frac{1}{x}ge e^{-x}$ without changing the field of definition for $x$?
$endgroup$
– Ivon
Jan 28 at 12:16
$begingroup$
How do you express the expression $1-xe^{-x}ge 0$ as $frac{1}{x}ge e^{-x}$ without changing the field of definition for $x$?
$endgroup$
– Ivon
Jan 28 at 12:16
$begingroup$
To go from $e^x geq x$ to an inequality about $1/x$ and $e^{-x}$ involves multiplying $e^{-x}/x$ on both sides which is negative or positive depending on $x$ and respectively affects the inequality
$endgroup$
– Naweed G. Seldon
Jan 28 at 12:17
$begingroup$
To go from $e^x geq x$ to an inequality about $1/x$ and $e^{-x}$ involves multiplying $e^{-x}/x$ on both sides which is negative or positive depending on $x$ and respectively affects the inequality
$endgroup$
– Naweed G. Seldon
Jan 28 at 12:17
$begingroup$
Seem my answer to another question at this link.
$endgroup$
– TonyK
Jan 28 at 12:21
$begingroup$
Seem my answer to another question at this link.
$endgroup$
– TonyK
Jan 28 at 12:21
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Rearranging $1-xe^{-x}ge 0$ gives the equivalent $xe^{-x}le 1$. For $gge 0$, this implies $xe^{-x}gle g$. The choice $g=e^x$ is non-negative as required, so we can deduce $xle e^x$. But if you try to instead use $g=1/x$ to obtain $e^{-x}le 1/x$, the problem is this $g$ isn't non-negative for all $x$.
answered Jan 28 at 12:28
J.G.J.G.
32.2k23250
$endgroup$
1
$begingroup$
... and $g$ isn't even defined for $x=0$
$endgroup$
– Hagen von Eitzen
Jan 28 at 14:31
add a comment |
$begingroup$
You need to be careful when you multiply inequalities by $x$ as multiplying with negative numbers reverses the sign. That's why the first one is only true for positive numbers while you "made" the second one always true by multiplying with $x$ (which is negative if $x < 0$). The same happens with $x geq 0$ and $x^2 geq 0$.
EDIT: In fact, you did this twice. First you devided by $x$ and then you multiplied by $x$. Hence the function is always positive (the two mistakes cancel).
answered Jan 28 at 12:15
KlausKlaus
2,782113
$endgroup$
add a comment |
$begingroup$
"I came to the expression $1-xe^{-x} geq 0$ which can be expressed as $frac{1}{x} geq e^{-x}$ or as $e^{x} geq x$"
They only express the same thing if $x>0$.
answered Jan 28 at 12:18
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
Hint:
Notice that $f: 1 - xe^{-x}$ is defined for $f: mathbb{R} to mathbb{R}$.
also, $frac{1}{x} geq e^{-x}$ is defined for every $x in mathbb{R} setminus {0} $
but $e^ x geq x$ is defined for every $x in mathbb{R}$
answered Jan 28 at 12:20
JnevenJneven
951322
$endgroup$
$begingroup$
So for avoiding this mistake, when reducing an inequality, should I keep the same domain?
$endgroup$
– Edoardo Grassi
Jan 28 at 12:27
$begingroup$
you should make sure that there exists no $x$ such that expression (1) is defined for but expression (2) isn't defined for. for this one, as instance, for $x =0$ expression (1): $frac{1}{x} geq e^{-x}$ isn't define for, but for $x =0$, expression (2): $e^x geq x $ IS DEFINED FOR
$endgroup$
– Jneven
Jan 28 at 13:38
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Rearranging $1-xe^{-x}ge 0$ gives the equivalent $xe^{-x}le 1$. For $gge 0$, this implies $xe^{-x}gle g$. The choice $g=e^x$ is non-negative as required, so we can deduce $xle e^x$. But if you try to instead use $g=1/x$ to obtain $e^{-x}le 1/x$, the problem is this $g$ isn't non-negative for all $x$.
answered Jan 28 at 12:28
J.G.J.G.
32.2k23250
$endgroup$
1
$begingroup$
... and $g$ isn't even defined for $x=0$
$endgroup$
– Hagen von Eitzen
Jan 28 at 14:31
add a comment |
$begingroup$
Rearranging $1-xe^{-x}ge 0$ gives the equivalent $xe^{-x}le 1$. For $gge 0$, this implies $xe^{-x}gle g$. The choice $g=e^x$ is non-negative as required, so we can deduce $xle e^x$. But if you try to instead use $g=1/x$ to obtain $e^{-x}le 1/x$, the problem is this $g$ isn't non-negative for all $x$.
answered Jan 28 at 12:28
J.G.J.G.
32.2k23250
$endgroup$
1
$begingroup$
... and $g$ isn't even defined for $x=0$
$endgroup$
– Hagen von Eitzen
Jan 28 at 14:31
add a comment |
$begingroup$
Rearranging $1-xe^{-x}ge 0$ gives the equivalent $xe^{-x}le 1$. For $gge 0$, this implies $xe^{-x}gle g$. The choice $g=e^x$ is non-negative as required, so we can deduce $xle e^x$. But if you try to instead use $g=1/x$ to obtain $e^{-x}le 1/x$, the problem is this $g$ isn't non-negative for all $x$.
answered Jan 28 at 12:28
J.G.J.G.
32.2k23250
$endgroup$
Rearranging $1-xe^{-x}ge 0$ gives the equivalent $xe^{-x}le 1$. For $gge 0$, this implies $xe^{-x}gle g$. The choice $g=e^x$ is non-negative as required, so we can deduce $xle e^x$. But if you try to instead use $g=1/x$ to obtain $e^{-x}le 1/x$, the problem is this $g$ isn't non-negative for all $x$.
answered Jan 28 at 12:28
J.G.J.G.
32.2k23250
answered Jan 28 at 12:28
J.G.J.G.
32.2k23250
answered Jan 28 at 12:28
J.G.J.G.
32.2k23250
answered Jan 28 at 12:28
J.G.J.G.
32.2k23250
32.2k23250
1
$begingroup$
... and $g$ isn't even defined for $x=0$
$endgroup$
– Hagen von Eitzen
Jan 28 at 14:31
add a comment |
1
$begingroup$
... and $g$ isn't even defined for $x=0$
$endgroup$
– Hagen von Eitzen
Jan 28 at 14:31
1
1
$begingroup$
... and $g$ isn't even defined for $x=0$
$endgroup$
– Hagen von Eitzen
Jan 28 at 14:31
$begingroup$
... and $g$ isn't even defined for $x=0$
$endgroup$
– Hagen von Eitzen
Jan 28 at 14:31
add a comment |
$begingroup$
You need to be careful when you multiply inequalities by $x$ as multiplying with negative numbers reverses the sign. That's why the first one is only true for positive numbers while you "made" the second one always true by multiplying with $x$ (which is negative if $x < 0$). The same happens with $x geq 0$ and $x^2 geq 0$.
EDIT: In fact, you did this twice. First you devided by $x$ and then you multiplied by $x$. Hence the function is always positive (the two mistakes cancel).
answered Jan 28 at 12:15
KlausKlaus
2,782113
$endgroup$
add a comment |
$begingroup$
You need to be careful when you multiply inequalities by $x$ as multiplying with negative numbers reverses the sign. That's why the first one is only true for positive numbers while you "made" the second one always true by multiplying with $x$ (which is negative if $x < 0$). The same happens with $x geq 0$ and $x^2 geq 0$.
EDIT: In fact, you did this twice. First you devided by $x$ and then you multiplied by $x$. Hence the function is always positive (the two mistakes cancel).
answered Jan 28 at 12:15
KlausKlaus
2,782113
$endgroup$
add a comment |
$begingroup$
You need to be careful when you multiply inequalities by $x$ as multiplying with negative numbers reverses the sign. That's why the first one is only true for positive numbers while you "made" the second one always true by multiplying with $x$ (which is negative if $x < 0$). The same happens with $x geq 0$ and $x^2 geq 0$.
EDIT: In fact, you did this twice. First you devided by $x$ and then you multiplied by $x$. Hence the function is always positive (the two mistakes cancel).
answered Jan 28 at 12:15
KlausKlaus
2,782113
$endgroup$
You need to be careful when you multiply inequalities by $x$ as multiplying with negative numbers reverses the sign. That's why the first one is only true for positive numbers while you "made" the second one always true by multiplying with $x$ (which is negative if $x < 0$). The same happens with $x geq 0$ and $x^2 geq 0$.
EDIT: In fact, you did this twice. First you devided by $x$ and then you multiplied by $x$. Hence the function is always positive (the two mistakes cancel).
answered Jan 28 at 12:15
KlausKlaus
2,782113
answered Jan 28 at 12:15
KlausKlaus
2,782113
answered Jan 28 at 12:15
KlausKlaus
2,782113
answered Jan 28 at 12:15
KlausKlaus
2,782113
2,782113
add a comment |
add a comment |
$begingroup$
"I came to the expression $1-xe^{-x} geq 0$ which can be expressed as $frac{1}{x} geq e^{-x}$ or as $e^{x} geq x$"
They only express the same thing if $x>0$.
answered Jan 28 at 12:18
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
"I came to the expression $1-xe^{-x} geq 0$ which can be expressed as $frac{1}{x} geq e^{-x}$ or as $e^{x} geq x$"
They only express the same thing if $x>0$.
answered Jan 28 at 12:18
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
"I came to the expression $1-xe^{-x} geq 0$ which can be expressed as $frac{1}{x} geq e^{-x}$ or as $e^{x} geq x$"
They only express the same thing if $x>0$.
answered Jan 28 at 12:18
drhabdrhab
104k545136
$endgroup$
"I came to the expression $1-xe^{-x} geq 0$ which can be expressed as $frac{1}{x} geq e^{-x}$ or as $e^{x} geq x$"
They only express the same thing if $x>0$.
answered Jan 28 at 12:18
drhabdrhab
104k545136
answered Jan 28 at 12:18
drhabdrhab
104k545136
answered Jan 28 at 12:18
drhabdrhab
104k545136
answered Jan 28 at 12:18
drhabdrhab
104k545136
104k545136
add a comment |
add a comment |
$begingroup$
Hint:
Notice that $f: 1 - xe^{-x}$ is defined for $f: mathbb{R} to mathbb{R}$.
also, $frac{1}{x} geq e^{-x}$ is defined for every $x in mathbb{R} setminus {0} $
but $e^ x geq x$ is defined for every $x in mathbb{R}$
answered Jan 28 at 12:20
JnevenJneven
951322
$endgroup$
$begingroup$
So for avoiding this mistake, when reducing an inequality, should I keep the same domain?
$endgroup$
– Edoardo Grassi
Jan 28 at 12:27
$begingroup$
you should make sure that there exists no $x$ such that expression (1) is defined for but expression (2) isn't defined for. for this one, as instance, for $x =0$ expression (1): $frac{1}{x} geq e^{-x}$ isn't define for, but for $x =0$, expression (2): $e^x geq x $ IS DEFINED FOR
$endgroup$
– Jneven
Jan 28 at 13:38
add a comment |
$begingroup$
Hint:
Notice that $f: 1 - xe^{-x}$ is defined for $f: mathbb{R} to mathbb{R}$.
also, $frac{1}{x} geq e^{-x}$ is defined for every $x in mathbb{R} setminus {0} $
but $e^ x geq x$ is defined for every $x in mathbb{R}$
answered Jan 28 at 12:20
JnevenJneven
951322
$endgroup$
$begingroup$
So for avoiding this mistake, when reducing an inequality, should I keep the same domain?
$endgroup$
– Edoardo Grassi
Jan 28 at 12:27
$begingroup$
you should make sure that there exists no $x$ such that expression (1) is defined for but expression (2) isn't defined for. for this one, as instance, for $x =0$ expression (1): $frac{1}{x} geq e^{-x}$ isn't define for, but for $x =0$, expression (2): $e^x geq x $ IS DEFINED FOR
$endgroup$
– Jneven
Jan 28 at 13:38
add a comment |
$begingroup$
Hint:
Notice that $f: 1 - xe^{-x}$ is defined for $f: mathbb{R} to mathbb{R}$.
also, $frac{1}{x} geq e^{-x}$ is defined for every $x in mathbb{R} setminus {0} $
but $e^ x geq x$ is defined for every $x in mathbb{R}$
answered Jan 28 at 12:20
JnevenJneven
951322
$endgroup$
Hint:
Notice that $f: 1 - xe^{-x}$ is defined for $f: mathbb{R} to mathbb{R}$.
also, $frac{1}{x} geq e^{-x}$ is defined for every $x in mathbb{R} setminus {0} $
but $e^ x geq x$ is defined for every $x in mathbb{R}$
answered Jan 28 at 12:20
JnevenJneven
951322
answered Jan 28 at 12:20
JnevenJneven
951322
answered Jan 28 at 12:20
JnevenJneven
951322
answered Jan 28 at 12:20
JnevenJneven
951322
951322
$begingroup$
So for avoiding this mistake, when reducing an inequality, should I keep the same domain?
$endgroup$
– Edoardo Grassi
Jan 28 at 12:27
$begingroup$
you should make sure that there exists no $x$ such that expression (1) is defined for but expression (2) isn't defined for. for this one, as instance, for $x =0$ expression (1): $frac{1}{x} geq e^{-x}$ isn't define for, but for $x =0$, expression (2): $e^x geq x $ IS DEFINED FOR
$endgroup$
– Jneven
Jan 28 at 13:38
add a comment |
$begingroup$
So for avoiding this mistake, when reducing an inequality, should I keep the same domain?
$endgroup$
– Edoardo Grassi
Jan 28 at 12:27
$begingroup$
you should make sure that there exists no $x$ such that expression (1) is defined for but expression (2) isn't defined for. for this one, as instance, for $x =0$ expression (1): $frac{1}{x} geq e^{-x}$ isn't define for, but for $x =0$, expression (2): $e^x geq x $ IS DEFINED FOR
$endgroup$
– Jneven
Jan 28 at 13:38
$begingroup$
So for avoiding this mistake, when reducing an inequality, should I keep the same domain?
$endgroup$
– Edoardo Grassi
Jan 28 at 12:27
$begingroup$
So for avoiding this mistake, when reducing an inequality, should I keep the same domain?
$endgroup$
– Edoardo Grassi
Jan 28 at 12:27
$begingroup$
you should make sure that there exists no $x$ such that expression (1) is defined for but expression (2) isn't defined for. for this one, as instance, for $x =0$ expression (1): $frac{1}{x} geq e^{-x}$ isn't define for, but for $x =0$, expression (2): $e^x geq x $ IS DEFINED FOR
$endgroup$
– Jneven
Jan 28 at 13:38
$begingroup$
you should make sure that there exists no $x$ such that expression (1) is defined for but expression (2) isn't defined for. for this one, as instance, for $x =0$ expression (1): $frac{1}{x} geq e^{-x}$ isn't define for, but for $x =0$, expression (2): $e^x geq x $ IS DEFINED FOR
$endgroup$
– Jneven
Jan 28 at 13:38
add a comment |
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If you divide an inequality by a negative number the inequality gets reversed.
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– Kavi Rama Murthy
Jan 28 at 12:15
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How do you express the expression $1-xe^{-x}ge 0$ as $frac{1}{x}ge e^{-x}$ without changing the field of definition for $x$?
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– Ivon
Jan 28 at 12:16
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To go from $e^x geq x$ to an inequality about $1/x$ and $e^{-x}$ involves multiplying $e^{-x}/x$ on both sides which is negative or positive depending on $x$ and respectively affects the inequality
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– Naweed G. Seldon
Jan 28 at 12:17
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Seem my answer to another question at this link.
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– TonyK
Jan 28 at 12:21