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A non constant function satisfying the condition of Rolle's theorem cannot be monotonic.
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I'm reading a book of elementary calculus there a found the statement "A non constant function satisfying the condition of Rolle's theorem cannot be monotonic."
But i'm finding myself unable to prove it.I need help in proving this
calculus real-analysis means
asked Aug 6 at 21:43
PK Styles
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I'm reading a book of elementary calculus there a found the statement "A non constant function satisfying the condition of Rolle's theorem cannot be monotonic."
But i'm finding myself unable to prove it.I need help in proving this
calculus real-analysis means
asked Aug 6 at 21:43
PK Styles
1,394624
3
Well $f(a) = f(b)$. I'm reminded of the phrase "What goes up must come down".
– Rhys Steele
Aug 6 at 21:49
1
As written, it seems technically false. The function may be constant on the interval $[a,b]$ of Rolle's theorem but monotonically strictly increasing outside of that. However, if we restrict to being non-constant on $[a,b]$, the proof of Rolle's theorem on wiki shows that the derivative must change sign in the given region, which is sufficient.
– AdamW
Aug 6 at 21:57
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up vote
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down vote
favorite
I'm reading a book of elementary calculus there a found the statement "A non constant function satisfying the condition of Rolle's theorem cannot be monotonic."
But i'm finding myself unable to prove it.I need help in proving this
calculus real-analysis means
asked Aug 6 at 21:43
PK Styles
1,394624
I'm reading a book of elementary calculus there a found the statement "A non constant function satisfying the condition of Rolle's theorem cannot be monotonic."
But i'm finding myself unable to prove it.I need help in proving this
calculus real-analysis means
asked Aug 6 at 21:43
PK Styles
1,394624
asked Aug 6 at 21:43
PK Styles
1,394624
asked Aug 6 at 21:43
PK Styles
1,394624
asked Aug 6 at 21:43
PK Styles
1,394624
1,394624
3
Well $f(a) = f(b)$. I'm reminded of the phrase "What goes up must come down".
– Rhys Steele
Aug 6 at 21:49
1
As written, it seems technically false. The function may be constant on the interval $[a,b]$ of Rolle's theorem but monotonically strictly increasing outside of that. However, if we restrict to being non-constant on $[a,b]$, the proof of Rolle's theorem on wiki shows that the derivative must change sign in the given region, which is sufficient.
– AdamW
Aug 6 at 21:57
add a comment |Â
3
Well $f(a) = f(b)$. I'm reminded of the phrase "What goes up must come down".
– Rhys Steele
Aug 6 at 21:49
1
As written, it seems technically false. The function may be constant on the interval $[a,b]$ of Rolle's theorem but monotonically strictly increasing outside of that. However, if we restrict to being non-constant on $[a,b]$, the proof of Rolle's theorem on wiki shows that the derivative must change sign in the given region, which is sufficient.
– AdamW
Aug 6 at 21:57
3
3
Well $f(a) = f(b)$. I'm reminded of the phrase "What goes up must come down".
– Rhys Steele
Aug 6 at 21:49
Well $f(a) = f(b)$. I'm reminded of the phrase "What goes up must come down".
– Rhys Steele
Aug 6 at 21:49
1
1
As written, it seems technically false. The function may be constant on the interval $[a,b]$ of Rolle's theorem but monotonically strictly increasing outside of that. However, if we restrict to being non-constant on $[a,b]$, the proof of Rolle's theorem on wiki shows that the derivative must change sign in the given region, which is sufficient.
– AdamW
Aug 6 at 21:57
As written, it seems technically false. The function may be constant on the interval $[a,b]$ of Rolle's theorem but monotonically strictly increasing outside of that. However, if we restrict to being non-constant on $[a,b]$, the proof of Rolle's theorem on wiki shows that the derivative must change sign in the given region, which is sufficient.
– AdamW
Aug 6 at 21:57
add a comment |Â
2 Answers
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We have $f$ continuous on $[a,b]$ and differentiable on $(a,b)$ with $f(a)=f(b)$.
We assuming that $f$ is not constant hence there exists $x_0in(a,b)$ such that $f(x_0)ne f(a)$(I will assume that $f(x_0)> f(a)$, but it is all the same if it is $<$).
So we have $a<x_0<b$ but $f(a)<f(x_0)>f(b)$, i.e. $f(a)<f(x_0)$ and $f(x_0)>f(b)$.
answered Aug 6 at 22:01
Holo
4,2662629
What does $f(a)<f(x_0)>f(b)$ mean?
– Nosrati
Aug 7 at 7:10
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Without loss of generality, let's use a monotonically increasing function. It means that for any $y>x$ you have $f(y)ge f(x)$. Now choose the ends of the interval, so you get $f(b)ge f(a)$. Since we want Rolles' theorem conditions, $f(b)=f(a)$. Obviously this is true for the constant function (you can always call a constant function monotonically increasing). Now let's assume that the function is not constant, so that means that there exists at least one $c$ in $(a,b)$ where $f(c)ne f(a)$. Since we use monotonically increasing function, it means that $f(c)gt f(a)$ (strict inequality). Now apply monotonic definition between $b$ and $c$ and you get $f(b)ge f(c) gt f(a)$, or $f(b)>f(a)$. This does not obey the Rolles' theorem definition.
answered Aug 6 at 21:59
Andrei
7,5852822
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
We have $f$ continuous on $[a,b]$ and differentiable on $(a,b)$ with $f(a)=f(b)$.
We assuming that $f$ is not constant hence there exists $x_0in(a,b)$ such that $f(x_0)ne f(a)$(I will assume that $f(x_0)> f(a)$, but it is all the same if it is $<$).
So we have $a<x_0<b$ but $f(a)<f(x_0)>f(b)$, i.e. $f(a)<f(x_0)$ and $f(x_0)>f(b)$.
answered Aug 6 at 22:01
Holo
4,2662629
What does $f(a)<f(x_0)>f(b)$ mean?
– Nosrati
Aug 7 at 7:10
add a comment |Â
up vote
1
down vote
We have $f$ continuous on $[a,b]$ and differentiable on $(a,b)$ with $f(a)=f(b)$.
We assuming that $f$ is not constant hence there exists $x_0in(a,b)$ such that $f(x_0)ne f(a)$(I will assume that $f(x_0)> f(a)$, but it is all the same if it is $<$).
So we have $a<x_0<b$ but $f(a)<f(x_0)>f(b)$, i.e. $f(a)<f(x_0)$ and $f(x_0)>f(b)$.
answered Aug 6 at 22:01
Holo
4,2662629
What does $f(a)<f(x_0)>f(b)$ mean?
– Nosrati
Aug 7 at 7:10
add a comment |Â
up vote
1
down vote
up vote
1
down vote
We have $f$ continuous on $[a,b]$ and differentiable on $(a,b)$ with $f(a)=f(b)$.
We assuming that $f$ is not constant hence there exists $x_0in(a,b)$ such that $f(x_0)ne f(a)$(I will assume that $f(x_0)> f(a)$, but it is all the same if it is $<$).
So we have $a<x_0<b$ but $f(a)<f(x_0)>f(b)$, i.e. $f(a)<f(x_0)$ and $f(x_0)>f(b)$.
answered Aug 6 at 22:01
Holo
4,2662629
We have $f$ continuous on $[a,b]$ and differentiable on $(a,b)$ with $f(a)=f(b)$.
We assuming that $f$ is not constant hence there exists $x_0in(a,b)$ such that $f(x_0)ne f(a)$(I will assume that $f(x_0)> f(a)$, but it is all the same if it is $<$).
So we have $a<x_0<b$ but $f(a)<f(x_0)>f(b)$, i.e. $f(a)<f(x_0)$ and $f(x_0)>f(b)$.
answered Aug 6 at 22:01
Holo
4,2662629
edited Aug 7 at 10:54
answered Aug 6 at 22:01
Holo
4,2662629
answered Aug 6 at 22:01
Holo
4,2662629
answered Aug 6 at 22:01
Holo
4,2662629
4,2662629
What does $f(a)<f(x_0)>f(b)$ mean?
– Nosrati
Aug 7 at 7:10
add a comment |Â
What does $f(a)<f(x_0)>f(b)$ mean?
– Nosrati
Aug 7 at 7:10
What does $f(a)<f(x_0)>f(b)$ mean?
– Nosrati
Aug 7 at 7:10
What does $f(a)<f(x_0)>f(b)$ mean?
– Nosrati
Aug 7 at 7:10
add a comment |Â
up vote
0
down vote
Without loss of generality, let's use a monotonically increasing function. It means that for any $y>x$ you have $f(y)ge f(x)$. Now choose the ends of the interval, so you get $f(b)ge f(a)$. Since we want Rolles' theorem conditions, $f(b)=f(a)$. Obviously this is true for the constant function (you can always call a constant function monotonically increasing). Now let's assume that the function is not constant, so that means that there exists at least one $c$ in $(a,b)$ where $f(c)ne f(a)$. Since we use monotonically increasing function, it means that $f(c)gt f(a)$ (strict inequality). Now apply monotonic definition between $b$ and $c$ and you get $f(b)ge f(c) gt f(a)$, or $f(b)>f(a)$. This does not obey the Rolles' theorem definition.
answered Aug 6 at 21:59
Andrei
7,5852822
add a comment |Â
up vote
0
down vote
Without loss of generality, let's use a monotonically increasing function. It means that for any $y>x$ you have $f(y)ge f(x)$. Now choose the ends of the interval, so you get $f(b)ge f(a)$. Since we want Rolles' theorem conditions, $f(b)=f(a)$. Obviously this is true for the constant function (you can always call a constant function monotonically increasing). Now let's assume that the function is not constant, so that means that there exists at least one $c$ in $(a,b)$ where $f(c)ne f(a)$. Since we use monotonically increasing function, it means that $f(c)gt f(a)$ (strict inequality). Now apply monotonic definition between $b$ and $c$ and you get $f(b)ge f(c) gt f(a)$, or $f(b)>f(a)$. This does not obey the Rolles' theorem definition.
answered Aug 6 at 21:59
Andrei
7,5852822
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Without loss of generality, let's use a monotonically increasing function. It means that for any $y>x$ you have $f(y)ge f(x)$. Now choose the ends of the interval, so you get $f(b)ge f(a)$. Since we want Rolles' theorem conditions, $f(b)=f(a)$. Obviously this is true for the constant function (you can always call a constant function monotonically increasing). Now let's assume that the function is not constant, so that means that there exists at least one $c$ in $(a,b)$ where $f(c)ne f(a)$. Since we use monotonically increasing function, it means that $f(c)gt f(a)$ (strict inequality). Now apply monotonic definition between $b$ and $c$ and you get $f(b)ge f(c) gt f(a)$, or $f(b)>f(a)$. This does not obey the Rolles' theorem definition.
answered Aug 6 at 21:59
Andrei
7,5852822
Without loss of generality, let's use a monotonically increasing function. It means that for any $y>x$ you have $f(y)ge f(x)$. Now choose the ends of the interval, so you get $f(b)ge f(a)$. Since we want Rolles' theorem conditions, $f(b)=f(a)$. Obviously this is true for the constant function (you can always call a constant function monotonically increasing). Now let's assume that the function is not constant, so that means that there exists at least one $c$ in $(a,b)$ where $f(c)ne f(a)$. Since we use monotonically increasing function, it means that $f(c)gt f(a)$ (strict inequality). Now apply monotonic definition between $b$ and $c$ and you get $f(b)ge f(c) gt f(a)$, or $f(b)>f(a)$. This does not obey the Rolles' theorem definition.
answered Aug 6 at 21:59
Andrei
7,5852822
answered Aug 6 at 21:59
Andrei
7,5852822
answered Aug 6 at 21:59
Andrei
7,5852822
answered Aug 6 at 21:59
Andrei
7,5852822
7,5852822
add a comment |Â
add a comment |Â
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3
Well $f(a) = f(b)$. I'm reminded of the phrase "What goes up must come down".
– Rhys Steele
Aug 6 at 21:49
1
As written, it seems technically false. The function may be constant on the interval $[a,b]$ of Rolle's theorem but monotonically strictly increasing outside of that. However, if we restrict to being non-constant on $[a,b]$, the proof of Rolle's theorem on wiki shows that the derivative must change sign in the given region, which is sufficient.
– AdamW
Aug 6 at 21:57