How to show $a$ and $b$ are different??
$begingroup$
We have a continuously differenciable function $f:[0,1]rightarrow[0,1]$ such that $f(0)=0, f(1)=1$. We need to show that there exists different real numbers $a,bin(0,1)$ such that $f'(a)f'(b)=1.$
I used the mean value theorem as follows: There exists $alphain(0,1)$ such that $(fcirc f)'(alpha)=1$, that is to say, $f'(f(alpha))f'(alpha)=1$. Let $a=f(alpha)$ and $b=alpha$. Then, $f'(a)f'(b)=1$. But I didn't use the continuity of the derivative, and a don't know how to show that $aneq b$.
Thanks, any help will be appreciated.
calculus analysis
asked Nov 26 '18 at 21:58
DIEGO R.DIEGO R.
279114
$endgroup$
add a comment |
$begingroup$
We have a continuously differenciable function $f:[0,1]rightarrow[0,1]$ such that $f(0)=0, f(1)=1$. We need to show that there exists different real numbers $a,bin(0,1)$ such that $f'(a)f'(b)=1.$
I used the mean value theorem as follows: There exists $alphain(0,1)$ such that $(fcirc f)'(alpha)=1$, that is to say, $f'(f(alpha))f'(alpha)=1$. Let $a=f(alpha)$ and $b=alpha$. Then, $f'(a)f'(b)=1$. But I didn't use the continuity of the derivative, and a don't know how to show that $aneq b$.
Thanks, any help will be appreciated.
calculus analysis
asked Nov 26 '18 at 21:58
DIEGO R.DIEGO R.
279114
$endgroup$
$begingroup$
I think you might need to change your approach slightly, since if $f(x) = x,$ then $a=b$ no matter what $alpha$ is in your construction.
$endgroup$
– MisterRiemann
Nov 26 '18 at 22:03
add a comment |
$begingroup$
We have a continuously differenciable function $f:[0,1]rightarrow[0,1]$ such that $f(0)=0, f(1)=1$. We need to show that there exists different real numbers $a,bin(0,1)$ such that $f'(a)f'(b)=1.$
I used the mean value theorem as follows: There exists $alphain(0,1)$ such that $(fcirc f)'(alpha)=1$, that is to say, $f'(f(alpha))f'(alpha)=1$. Let $a=f(alpha)$ and $b=alpha$. Then, $f'(a)f'(b)=1$. But I didn't use the continuity of the derivative, and a don't know how to show that $aneq b$.
Thanks, any help will be appreciated.
calculus analysis
asked Nov 26 '18 at 21:58
DIEGO R.DIEGO R.
279114
$endgroup$
We have a continuously differenciable function $f:[0,1]rightarrow[0,1]$ such that $f(0)=0, f(1)=1$. We need to show that there exists different real numbers $a,bin(0,1)$ such that $f'(a)f'(b)=1.$
I used the mean value theorem as follows: There exists $alphain(0,1)$ such that $(fcirc f)'(alpha)=1$, that is to say, $f'(f(alpha))f'(alpha)=1$. Let $a=f(alpha)$ and $b=alpha$. Then, $f'(a)f'(b)=1$. But I didn't use the continuity of the derivative, and a don't know how to show that $aneq b$.
Thanks, any help will be appreciated.
calculus analysis
calculus analysis
asked Nov 26 '18 at 21:58
DIEGO R.DIEGO R.
279114
asked Nov 26 '18 at 21:58
DIEGO R.DIEGO R.
279114
edited Nov 26 '18 at 22:30
DIEGO R.
asked Nov 26 '18 at 21:58
DIEGO R.DIEGO R.
279114
asked Nov 26 '18 at 21:58
DIEGO R.DIEGO R.
279114
asked Nov 26 '18 at 21:58
DIEGO R.DIEGO R.
279114
279114
$begingroup$
I think you might need to change your approach slightly, since if $f(x) = x,$ then $a=b$ no matter what $alpha$ is in your construction.
$endgroup$
– MisterRiemann
Nov 26 '18 at 22:03
add a comment |
$begingroup$
I think you might need to change your approach slightly, since if $f(x) = x,$ then $a=b$ no matter what $alpha$ is in your construction.
$endgroup$
– MisterRiemann
Nov 26 '18 at 22:03
$begingroup$
I think you might need to change your approach slightly, since if $f(x) = x,$ then $a=b$ no matter what $alpha$ is in your construction.
$endgroup$
– MisterRiemann
Nov 26 '18 at 22:03
$begingroup$
I think you might need to change your approach slightly, since if $f(x) = x,$ then $a=b$ no matter what $alpha$ is in your construction.
$endgroup$
– MisterRiemann
Nov 26 '18 at 22:03
add a comment |
1 Answer
1
active
oldest
votes
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All we need is to find a point where the graph crosses the diagonal line $x+y=1.$ Then we will apply MVT on both sides of that point.
Take
$$ g(x) = x + f(x) - 1. $$
We get $$ g(0) = -1, ; ; ; g(1) = 1. $$
By the Intermediate Value Theorem, for some $0 < t < 1$ we get $g(t) = 0,$ meaning
$ t + f(t) - 1 = 0 $ or
$$ f(t) = 1-t $$
Note that $1-t neq 0,$ indeed $0 < 1-t < 1.$
By the mean value theorem, there is a value $a$ between $0$ and $t$ so that $$f'(a) = frac{1-t}{t}$$
There is also a value $b$ between $t$ and $1$ such that
$$f'(b) = frac{1-(1-t)}{1-t} =frac{t}{1-t} $$
answered Nov 26 '18 at 22:18
Will JagyWill Jagy
102k5101200
$endgroup$
$begingroup$
Well, but do not we need the hypotesis about continuity of the derivative?
$endgroup$
– DIEGO R.
Nov 26 '18 at 22:23
$begingroup$
@DIEGOR. if your question had said "continuously differentiable" I would probably have tried to use that. However, you wrote "continuous differenciable" and I was not sure what was going on. I decided it meant "continuous and differentiable" Anyway, you just need enough to call on the Mean Value Theorem.
$endgroup$
– Will Jagy
Nov 26 '18 at 22:27
$begingroup$
@WillJagy: I wondered about that as well, but since differentiable implies continuous, I figured the OP was not being redundant and hence assumed "continuously differentiable" was intended.
$endgroup$
– Robert Lewis
Nov 26 '18 at 22:32
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Yes I have edited it. I think that we do not need the continuity of derivative according with this proof. Thanks
$endgroup$
– DIEGO R.
Nov 26 '18 at 22:32
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
All we need is to find a point where the graph crosses the diagonal line $x+y=1.$ Then we will apply MVT on both sides of that point.
Take
$$ g(x) = x + f(x) - 1. $$
We get $$ g(0) = -1, ; ; ; g(1) = 1. $$
By the Intermediate Value Theorem, for some $0 < t < 1$ we get $g(t) = 0,$ meaning
$ t + f(t) - 1 = 0 $ or
$$ f(t) = 1-t $$
Note that $1-t neq 0,$ indeed $0 < 1-t < 1.$
By the mean value theorem, there is a value $a$ between $0$ and $t$ so that $$f'(a) = frac{1-t}{t}$$
There is also a value $b$ between $t$ and $1$ such that
$$f'(b) = frac{1-(1-t)}{1-t} =frac{t}{1-t} $$
answered Nov 26 '18 at 22:18
Will JagyWill Jagy
102k5101200
$endgroup$
$begingroup$
Well, but do not we need the hypotesis about continuity of the derivative?
$endgroup$
– DIEGO R.
Nov 26 '18 at 22:23
$begingroup$
@DIEGOR. if your question had said "continuously differentiable" I would probably have tried to use that. However, you wrote "continuous differenciable" and I was not sure what was going on. I decided it meant "continuous and differentiable" Anyway, you just need enough to call on the Mean Value Theorem.
$endgroup$
– Will Jagy
Nov 26 '18 at 22:27
$begingroup$
@WillJagy: I wondered about that as well, but since differentiable implies continuous, I figured the OP was not being redundant and hence assumed "continuously differentiable" was intended.
$endgroup$
– Robert Lewis
Nov 26 '18 at 22:32
$begingroup$
Yes I have edited it. I think that we do not need the continuity of derivative according with this proof. Thanks
$endgroup$
– DIEGO R.
Nov 26 '18 at 22:32
add a comment |
$begingroup$
All we need is to find a point where the graph crosses the diagonal line $x+y=1.$ Then we will apply MVT on both sides of that point.
Take
$$ g(x) = x + f(x) - 1. $$
We get $$ g(0) = -1, ; ; ; g(1) = 1. $$
By the Intermediate Value Theorem, for some $0 < t < 1$ we get $g(t) = 0,$ meaning
$ t + f(t) - 1 = 0 $ or
$$ f(t) = 1-t $$
Note that $1-t neq 0,$ indeed $0 < 1-t < 1.$
By the mean value theorem, there is a value $a$ between $0$ and $t$ so that $$f'(a) = frac{1-t}{t}$$
There is also a value $b$ between $t$ and $1$ such that
$$f'(b) = frac{1-(1-t)}{1-t} =frac{t}{1-t} $$
answered Nov 26 '18 at 22:18
Will JagyWill Jagy
102k5101200
$endgroup$
$begingroup$
Well, but do not we need the hypotesis about continuity of the derivative?
$endgroup$
– DIEGO R.
Nov 26 '18 at 22:23
$begingroup$
@DIEGOR. if your question had said "continuously differentiable" I would probably have tried to use that. However, you wrote "continuous differenciable" and I was not sure what was going on. I decided it meant "continuous and differentiable" Anyway, you just need enough to call on the Mean Value Theorem.
$endgroup$
– Will Jagy
Nov 26 '18 at 22:27
$begingroup$
@WillJagy: I wondered about that as well, but since differentiable implies continuous, I figured the OP was not being redundant and hence assumed "continuously differentiable" was intended.
$endgroup$
– Robert Lewis
Nov 26 '18 at 22:32
$begingroup$
Yes I have edited it. I think that we do not need the continuity of derivative according with this proof. Thanks
$endgroup$
– DIEGO R.
Nov 26 '18 at 22:32
add a comment |
$begingroup$
All we need is to find a point where the graph crosses the diagonal line $x+y=1.$ Then we will apply MVT on both sides of that point.
Take
$$ g(x) = x + f(x) - 1. $$
We get $$ g(0) = -1, ; ; ; g(1) = 1. $$
By the Intermediate Value Theorem, for some $0 < t < 1$ we get $g(t) = 0,$ meaning
$ t + f(t) - 1 = 0 $ or
$$ f(t) = 1-t $$
Note that $1-t neq 0,$ indeed $0 < 1-t < 1.$
By the mean value theorem, there is a value $a$ between $0$ and $t$ so that $$f'(a) = frac{1-t}{t}$$
There is also a value $b$ between $t$ and $1$ such that
$$f'(b) = frac{1-(1-t)}{1-t} =frac{t}{1-t} $$
answered Nov 26 '18 at 22:18
Will JagyWill Jagy
102k5101200
$endgroup$
All we need is to find a point where the graph crosses the diagonal line $x+y=1.$ Then we will apply MVT on both sides of that point.
Take
$$ g(x) = x + f(x) - 1. $$
We get $$ g(0) = -1, ; ; ; g(1) = 1. $$
By the Intermediate Value Theorem, for some $0 < t < 1$ we get $g(t) = 0,$ meaning
$ t + f(t) - 1 = 0 $ or
$$ f(t) = 1-t $$
Note that $1-t neq 0,$ indeed $0 < 1-t < 1.$
By the mean value theorem, there is a value $a$ between $0$ and $t$ so that $$f'(a) = frac{1-t}{t}$$
There is also a value $b$ between $t$ and $1$ such that
$$f'(b) = frac{1-(1-t)}{1-t} =frac{t}{1-t} $$
answered Nov 26 '18 at 22:18
Will JagyWill Jagy
102k5101200
edited Nov 26 '18 at 22:55
answered Nov 26 '18 at 22:18
Will JagyWill Jagy
102k5101200
answered Nov 26 '18 at 22:18
Will JagyWill Jagy
102k5101200
answered Nov 26 '18 at 22:18
Will JagyWill Jagy
102k5101200
102k5101200
$begingroup$
Well, but do not we need the hypotesis about continuity of the derivative?
$endgroup$
– DIEGO R.
Nov 26 '18 at 22:23
$begingroup$
@DIEGOR. if your question had said "continuously differentiable" I would probably have tried to use that. However, you wrote "continuous differenciable" and I was not sure what was going on. I decided it meant "continuous and differentiable" Anyway, you just need enough to call on the Mean Value Theorem.
$endgroup$
– Will Jagy
Nov 26 '18 at 22:27
$begingroup$
@WillJagy: I wondered about that as well, but since differentiable implies continuous, I figured the OP was not being redundant and hence assumed "continuously differentiable" was intended.
$endgroup$
– Robert Lewis
Nov 26 '18 at 22:32
$begingroup$
Yes I have edited it. I think that we do not need the continuity of derivative according with this proof. Thanks
$endgroup$
– DIEGO R.
Nov 26 '18 at 22:32
add a comment |
$begingroup$
Well, but do not we need the hypotesis about continuity of the derivative?
$endgroup$
– DIEGO R.
Nov 26 '18 at 22:23
$begingroup$
@DIEGOR. if your question had said "continuously differentiable" I would probably have tried to use that. However, you wrote "continuous differenciable" and I was not sure what was going on. I decided it meant "continuous and differentiable" Anyway, you just need enough to call on the Mean Value Theorem.
$endgroup$
– Will Jagy
Nov 26 '18 at 22:27
$begingroup$
@WillJagy: I wondered about that as well, but since differentiable implies continuous, I figured the OP was not being redundant and hence assumed "continuously differentiable" was intended.
$endgroup$
– Robert Lewis
Nov 26 '18 at 22:32
$begingroup$
Yes I have edited it. I think that we do not need the continuity of derivative according with this proof. Thanks
$endgroup$
– DIEGO R.
Nov 26 '18 at 22:32
$begingroup$
Well, but do not we need the hypotesis about continuity of the derivative?
$endgroup$
– DIEGO R.
Nov 26 '18 at 22:23
$begingroup$
Well, but do not we need the hypotesis about continuity of the derivative?
$endgroup$
– DIEGO R.
Nov 26 '18 at 22:23
$begingroup$
@DIEGOR. if your question had said "continuously differentiable" I would probably have tried to use that. However, you wrote "continuous differenciable" and I was not sure what was going on. I decided it meant "continuous and differentiable" Anyway, you just need enough to call on the Mean Value Theorem.
$endgroup$
– Will Jagy
Nov 26 '18 at 22:27
$begingroup$
@DIEGOR. if your question had said "continuously differentiable" I would probably have tried to use that. However, you wrote "continuous differenciable" and I was not sure what was going on. I decided it meant "continuous and differentiable" Anyway, you just need enough to call on the Mean Value Theorem.
$endgroup$
– Will Jagy
Nov 26 '18 at 22:27
$begingroup$
@WillJagy: I wondered about that as well, but since differentiable implies continuous, I figured the OP was not being redundant and hence assumed "continuously differentiable" was intended.
$endgroup$
– Robert Lewis
Nov 26 '18 at 22:32
$begingroup$
@WillJagy: I wondered about that as well, but since differentiable implies continuous, I figured the OP was not being redundant and hence assumed "continuously differentiable" was intended.
$endgroup$
– Robert Lewis
Nov 26 '18 at 22:32
$begingroup$
Yes I have edited it. I think that we do not need the continuity of derivative according with this proof. Thanks
$endgroup$
– DIEGO R.
Nov 26 '18 at 22:32
$begingroup$
Yes I have edited it. I think that we do not need the continuity of derivative according with this proof. Thanks
$endgroup$
– DIEGO R.
Nov 26 '18 at 22:32
add a comment |
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$begingroup$
I think you might need to change your approach slightly, since if $f(x) = x,$ then $a=b$ no matter what $alpha$ is in your construction.
$endgroup$
– MisterRiemann
Nov 26 '18 at 22:03