Mean Value Theorem at infinity
Suppose $f: (a,b) rightarrow mathbb{R}$ is differentiable with $lim_{x rightarrow a} f(x) = infty$ and $lim_{x rightarrow b} f(x) = -infty$. Suppose also that there are $c_1 < c_2 in (a, b)$ with $f(c_1) leq f(c_2)$. For any $lambda < 0$, there is $x_0 in (a, b)$ with $f'(x_0) = lambda$.
We say that $f(x)rightarrow infty$ as $xrightarrow c$ if for every $Binmathbb{R}$ there is $delta>0$ so that $0<lvert x-crvert<deltaRightarrow f(x)>B$.
We say that $f(x)rightarrow -infty$ as $xrightarrow c$ if for every $Binmathbb{R}$ there is $delta>0$ so that $0<lvert x-crvert<deltaRightarrow f(x)<B$.
My attempt: so my idea for the problem is that I can just consider halfof the graph and pick a point $alpha$ that is nearby to $a$, then consider the interval $[alpha, c_2]$, we can get the slope with the slope formula and we get $frac{f(alpha) - f(c_2)}{alpha - c_2} > frac{B - f(c_2)}{alpha - c_2} geq lambda + 1 > lambda$.
My question: I am not sure how to pick a point that is nearby to $a$, will I be picking a point that is within the delta neighbourhood? Also, I don't know how to prove $lambda < 0$.
real-analysis derivatives
asked Nov 19 '18 at 6:25
HD5450
64
add a comment |
Suppose $f: (a,b) rightarrow mathbb{R}$ is differentiable with $lim_{x rightarrow a} f(x) = infty$ and $lim_{x rightarrow b} f(x) = -infty$. Suppose also that there are $c_1 < c_2 in (a, b)$ with $f(c_1) leq f(c_2)$. For any $lambda < 0$, there is $x_0 in (a, b)$ with $f'(x_0) = lambda$.
We say that $f(x)rightarrow infty$ as $xrightarrow c$ if for every $Binmathbb{R}$ there is $delta>0$ so that $0<lvert x-crvert<deltaRightarrow f(x)>B$.
We say that $f(x)rightarrow -infty$ as $xrightarrow c$ if for every $Binmathbb{R}$ there is $delta>0$ so that $0<lvert x-crvert<deltaRightarrow f(x)<B$.
My attempt: so my idea for the problem is that I can just consider halfof the graph and pick a point $alpha$ that is nearby to $a$, then consider the interval $[alpha, c_2]$, we can get the slope with the slope formula and we get $frac{f(alpha) - f(c_2)}{alpha - c_2} > frac{B - f(c_2)}{alpha - c_2} geq lambda + 1 > lambda$.
My question: I am not sure how to pick a point that is nearby to $a$, will I be picking a point that is within the delta neighbourhood? Also, I don't know how to prove $lambda < 0$.
real-analysis derivatives
asked Nov 19 '18 at 6:25
HD5450
64
add a comment |
Suppose $f: (a,b) rightarrow mathbb{R}$ is differentiable with $lim_{x rightarrow a} f(x) = infty$ and $lim_{x rightarrow b} f(x) = -infty$. Suppose also that there are $c_1 < c_2 in (a, b)$ with $f(c_1) leq f(c_2)$. For any $lambda < 0$, there is $x_0 in (a, b)$ with $f'(x_0) = lambda$.
We say that $f(x)rightarrow infty$ as $xrightarrow c$ if for every $Binmathbb{R}$ there is $delta>0$ so that $0<lvert x-crvert<deltaRightarrow f(x)>B$.
We say that $f(x)rightarrow -infty$ as $xrightarrow c$ if for every $Binmathbb{R}$ there is $delta>0$ so that $0<lvert x-crvert<deltaRightarrow f(x)<B$.
My attempt: so my idea for the problem is that I can just consider halfof the graph and pick a point $alpha$ that is nearby to $a$, then consider the interval $[alpha, c_2]$, we can get the slope with the slope formula and we get $frac{f(alpha) - f(c_2)}{alpha - c_2} > frac{B - f(c_2)}{alpha - c_2} geq lambda + 1 > lambda$.
My question: I am not sure how to pick a point that is nearby to $a$, will I be picking a point that is within the delta neighbourhood? Also, I don't know how to prove $lambda < 0$.
real-analysis derivatives
asked Nov 19 '18 at 6:25
HD5450
64
Suppose $f: (a,b) rightarrow mathbb{R}$ is differentiable with $lim_{x rightarrow a} f(x) = infty$ and $lim_{x rightarrow b} f(x) = -infty$. Suppose also that there are $c_1 < c_2 in (a, b)$ with $f(c_1) leq f(c_2)$. For any $lambda < 0$, there is $x_0 in (a, b)$ with $f'(x_0) = lambda$.
We say that $f(x)rightarrow infty$ as $xrightarrow c$ if for every $Binmathbb{R}$ there is $delta>0$ so that $0<lvert x-crvert<deltaRightarrow f(x)>B$.
We say that $f(x)rightarrow -infty$ as $xrightarrow c$ if for every $Binmathbb{R}$ there is $delta>0$ so that $0<lvert x-crvert<deltaRightarrow f(x)<B$.
My attempt: so my idea for the problem is that I can just consider halfof the graph and pick a point $alpha$ that is nearby to $a$, then consider the interval $[alpha, c_2]$, we can get the slope with the slope formula and we get $frac{f(alpha) - f(c_2)}{alpha - c_2} > frac{B - f(c_2)}{alpha - c_2} geq lambda + 1 > lambda$.
My question: I am not sure how to pick a point that is nearby to $a$, will I be picking a point that is within the delta neighbourhood? Also, I don't know how to prove $lambda < 0$.
real-analysis derivatives
real-analysis derivatives
asked Nov 19 '18 at 6:25
HD5450
64
asked Nov 19 '18 at 6:25
HD5450
64
asked Nov 19 '18 at 6:25
HD5450
64
asked Nov 19 '18 at 6:25
HD5450
64
asked Nov 19 '18 at 6:25
HD5450
64
64
add a comment |
add a comment |
1 Answer
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All derivatives have IVP. So it is enough to show that there exists points $x,y$ with $f'(x) >lambda$ and $f'(y) <lambda$. If $f'<0$ at every point we have a contradiction to the hypothesis that $f(c_1) leq f(c_2)$. So $f' geq 0 >lambda$ at some point. Now suppose $f'(y) geq lambda$ for all $y$. Then $f(b-r)-f(a+r) geq (b-a-2r)lambda$ for all $r>0$ sufficiently small (by MVT). Letting $r to 0$ we a get contradiction to the fact that $f(b-)=-infty$ and $f(a+)=infty$. This completes teh proof.
answered Nov 19 '18 at 6:43
Kavi Rama Murthy
50.2k31854
How do you know that $f(b - r) - f(a + r) geq (b - a - 2r)lambda$?
– HD5450
Nov 19 '18 at 6:49
@HD5450 I am using Mean Value Theorem. $f(b-r)-f(a-r)=f'(xi) (b-a-2r)$ for some $xi$ and my assumption is $f'(y) geq lambda $ for all $y$. Take $y=xi$.
– Kavi Rama Murthy
Nov 19 '18 at 7:17
How do we know that $f(b - r)$ and $f(a + r)$ are defined
– HD5450
Nov 19 '18 at 7:21
@HD5450 Your function id defined on $(a,b)$. $a+r$ and $b-r$ are points in this interval.
– Kavi Rama Murthy
Nov 19 '18 at 7:23
add a comment |
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All derivatives have IVP. So it is enough to show that there exists points $x,y$ with $f'(x) >lambda$ and $f'(y) <lambda$. If $f'<0$ at every point we have a contradiction to the hypothesis that $f(c_1) leq f(c_2)$. So $f' geq 0 >lambda$ at some point. Now suppose $f'(y) geq lambda$ for all $y$. Then $f(b-r)-f(a+r) geq (b-a-2r)lambda$ for all $r>0$ sufficiently small (by MVT). Letting $r to 0$ we a get contradiction to the fact that $f(b-)=-infty$ and $f(a+)=infty$. This completes teh proof.
answered Nov 19 '18 at 6:43
Kavi Rama Murthy
50.2k31854
How do you know that $f(b - r) - f(a + r) geq (b - a - 2r)lambda$?
– HD5450
Nov 19 '18 at 6:49
@HD5450 I am using Mean Value Theorem. $f(b-r)-f(a-r)=f'(xi) (b-a-2r)$ for some $xi$ and my assumption is $f'(y) geq lambda $ for all $y$. Take $y=xi$.
– Kavi Rama Murthy
Nov 19 '18 at 7:17
How do we know that $f(b - r)$ and $f(a + r)$ are defined
– HD5450
Nov 19 '18 at 7:21
@HD5450 Your function id defined on $(a,b)$. $a+r$ and $b-r$ are points in this interval.
– Kavi Rama Murthy
Nov 19 '18 at 7:23
add a comment |
All derivatives have IVP. So it is enough to show that there exists points $x,y$ with $f'(x) >lambda$ and $f'(y) <lambda$. If $f'<0$ at every point we have a contradiction to the hypothesis that $f(c_1) leq f(c_2)$. So $f' geq 0 >lambda$ at some point. Now suppose $f'(y) geq lambda$ for all $y$. Then $f(b-r)-f(a+r) geq (b-a-2r)lambda$ for all $r>0$ sufficiently small (by MVT). Letting $r to 0$ we a get contradiction to the fact that $f(b-)=-infty$ and $f(a+)=infty$. This completes teh proof.
answered Nov 19 '18 at 6:43
Kavi Rama Murthy
50.2k31854
How do you know that $f(b - r) - f(a + r) geq (b - a - 2r)lambda$?
– HD5450
Nov 19 '18 at 6:49
@HD5450 I am using Mean Value Theorem. $f(b-r)-f(a-r)=f'(xi) (b-a-2r)$ for some $xi$ and my assumption is $f'(y) geq lambda $ for all $y$. Take $y=xi$.
– Kavi Rama Murthy
Nov 19 '18 at 7:17
How do we know that $f(b - r)$ and $f(a + r)$ are defined
– HD5450
Nov 19 '18 at 7:21
@HD5450 Your function id defined on $(a,b)$. $a+r$ and $b-r$ are points in this interval.
– Kavi Rama Murthy
Nov 19 '18 at 7:23
add a comment |
All derivatives have IVP. So it is enough to show that there exists points $x,y$ with $f'(x) >lambda$ and $f'(y) <lambda$. If $f'<0$ at every point we have a contradiction to the hypothesis that $f(c_1) leq f(c_2)$. So $f' geq 0 >lambda$ at some point. Now suppose $f'(y) geq lambda$ for all $y$. Then $f(b-r)-f(a+r) geq (b-a-2r)lambda$ for all $r>0$ sufficiently small (by MVT). Letting $r to 0$ we a get contradiction to the fact that $f(b-)=-infty$ and $f(a+)=infty$. This completes teh proof.
answered Nov 19 '18 at 6:43
Kavi Rama Murthy
50.2k31854
All derivatives have IVP. So it is enough to show that there exists points $x,y$ with $f'(x) >lambda$ and $f'(y) <lambda$. If $f'<0$ at every point we have a contradiction to the hypothesis that $f(c_1) leq f(c_2)$. So $f' geq 0 >lambda$ at some point. Now suppose $f'(y) geq lambda$ for all $y$. Then $f(b-r)-f(a+r) geq (b-a-2r)lambda$ for all $r>0$ sufficiently small (by MVT). Letting $r to 0$ we a get contradiction to the fact that $f(b-)=-infty$ and $f(a+)=infty$. This completes teh proof.
answered Nov 19 '18 at 6:43
Kavi Rama Murthy
50.2k31854
answered Nov 19 '18 at 6:43
Kavi Rama Murthy
50.2k31854
answered Nov 19 '18 at 6:43
Kavi Rama Murthy
50.2k31854
answered Nov 19 '18 at 6:43
Kavi Rama Murthy
50.2k31854
50.2k31854
How do you know that $f(b - r) - f(a + r) geq (b - a - 2r)lambda$?
– HD5450
Nov 19 '18 at 6:49
@HD5450 I am using Mean Value Theorem. $f(b-r)-f(a-r)=f'(xi) (b-a-2r)$ for some $xi$ and my assumption is $f'(y) geq lambda $ for all $y$. Take $y=xi$.
– Kavi Rama Murthy
Nov 19 '18 at 7:17
How do we know that $f(b - r)$ and $f(a + r)$ are defined
– HD5450
Nov 19 '18 at 7:21
@HD5450 Your function id defined on $(a,b)$. $a+r$ and $b-r$ are points in this interval.
– Kavi Rama Murthy
Nov 19 '18 at 7:23
add a comment |
How do you know that $f(b - r) - f(a + r) geq (b - a - 2r)lambda$?
– HD5450
Nov 19 '18 at 6:49
@HD5450 I am using Mean Value Theorem. $f(b-r)-f(a-r)=f'(xi) (b-a-2r)$ for some $xi$ and my assumption is $f'(y) geq lambda $ for all $y$. Take $y=xi$.
– Kavi Rama Murthy
Nov 19 '18 at 7:17
How do we know that $f(b - r)$ and $f(a + r)$ are defined
– HD5450
Nov 19 '18 at 7:21
@HD5450 Your function id defined on $(a,b)$. $a+r$ and $b-r$ are points in this interval.
– Kavi Rama Murthy
Nov 19 '18 at 7:23
How do you know that $f(b - r) - f(a + r) geq (b - a - 2r)lambda$?
– HD5450
Nov 19 '18 at 6:49
How do you know that $f(b - r) - f(a + r) geq (b - a - 2r)lambda$?
– HD5450
Nov 19 '18 at 6:49
@HD5450 I am using Mean Value Theorem. $f(b-r)-f(a-r)=f'(xi) (b-a-2r)$ for some $xi$ and my assumption is $f'(y) geq lambda $ for all $y$. Take $y=xi$.
– Kavi Rama Murthy
Nov 19 '18 at 7:17
@HD5450 I am using Mean Value Theorem. $f(b-r)-f(a-r)=f'(xi) (b-a-2r)$ for some $xi$ and my assumption is $f'(y) geq lambda $ for all $y$. Take $y=xi$.
– Kavi Rama Murthy
Nov 19 '18 at 7:17
How do we know that $f(b - r)$ and $f(a + r)$ are defined
– HD5450
Nov 19 '18 at 7:21
How do we know that $f(b - r)$ and $f(a + r)$ are defined
– HD5450
Nov 19 '18 at 7:21
@HD5450 Your function id defined on $(a,b)$. $a+r$ and $b-r$ are points in this interval.
– Kavi Rama Murthy
Nov 19 '18 at 7:23
@HD5450 Your function id defined on $(a,b)$. $a+r$ and $b-r$ are points in this interval.
– Kavi Rama Murthy
Nov 19 '18 at 7:23
add a comment |
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