Mixing
Extending Rolle's Theorem to Infinity
Clash Royale CLAN TAG#URR8PPP
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Let $f : mathbbR to mathbbR$ be a function which is differentiable everywhere on $mathbbR$. If we know that
$$lim_xto-inftyf(x) = L = lim_xtoinftyf(x)$$
How would one go about showing that there must exist $c in mathbbR$, such that $f'(c) = 0$ for all cases at once?
calculus real-analysis
asked 5 hours ago
Eetu Koskela
206
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up vote
1
down vote
favorite
Let $f : mathbbR to mathbbR$ be a function which is differentiable everywhere on $mathbbR$. If we know that
$$lim_xto-inftyf(x) = L = lim_xtoinftyf(x)$$
How would one go about showing that there must exist $c in mathbbR$, such that $f'(c) = 0$ for all cases at once?
calculus real-analysis
asked 5 hours ago
Eetu Koskela
206
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $f : mathbbR to mathbbR$ be a function which is differentiable everywhere on $mathbbR$. If we know that
$$lim_xto-inftyf(x) = L = lim_xtoinftyf(x)$$
How would one go about showing that there must exist $c in mathbbR$, such that $f'(c) = 0$ for all cases at once?
calculus real-analysis
asked 5 hours ago
Eetu Koskela
206
Let $f : mathbbR to mathbbR$ be a function which is differentiable everywhere on $mathbbR$. If we know that
$$lim_xto-inftyf(x) = L = lim_xtoinftyf(x)$$
How would one go about showing that there must exist $c in mathbbR$, such that $f'(c) = 0$ for all cases at once?
calculus real-analysis
calculus real-analysis
asked 5 hours ago
Eetu Koskela
206
asked 5 hours ago
Eetu Koskela
206
asked 5 hours ago
Eetu Koskela
206
asked 5 hours ago
Eetu Koskela
206
asked 5 hours ago
Eetu Koskela
206
206
add a comment |Â
add a comment |Â
3 Answers
3
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up vote
4
down vote
Consider
$$
g(x)=f(tan x)
$$
defined over $(-pi/2,pi/2)$. Then we can extend it to $[-pi/2,pi/2]$ by setting
$$
g(-pi/2)=g(pi/2)=L
$$
The derivative of $g$ exists over $(-pi/2,pi/2)$ and is equal to
$$
g'(x)=fracf'(tan x)cos^2x
$$
By Rolle's theorem, there exists $c$ such that $g'(c)=0$, so that $f'(tan c)=0$.
You can do similarly for a function $f$ defined over $[a,infty)$, differentiable over $(a,infty)$ and such that
$$
f(a)=lim_xtoinftyf(x)
$$
Define $g(x)=f(tan x)$ over $[arctan a,pi/2)$ and the proof will be the same as before.
answered 5 hours ago
egreg
172k1283194
add a comment |Â
up vote
3
down vote
If $f=L$ everywhere, the conclusion is clear. So let's assume $f(x_0)ne L$ for some $x_0.$ WLOG, $f(x_0)>L.$ Because of the given limit conditions, there exists $a>|x_0|$ such that both $f(-a),f(a) < f(x_0).$ Then the maximum value of $f$ on $[-a,a]$ can't occur at either of the end points. It thus occurs at an interior point $c,$ and at that $c$ we have $ f'(c)=0.$
answered 4 hours ago
zhw.
69.9k42974
add a comment |Â
up vote
1
down vote
Derivatives satisfy the intermediate value property.
If $f'(c) not= 0$ for all $c$, then either $f'(c) > 0$ for all $c$ or $f'(c) < 0$ for all $c$.
In either case, $f$ is strictly monotone ruling out $displaystyle lim_x to -infty f(x) = lim_x to infty f(x)$.
answered 4 hours ago
Umberto P.
37.3k12962
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Consider
$$
g(x)=f(tan x)
$$
defined over $(-pi/2,pi/2)$. Then we can extend it to $[-pi/2,pi/2]$ by setting
$$
g(-pi/2)=g(pi/2)=L
$$
The derivative of $g$ exists over $(-pi/2,pi/2)$ and is equal to
$$
g'(x)=fracf'(tan x)cos^2x
$$
By Rolle's theorem, there exists $c$ such that $g'(c)=0$, so that $f'(tan c)=0$.
You can do similarly for a function $f$ defined over $[a,infty)$, differentiable over $(a,infty)$ and such that
$$
f(a)=lim_xtoinftyf(x)
$$
Define $g(x)=f(tan x)$ over $[arctan a,pi/2)$ and the proof will be the same as before.
answered 5 hours ago
egreg
172k1283194
add a comment |Â
up vote
4
down vote
Consider
$$
g(x)=f(tan x)
$$
defined over $(-pi/2,pi/2)$. Then we can extend it to $[-pi/2,pi/2]$ by setting
$$
g(-pi/2)=g(pi/2)=L
$$
The derivative of $g$ exists over $(-pi/2,pi/2)$ and is equal to
$$
g'(x)=fracf'(tan x)cos^2x
$$
By Rolle's theorem, there exists $c$ such that $g'(c)=0$, so that $f'(tan c)=0$.
You can do similarly for a function $f$ defined over $[a,infty)$, differentiable over $(a,infty)$ and such that
$$
f(a)=lim_xtoinftyf(x)
$$
Define $g(x)=f(tan x)$ over $[arctan a,pi/2)$ and the proof will be the same as before.
answered 5 hours ago
egreg
172k1283194
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Consider
$$
g(x)=f(tan x)
$$
defined over $(-pi/2,pi/2)$. Then we can extend it to $[-pi/2,pi/2]$ by setting
$$
g(-pi/2)=g(pi/2)=L
$$
The derivative of $g$ exists over $(-pi/2,pi/2)$ and is equal to
$$
g'(x)=fracf'(tan x)cos^2x
$$
By Rolle's theorem, there exists $c$ such that $g'(c)=0$, so that $f'(tan c)=0$.
You can do similarly for a function $f$ defined over $[a,infty)$, differentiable over $(a,infty)$ and such that
$$
f(a)=lim_xtoinftyf(x)
$$
Define $g(x)=f(tan x)$ over $[arctan a,pi/2)$ and the proof will be the same as before.
answered 5 hours ago
egreg
172k1283194
Consider
$$
g(x)=f(tan x)
$$
defined over $(-pi/2,pi/2)$. Then we can extend it to $[-pi/2,pi/2]$ by setting
$$
g(-pi/2)=g(pi/2)=L
$$
The derivative of $g$ exists over $(-pi/2,pi/2)$ and is equal to
$$
g'(x)=fracf'(tan x)cos^2x
$$
By Rolle's theorem, there exists $c$ such that $g'(c)=0$, so that $f'(tan c)=0$.
You can do similarly for a function $f$ defined over $[a,infty)$, differentiable over $(a,infty)$ and such that
$$
f(a)=lim_xtoinftyf(x)
$$
Define $g(x)=f(tan x)$ over $[arctan a,pi/2)$ and the proof will be the same as before.
answered 5 hours ago
egreg
172k1283194
answered 5 hours ago
egreg
172k1283194
answered 5 hours ago
egreg
172k1283194
answered 5 hours ago
egreg
172k1283194
172k1283194
add a comment |Â
add a comment |Â
up vote
3
down vote
If $f=L$ everywhere, the conclusion is clear. So let's assume $f(x_0)ne L$ for some $x_0.$ WLOG, $f(x_0)>L.$ Because of the given limit conditions, there exists $a>|x_0|$ such that both $f(-a),f(a) < f(x_0).$ Then the maximum value of $f$ on $[-a,a]$ can't occur at either of the end points. It thus occurs at an interior point $c,$ and at that $c$ we have $ f'(c)=0.$
answered 4 hours ago
zhw.
69.9k42974
add a comment |Â
up vote
3
down vote
If $f=L$ everywhere, the conclusion is clear. So let's assume $f(x_0)ne L$ for some $x_0.$ WLOG, $f(x_0)>L.$ Because of the given limit conditions, there exists $a>|x_0|$ such that both $f(-a),f(a) < f(x_0).$ Then the maximum value of $f$ on $[-a,a]$ can't occur at either of the end points. It thus occurs at an interior point $c,$ and at that $c$ we have $ f'(c)=0.$
answered 4 hours ago
zhw.
69.9k42974
add a comment |Â
up vote
3
down vote
up vote
3
down vote
If $f=L$ everywhere, the conclusion is clear. So let's assume $f(x_0)ne L$ for some $x_0.$ WLOG, $f(x_0)>L.$ Because of the given limit conditions, there exists $a>|x_0|$ such that both $f(-a),f(a) < f(x_0).$ Then the maximum value of $f$ on $[-a,a]$ can't occur at either of the end points. It thus occurs at an interior point $c,$ and at that $c$ we have $ f'(c)=0.$
answered 4 hours ago
zhw.
69.9k42974
If $f=L$ everywhere, the conclusion is clear. So let's assume $f(x_0)ne L$ for some $x_0.$ WLOG, $f(x_0)>L.$ Because of the given limit conditions, there exists $a>|x_0|$ such that both $f(-a),f(a) < f(x_0).$ Then the maximum value of $f$ on $[-a,a]$ can't occur at either of the end points. It thus occurs at an interior point $c,$ and at that $c$ we have $ f'(c)=0.$
answered 4 hours ago
zhw.
69.9k42974
answered 4 hours ago
zhw.
69.9k42974
answered 4 hours ago
zhw.
69.9k42974
answered 4 hours ago
zhw.
69.9k42974
69.9k42974
add a comment |Â
add a comment |Â
up vote
1
down vote
Derivatives satisfy the intermediate value property.
If $f'(c) not= 0$ for all $c$, then either $f'(c) > 0$ for all $c$ or $f'(c) < 0$ for all $c$.
In either case, $f$ is strictly monotone ruling out $displaystyle lim_x to -infty f(x) = lim_x to infty f(x)$.
answered 4 hours ago
Umberto P.
37.3k12962
add a comment |Â
up vote
1
down vote
Derivatives satisfy the intermediate value property.
If $f'(c) not= 0$ for all $c$, then either $f'(c) > 0$ for all $c$ or $f'(c) < 0$ for all $c$.
In either case, $f$ is strictly monotone ruling out $displaystyle lim_x to -infty f(x) = lim_x to infty f(x)$.
answered 4 hours ago
Umberto P.
37.3k12962
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Derivatives satisfy the intermediate value property.
If $f'(c) not= 0$ for all $c$, then either $f'(c) > 0$ for all $c$ or $f'(c) < 0$ for all $c$.
In either case, $f$ is strictly monotone ruling out $displaystyle lim_x to -infty f(x) = lim_x to infty f(x)$.
answered 4 hours ago
Umberto P.
37.3k12962
Derivatives satisfy the intermediate value property.
If $f'(c) not= 0$ for all $c$, then either $f'(c) > 0$ for all $c$ or $f'(c) < 0$ for all $c$.
In either case, $f$ is strictly monotone ruling out $displaystyle lim_x to -infty f(x) = lim_x to infty f(x)$.
answered 4 hours ago
Umberto P.
37.3k12962
answered 4 hours ago
Umberto P.
37.3k12962
answered 4 hours ago
Umberto P.
37.3k12962
answered 4 hours ago
Umberto P.
37.3k12962
37.3k12962
add a comment |Â
add a comment |Â
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