Mixing
Prove that there exists at least one $x_0inmathbb{R}$, such that $f(x_0)+f''(x_0)=0$
Let $f:mathbb{R}tomathbb{R}$ be a function, two times
differentiable with $left|f(x)right|leq1,forall xinmathbb{R}$
and $f^2(0)+left(f'(0)right)^2=4$. Prove that there exists at least
one $x_0inmathbb{R}$, such that $f(x_0)+f''(x_0)=0$.
I have tried the usual Rolle method by $e^x$, but didn't go far... Any help available?
calculus functions inequality
asked Mar 6 '16 at 15:23
Jason
1,3091025
add a comment |
Let $f:mathbb{R}tomathbb{R}$ be a function, two times
differentiable with $left|f(x)right|leq1,forall xinmathbb{R}$
and $f^2(0)+left(f'(0)right)^2=4$. Prove that there exists at least
one $x_0inmathbb{R}$, such that $f(x_0)+f''(x_0)=0$.
I have tried the usual Rolle method by $e^x$, but didn't go far... Any help available?
calculus functions inequality
asked Mar 6 '16 at 15:23
Jason
1,3091025
add a comment |
Let $f:mathbb{R}tomathbb{R}$ be a function, two times
differentiable with $left|f(x)right|leq1,forall xinmathbb{R}$
and $f^2(0)+left(f'(0)right)^2=4$. Prove that there exists at least
one $x_0inmathbb{R}$, such that $f(x_0)+f''(x_0)=0$.
I have tried the usual Rolle method by $e^x$, but didn't go far... Any help available?
calculus functions inequality
asked Mar 6 '16 at 15:23
Jason
1,3091025
Let $f:mathbb{R}tomathbb{R}$ be a function, two times
differentiable with $left|f(x)right|leq1,forall xinmathbb{R}$
and $f^2(0)+left(f'(0)right)^2=4$. Prove that there exists at least
one $x_0inmathbb{R}$, such that $f(x_0)+f''(x_0)=0$.
I have tried the usual Rolle method by $e^x$, but didn't go far... Any help available?
calculus functions inequality
calculus functions inequality
asked Mar 6 '16 at 15:23
Jason
1,3091025
asked Mar 6 '16 at 15:23
Jason
1,3091025
asked Mar 6 '16 at 15:23
Jason
1,3091025
asked Mar 6 '16 at 15:23
Jason
1,3091025
asked Mar 6 '16 at 15:23
Jason
1,3091025
1,3091025
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Let's define $g(x) = f(x)^2 + f'(x)^2$. We have:
- $g'(x) = 2f'(x)(f(x)+f''(x))$
- $g(0) = 4$
We will find a local maximum of $g$, noted $x_0$, which verifies $g(x_0) ge 4$. We shall distinguish two options:
$g(0) = 4 ge g(x) forall x in mathbb{R}$, in which case $0$ is the local maximum that we are trying to find.
There exists $y in mathbb{R} setminus {0}$ with $g(y) > 4$. Let's supose that $y > 0$ (otherwise apply the following reasoning to $-g$) and try to find a number $x' > y$ which verifies $g(x') = 4$. If there isn't such a number, then we would have $g(x) > 4$ for every $x > y$. Thus, $left| f'(x) right| > sqrt 3$ for every $x > y$. Using the mean value theorem on $x > max{2,y}$ we get a contradiction:
$$ 2 ge left| f(x) - f(0)right| = left| f'(xi_x) x right| > 2 sqrt 3 > 2 $$
As a consequence, we can take $x' > y$ with $g(x') = 4$. The function $g$ has a global maximum in $[0, x']$ because it is continuous. Let $x_0$ be that global maximum. Then, $g(x_0) ge g(y) > 4$ and, consequently, $x_0 ne 0, x'$. Hence, $x_0$ is a local maximum of $g$ with $g(x_0) ge 4$.
Finally, since $x_0$ is a local extrema we have:
$$ 0 = g'(x_0) = 2f'(x_0)(f(x_0)+f''(x_0)) $$
Furthermore, $f'(x_0)^2 ge 4 - f(x_0)^2 ge 3 > 0$ and, thus, $f(x_0)+f''(x_0) = 0$.
answered Mar 6 '16 at 16:52
andreshp
15019
2
Can you clarify your arguments for the contradiction? Specifically, why must we find such a number $x_n$ and why if $f'$ is not bounded, then $f$ is also not bounded?
– Jason
Mar 6 '16 at 17:02
1
You are assuming that if $g$ has no global maximum then it is unbounded. But $g$ may be bounded and NOT have a global maximum. For example, $g(x)=tan ^{-1} x$
– Matematleta
Mar 6 '16 at 17:21
@Chilango Do you have any proof for my exercise? Because nothing has been done yet...
– Jason
Mar 6 '16 at 18:13
Thanks for the observation. I repaired the solution. Note that the function $f(x) = sin(2e^x -1)$ verifies the hypothesis and the derivative is not bounded, that was a mistake.
– andreshp
Mar 6 '16 at 19:18
1
You can't use Rolle's theorem there because you need $g(x_0) ge 4$ and in order to affirm that you should $x_0$ as the global maximum of $[0,x']$.
– andreshp
Mar 6 '16 at 21:53
|
show 3 more comments
Try to explore the extrema of
$$
g(x)=f(x)^2+f'(x)^2implies g'(x)=2f'(x)(f(x)+f''(x)).
$$
answered Mar 6 '16 at 15:48
LutzL
56.1k42054
What about another, more specific, hint?
– Jason
Mar 6 '16 at 16:30
add a comment |
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2 Answers
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2 Answers
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Let's define $g(x) = f(x)^2 + f'(x)^2$. We have:
- $g'(x) = 2f'(x)(f(x)+f''(x))$
- $g(0) = 4$
We will find a local maximum of $g$, noted $x_0$, which verifies $g(x_0) ge 4$. We shall distinguish two options:
$g(0) = 4 ge g(x) forall x in mathbb{R}$, in which case $0$ is the local maximum that we are trying to find.
There exists $y in mathbb{R} setminus {0}$ with $g(y) > 4$. Let's supose that $y > 0$ (otherwise apply the following reasoning to $-g$) and try to find a number $x' > y$ which verifies $g(x') = 4$. If there isn't such a number, then we would have $g(x) > 4$ for every $x > y$. Thus, $left| f'(x) right| > sqrt 3$ for every $x > y$. Using the mean value theorem on $x > max{2,y}$ we get a contradiction:
$$ 2 ge left| f(x) - f(0)right| = left| f'(xi_x) x right| > 2 sqrt 3 > 2 $$
As a consequence, we can take $x' > y$ with $g(x') = 4$. The function $g$ has a global maximum in $[0, x']$ because it is continuous. Let $x_0$ be that global maximum. Then, $g(x_0) ge g(y) > 4$ and, consequently, $x_0 ne 0, x'$. Hence, $x_0$ is a local maximum of $g$ with $g(x_0) ge 4$.
Finally, since $x_0$ is a local extrema we have:
$$ 0 = g'(x_0) = 2f'(x_0)(f(x_0)+f''(x_0)) $$
Furthermore, $f'(x_0)^2 ge 4 - f(x_0)^2 ge 3 > 0$ and, thus, $f(x_0)+f''(x_0) = 0$.
answered Mar 6 '16 at 16:52
andreshp
15019
2
Can you clarify your arguments for the contradiction? Specifically, why must we find such a number $x_n$ and why if $f'$ is not bounded, then $f$ is also not bounded?
– Jason
Mar 6 '16 at 17:02
1
You are assuming that if $g$ has no global maximum then it is unbounded. But $g$ may be bounded and NOT have a global maximum. For example, $g(x)=tan ^{-1} x$
– Matematleta
Mar 6 '16 at 17:21
@Chilango Do you have any proof for my exercise? Because nothing has been done yet...
– Jason
Mar 6 '16 at 18:13
Thanks for the observation. I repaired the solution. Note that the function $f(x) = sin(2e^x -1)$ verifies the hypothesis and the derivative is not bounded, that was a mistake.
– andreshp
Mar 6 '16 at 19:18
1
You can't use Rolle's theorem there because you need $g(x_0) ge 4$ and in order to affirm that you should $x_0$ as the global maximum of $[0,x']$.
– andreshp
Mar 6 '16 at 21:53
|
show 3 more comments
Let's define $g(x) = f(x)^2 + f'(x)^2$. We have:
- $g'(x) = 2f'(x)(f(x)+f''(x))$
- $g(0) = 4$
We will find a local maximum of $g$, noted $x_0$, which verifies $g(x_0) ge 4$. We shall distinguish two options:
$g(0) = 4 ge g(x) forall x in mathbb{R}$, in which case $0$ is the local maximum that we are trying to find.
There exists $y in mathbb{R} setminus {0}$ with $g(y) > 4$. Let's supose that $y > 0$ (otherwise apply the following reasoning to $-g$) and try to find a number $x' > y$ which verifies $g(x') = 4$. If there isn't such a number, then we would have $g(x) > 4$ for every $x > y$. Thus, $left| f'(x) right| > sqrt 3$ for every $x > y$. Using the mean value theorem on $x > max{2,y}$ we get a contradiction:
$$ 2 ge left| f(x) - f(0)right| = left| f'(xi_x) x right| > 2 sqrt 3 > 2 $$
As a consequence, we can take $x' > y$ with $g(x') = 4$. The function $g$ has a global maximum in $[0, x']$ because it is continuous. Let $x_0$ be that global maximum. Then, $g(x_0) ge g(y) > 4$ and, consequently, $x_0 ne 0, x'$. Hence, $x_0$ is a local maximum of $g$ with $g(x_0) ge 4$.
Finally, since $x_0$ is a local extrema we have:
$$ 0 = g'(x_0) = 2f'(x_0)(f(x_0)+f''(x_0)) $$
Furthermore, $f'(x_0)^2 ge 4 - f(x_0)^2 ge 3 > 0$ and, thus, $f(x_0)+f''(x_0) = 0$.
answered Mar 6 '16 at 16:52
andreshp
15019
2
Can you clarify your arguments for the contradiction? Specifically, why must we find such a number $x_n$ and why if $f'$ is not bounded, then $f$ is also not bounded?
– Jason
Mar 6 '16 at 17:02
1
You are assuming that if $g$ has no global maximum then it is unbounded. But $g$ may be bounded and NOT have a global maximum. For example, $g(x)=tan ^{-1} x$
– Matematleta
Mar 6 '16 at 17:21
@Chilango Do you have any proof for my exercise? Because nothing has been done yet...
– Jason
Mar 6 '16 at 18:13
Thanks for the observation. I repaired the solution. Note that the function $f(x) = sin(2e^x -1)$ verifies the hypothesis and the derivative is not bounded, that was a mistake.
– andreshp
Mar 6 '16 at 19:18
1
You can't use Rolle's theorem there because you need $g(x_0) ge 4$ and in order to affirm that you should $x_0$ as the global maximum of $[0,x']$.
– andreshp
Mar 6 '16 at 21:53
|
show 3 more comments
Let's define $g(x) = f(x)^2 + f'(x)^2$. We have:
- $g'(x) = 2f'(x)(f(x)+f''(x))$
- $g(0) = 4$
We will find a local maximum of $g$, noted $x_0$, which verifies $g(x_0) ge 4$. We shall distinguish two options:
$g(0) = 4 ge g(x) forall x in mathbb{R}$, in which case $0$ is the local maximum that we are trying to find.
There exists $y in mathbb{R} setminus {0}$ with $g(y) > 4$. Let's supose that $y > 0$ (otherwise apply the following reasoning to $-g$) and try to find a number $x' > y$ which verifies $g(x') = 4$. If there isn't such a number, then we would have $g(x) > 4$ for every $x > y$. Thus, $left| f'(x) right| > sqrt 3$ for every $x > y$. Using the mean value theorem on $x > max{2,y}$ we get a contradiction:
$$ 2 ge left| f(x) - f(0)right| = left| f'(xi_x) x right| > 2 sqrt 3 > 2 $$
As a consequence, we can take $x' > y$ with $g(x') = 4$. The function $g$ has a global maximum in $[0, x']$ because it is continuous. Let $x_0$ be that global maximum. Then, $g(x_0) ge g(y) > 4$ and, consequently, $x_0 ne 0, x'$. Hence, $x_0$ is a local maximum of $g$ with $g(x_0) ge 4$.
Finally, since $x_0$ is a local extrema we have:
$$ 0 = g'(x_0) = 2f'(x_0)(f(x_0)+f''(x_0)) $$
Furthermore, $f'(x_0)^2 ge 4 - f(x_0)^2 ge 3 > 0$ and, thus, $f(x_0)+f''(x_0) = 0$.
answered Mar 6 '16 at 16:52
andreshp
15019
Let's define $g(x) = f(x)^2 + f'(x)^2$. We have:
- $g'(x) = 2f'(x)(f(x)+f''(x))$
- $g(0) = 4$
We will find a local maximum of $g$, noted $x_0$, which verifies $g(x_0) ge 4$. We shall distinguish two options:
$g(0) = 4 ge g(x) forall x in mathbb{R}$, in which case $0$ is the local maximum that we are trying to find.
There exists $y in mathbb{R} setminus {0}$ with $g(y) > 4$. Let's supose that $y > 0$ (otherwise apply the following reasoning to $-g$) and try to find a number $x' > y$ which verifies $g(x') = 4$. If there isn't such a number, then we would have $g(x) > 4$ for every $x > y$. Thus, $left| f'(x) right| > sqrt 3$ for every $x > y$. Using the mean value theorem on $x > max{2,y}$ we get a contradiction:
$$ 2 ge left| f(x) - f(0)right| = left| f'(xi_x) x right| > 2 sqrt 3 > 2 $$
As a consequence, we can take $x' > y$ with $g(x') = 4$. The function $g$ has a global maximum in $[0, x']$ because it is continuous. Let $x_0$ be that global maximum. Then, $g(x_0) ge g(y) > 4$ and, consequently, $x_0 ne 0, x'$. Hence, $x_0$ is a local maximum of $g$ with $g(x_0) ge 4$.
Finally, since $x_0$ is a local extrema we have:
$$ 0 = g'(x_0) = 2f'(x_0)(f(x_0)+f''(x_0)) $$
Furthermore, $f'(x_0)^2 ge 4 - f(x_0)^2 ge 3 > 0$ and, thus, $f(x_0)+f''(x_0) = 0$.
answered Mar 6 '16 at 16:52
andreshp
15019
edited Mar 6 '16 at 19:23
answered Mar 6 '16 at 16:52
andreshp
15019
answered Mar 6 '16 at 16:52
andreshp
15019
answered Mar 6 '16 at 16:52
andreshp
15019
15019
2
Can you clarify your arguments for the contradiction? Specifically, why must we find such a number $x_n$ and why if $f'$ is not bounded, then $f$ is also not bounded?
– Jason
Mar 6 '16 at 17:02
1
You are assuming that if $g$ has no global maximum then it is unbounded. But $g$ may be bounded and NOT have a global maximum. For example, $g(x)=tan ^{-1} x$
– Matematleta
Mar 6 '16 at 17:21
@Chilango Do you have any proof for my exercise? Because nothing has been done yet...
– Jason
Mar 6 '16 at 18:13
Thanks for the observation. I repaired the solution. Note that the function $f(x) = sin(2e^x -1)$ verifies the hypothesis and the derivative is not bounded, that was a mistake.
– andreshp
Mar 6 '16 at 19:18
1
You can't use Rolle's theorem there because you need $g(x_0) ge 4$ and in order to affirm that you should $x_0$ as the global maximum of $[0,x']$.
– andreshp
Mar 6 '16 at 21:53
|
show 3 more comments
2
Can you clarify your arguments for the contradiction? Specifically, why must we find such a number $x_n$ and why if $f'$ is not bounded, then $f$ is also not bounded?
– Jason
Mar 6 '16 at 17:02
1
You are assuming that if $g$ has no global maximum then it is unbounded. But $g$ may be bounded and NOT have a global maximum. For example, $g(x)=tan ^{-1} x$
– Matematleta
Mar 6 '16 at 17:21
@Chilango Do you have any proof for my exercise? Because nothing has been done yet...
– Jason
Mar 6 '16 at 18:13
Thanks for the observation. I repaired the solution. Note that the function $f(x) = sin(2e^x -1)$ verifies the hypothesis and the derivative is not bounded, that was a mistake.
– andreshp
Mar 6 '16 at 19:18
1
You can't use Rolle's theorem there because you need $g(x_0) ge 4$ and in order to affirm that you should $x_0$ as the global maximum of $[0,x']$.
– andreshp
Mar 6 '16 at 21:53
2
2
Can you clarify your arguments for the contradiction? Specifically, why must we find such a number $x_n$ and why if $f'$ is not bounded, then $f$ is also not bounded?
– Jason
Mar 6 '16 at 17:02
Can you clarify your arguments for the contradiction? Specifically, why must we find such a number $x_n$ and why if $f'$ is not bounded, then $f$ is also not bounded?
– Jason
Mar 6 '16 at 17:02
1
1
You are assuming that if $g$ has no global maximum then it is unbounded. But $g$ may be bounded and NOT have a global maximum. For example, $g(x)=tan ^{-1} x$
– Matematleta
Mar 6 '16 at 17:21
You are assuming that if $g$ has no global maximum then it is unbounded. But $g$ may be bounded and NOT have a global maximum. For example, $g(x)=tan ^{-1} x$
– Matematleta
Mar 6 '16 at 17:21
@Chilango Do you have any proof for my exercise? Because nothing has been done yet...
– Jason
Mar 6 '16 at 18:13
@Chilango Do you have any proof for my exercise? Because nothing has been done yet...
– Jason
Mar 6 '16 at 18:13
Thanks for the observation. I repaired the solution. Note that the function $f(x) = sin(2e^x -1)$ verifies the hypothesis and the derivative is not bounded, that was a mistake.
– andreshp
Mar 6 '16 at 19:18
Thanks for the observation. I repaired the solution. Note that the function $f(x) = sin(2e^x -1)$ verifies the hypothesis and the derivative is not bounded, that was a mistake.
– andreshp
Mar 6 '16 at 19:18
1
1
You can't use Rolle's theorem there because you need $g(x_0) ge 4$ and in order to affirm that you should $x_0$ as the global maximum of $[0,x']$.
– andreshp
Mar 6 '16 at 21:53
You can't use Rolle's theorem there because you need $g(x_0) ge 4$ and in order to affirm that you should $x_0$ as the global maximum of $[0,x']$.
– andreshp
Mar 6 '16 at 21:53
|
show 3 more comments
Try to explore the extrema of
$$
g(x)=f(x)^2+f'(x)^2implies g'(x)=2f'(x)(f(x)+f''(x)).
$$
answered Mar 6 '16 at 15:48
LutzL
56.1k42054
What about another, more specific, hint?
– Jason
Mar 6 '16 at 16:30
add a comment |
Try to explore the extrema of
$$
g(x)=f(x)^2+f'(x)^2implies g'(x)=2f'(x)(f(x)+f''(x)).
$$
answered Mar 6 '16 at 15:48
LutzL
56.1k42054
What about another, more specific, hint?
– Jason
Mar 6 '16 at 16:30
add a comment |
Try to explore the extrema of
$$
g(x)=f(x)^2+f'(x)^2implies g'(x)=2f'(x)(f(x)+f''(x)).
$$
answered Mar 6 '16 at 15:48
LutzL
56.1k42054
Try to explore the extrema of
$$
g(x)=f(x)^2+f'(x)^2implies g'(x)=2f'(x)(f(x)+f''(x)).
$$
answered Mar 6 '16 at 15:48
LutzL
56.1k42054
answered Mar 6 '16 at 15:48
LutzL
56.1k42054
answered Mar 6 '16 at 15:48
LutzL
56.1k42054
answered Mar 6 '16 at 15:48
LutzL
56.1k42054
56.1k42054
What about another, more specific, hint?
– Jason
Mar 6 '16 at 16:30
add a comment |
What about another, more specific, hint?
– Jason
Mar 6 '16 at 16:30
What about another, more specific, hint?
– Jason
Mar 6 '16 at 16:30
What about another, more specific, hint?
– Jason
Mar 6 '16 at 16:30
add a comment |
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