Mixing
If $f′(x) ne 0$ for every $x ∈ (a, b)$, then $f$ is injective and onto an open interval $I ⊆ R$.
$begingroup$
Let $a$ and $b$ be real numbers such that $a < b$ and let $f : (a, b) → R$
be a differentiable function.
Show that:
If $f′(x) ne 0$ for every $x ∈ (a, b)$, then $f$ is injective and onto an open
interval $I ⊆ R$.
I have proved the injective part, however I am having trouble with the ''onto an open interval '' part.
real-analysis analysis
asked Jan 15 at 16:52
John CoxJohn Cox
83
$endgroup$
add a comment |
$begingroup$
Let $a$ and $b$ be real numbers such that $a < b$ and let $f : (a, b) → R$
be a differentiable function.
Show that:
If $f′(x) ne 0$ for every $x ∈ (a, b)$, then $f$ is injective and onto an open
interval $I ⊆ R$.
I have proved the injective part, however I am having trouble with the ''onto an open interval '' part.
real-analysis analysis
asked Jan 15 at 16:52
John CoxJohn Cox
83
$endgroup$
$begingroup$
Since $f$ is differentiable, it is also continuous. The image of a connected set, e.g. an open interval, under a continuous map is also connected. $f$ is strictly monotone since its derivative is strictly positive (or strictly negative), so the intermediate value theorem yields that the image of $(a,b)$ under $f$ is also an open interval.
$endgroup$
– Math1000
Jan 15 at 17:24
add a comment |
$begingroup$
Let $a$ and $b$ be real numbers such that $a < b$ and let $f : (a, b) → R$
be a differentiable function.
Show that:
If $f′(x) ne 0$ for every $x ∈ (a, b)$, then $f$ is injective and onto an open
interval $I ⊆ R$.
I have proved the injective part, however I am having trouble with the ''onto an open interval '' part.
real-analysis analysis
asked Jan 15 at 16:52
John CoxJohn Cox
83
$endgroup$
Let $a$ and $b$ be real numbers such that $a < b$ and let $f : (a, b) → R$
be a differentiable function.
Show that:
If $f′(x) ne 0$ for every $x ∈ (a, b)$, then $f$ is injective and onto an open
interval $I ⊆ R$.
I have proved the injective part, however I am having trouble with the ''onto an open interval '' part.
real-analysis analysis
real-analysis analysis
asked Jan 15 at 16:52
John CoxJohn Cox
83
asked Jan 15 at 16:52
John CoxJohn Cox
83
asked Jan 15 at 16:52
John CoxJohn Cox
83
asked Jan 15 at 16:52
John CoxJohn Cox
83
asked Jan 15 at 16:52
John CoxJohn Cox
83
83
$begingroup$
Since $f$ is differentiable, it is also continuous. The image of a connected set, e.g. an open interval, under a continuous map is also connected. $f$ is strictly monotone since its derivative is strictly positive (or strictly negative), so the intermediate value theorem yields that the image of $(a,b)$ under $f$ is also an open interval.
$endgroup$
– Math1000
Jan 15 at 17:24
add a comment |
$begingroup$
Since $f$ is differentiable, it is also continuous. The image of a connected set, e.g. an open interval, under a continuous map is also connected. $f$ is strictly monotone since its derivative is strictly positive (or strictly negative), so the intermediate value theorem yields that the image of $(a,b)$ under $f$ is also an open interval.
$endgroup$
– Math1000
Jan 15 at 17:24
$begingroup$
Since $f$ is differentiable, it is also continuous. The image of a connected set, e.g. an open interval, under a continuous map is also connected. $f$ is strictly monotone since its derivative is strictly positive (or strictly negative), so the intermediate value theorem yields that the image of $(a,b)$ under $f$ is also an open interval.
$endgroup$
– Math1000
Jan 15 at 17:24
$begingroup$
Since $f$ is differentiable, it is also continuous. The image of a connected set, e.g. an open interval, under a continuous map is also connected. $f$ is strictly monotone since its derivative is strictly positive (or strictly negative), so the intermediate value theorem yields that the image of $(a,b)$ under $f$ is also an open interval.
$endgroup$
– Math1000
Jan 15 at 17:24
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $I=fbigl((a,b)bigr)subseteq Bbb R$ be the image of $f$, let $u=inf I$, $v=sup I$ (where possibly $u=-infty$ and/or $v=+infty$). We need to show that $I=(u,v)$, that is three claims:
- If $u<y<v$, then $yin I$
- $unotin I$
- $vnotin I$
For the first, if $u<y<v$, then by the definition of $inf$ and $sup$, there exist $y_1,y_2in I$ with $u<y_1<y<y_2<v$. If $y_{i}=f(x_{i})$, we can use the IVT to find $x$ between $x_1$ and $x_2$ with $f(x)=y$.
For the second, note that for $xin (a,b)$ we conclude that $f(xi)<f(x)$ for $xiapprox x$ and on the "correct" side of $x$ (depending on the sign of $f'(x)$). Therefore, $I$ contains vaues $<f(x)$ and $f(x)neinf I$ for all $xin(a,b)$.
The third follows by the same argument.
answered Jan 15 at 17:33
Hagen von EitzenHagen von Eitzen
280k23272504
$endgroup$
add a comment |
$begingroup$
Hint:
To show that $f((a,b))$ is an intervall it is enough to show that $f$ does not have a minimum/maximum and that for any $s,s'in f((a,b))$ $f$ attains all values between $s$ and $s'$.
For the first: Injective continous function are strictly monotone.
For the second: intermediate value theorem.
answered Jan 15 at 17:30
triitrii
1315
$endgroup$
add a comment |
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2 Answers
2
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2 Answers
2
active
oldest
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$begingroup$
Let $I=fbigl((a,b)bigr)subseteq Bbb R$ be the image of $f$, let $u=inf I$, $v=sup I$ (where possibly $u=-infty$ and/or $v=+infty$). We need to show that $I=(u,v)$, that is three claims:
- If $u<y<v$, then $yin I$
- $unotin I$
- $vnotin I$
For the first, if $u<y<v$, then by the definition of $inf$ and $sup$, there exist $y_1,y_2in I$ with $u<y_1<y<y_2<v$. If $y_{i}=f(x_{i})$, we can use the IVT to find $x$ between $x_1$ and $x_2$ with $f(x)=y$.
For the second, note that for $xin (a,b)$ we conclude that $f(xi)<f(x)$ for $xiapprox x$ and on the "correct" side of $x$ (depending on the sign of $f'(x)$). Therefore, $I$ contains vaues $<f(x)$ and $f(x)neinf I$ for all $xin(a,b)$.
The third follows by the same argument.
answered Jan 15 at 17:33
Hagen von EitzenHagen von Eitzen
280k23272504
$endgroup$
add a comment |
$begingroup$
Let $I=fbigl((a,b)bigr)subseteq Bbb R$ be the image of $f$, let $u=inf I$, $v=sup I$ (where possibly $u=-infty$ and/or $v=+infty$). We need to show that $I=(u,v)$, that is three claims:
- If $u<y<v$, then $yin I$
- $unotin I$
- $vnotin I$
For the first, if $u<y<v$, then by the definition of $inf$ and $sup$, there exist $y_1,y_2in I$ with $u<y_1<y<y_2<v$. If $y_{i}=f(x_{i})$, we can use the IVT to find $x$ between $x_1$ and $x_2$ with $f(x)=y$.
For the second, note that for $xin (a,b)$ we conclude that $f(xi)<f(x)$ for $xiapprox x$ and on the "correct" side of $x$ (depending on the sign of $f'(x)$). Therefore, $I$ contains vaues $<f(x)$ and $f(x)neinf I$ for all $xin(a,b)$.
The third follows by the same argument.
answered Jan 15 at 17:33
Hagen von EitzenHagen von Eitzen
280k23272504
$endgroup$
add a comment |
$begingroup$
Let $I=fbigl((a,b)bigr)subseteq Bbb R$ be the image of $f$, let $u=inf I$, $v=sup I$ (where possibly $u=-infty$ and/or $v=+infty$). We need to show that $I=(u,v)$, that is three claims:
- If $u<y<v$, then $yin I$
- $unotin I$
- $vnotin I$
For the first, if $u<y<v$, then by the definition of $inf$ and $sup$, there exist $y_1,y_2in I$ with $u<y_1<y<y_2<v$. If $y_{i}=f(x_{i})$, we can use the IVT to find $x$ between $x_1$ and $x_2$ with $f(x)=y$.
For the second, note that for $xin (a,b)$ we conclude that $f(xi)<f(x)$ for $xiapprox x$ and on the "correct" side of $x$ (depending on the sign of $f'(x)$). Therefore, $I$ contains vaues $<f(x)$ and $f(x)neinf I$ for all $xin(a,b)$.
The third follows by the same argument.
answered Jan 15 at 17:33
Hagen von EitzenHagen von Eitzen
280k23272504
$endgroup$
Let $I=fbigl((a,b)bigr)subseteq Bbb R$ be the image of $f$, let $u=inf I$, $v=sup I$ (where possibly $u=-infty$ and/or $v=+infty$). We need to show that $I=(u,v)$, that is three claims:
- If $u<y<v$, then $yin I$
- $unotin I$
- $vnotin I$
For the first, if $u<y<v$, then by the definition of $inf$ and $sup$, there exist $y_1,y_2in I$ with $u<y_1<y<y_2<v$. If $y_{i}=f(x_{i})$, we can use the IVT to find $x$ between $x_1$ and $x_2$ with $f(x)=y$.
For the second, note that for $xin (a,b)$ we conclude that $f(xi)<f(x)$ for $xiapprox x$ and on the "correct" side of $x$ (depending on the sign of $f'(x)$). Therefore, $I$ contains vaues $<f(x)$ and $f(x)neinf I$ for all $xin(a,b)$.
The third follows by the same argument.
answered Jan 15 at 17:33
Hagen von EitzenHagen von Eitzen
280k23272504
answered Jan 15 at 17:33
Hagen von EitzenHagen von Eitzen
280k23272504
answered Jan 15 at 17:33
Hagen von EitzenHagen von Eitzen
280k23272504
answered Jan 15 at 17:33
Hagen von EitzenHagen von Eitzen
280k23272504
280k23272504
add a comment |
add a comment |
$begingroup$
Hint:
To show that $f((a,b))$ is an intervall it is enough to show that $f$ does not have a minimum/maximum and that for any $s,s'in f((a,b))$ $f$ attains all values between $s$ and $s'$.
For the first: Injective continous function are strictly monotone.
For the second: intermediate value theorem.
answered Jan 15 at 17:30
triitrii
1315
$endgroup$
add a comment |
$begingroup$
Hint:
To show that $f((a,b))$ is an intervall it is enough to show that $f$ does not have a minimum/maximum and that for any $s,s'in f((a,b))$ $f$ attains all values between $s$ and $s'$.
For the first: Injective continous function are strictly monotone.
For the second: intermediate value theorem.
answered Jan 15 at 17:30
triitrii
1315
$endgroup$
add a comment |
$begingroup$
Hint:
To show that $f((a,b))$ is an intervall it is enough to show that $f$ does not have a minimum/maximum and that for any $s,s'in f((a,b))$ $f$ attains all values between $s$ and $s'$.
For the first: Injective continous function are strictly monotone.
For the second: intermediate value theorem.
answered Jan 15 at 17:30
triitrii
1315
$endgroup$
Hint:
To show that $f((a,b))$ is an intervall it is enough to show that $f$ does not have a minimum/maximum and that for any $s,s'in f((a,b))$ $f$ attains all values between $s$ and $s'$.
For the first: Injective continous function are strictly monotone.
For the second: intermediate value theorem.
answered Jan 15 at 17:30
triitrii
1315
answered Jan 15 at 17:30
triitrii
1315
answered Jan 15 at 17:30
triitrii
1315
answered Jan 15 at 17:30
triitrii
1315
1315
add a comment |
add a comment |
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$begingroup$
Since $f$ is differentiable, it is also continuous. The image of a connected set, e.g. an open interval, under a continuous map is also connected. $f$ is strictly monotone since its derivative is strictly positive (or strictly negative), so the intermediate value theorem yields that the image of $(a,b)$ under $f$ is also an open interval.
$endgroup$
– Math1000
Jan 15 at 17:24