Prove that there exists at least one $x_0inmathbb{R}$, such that $f(x_0)+f''(x_0)=0$












2















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?










share|cite|improve this question



























    2















    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?










    share|cite|improve this question

























      2












      2








      2


      5






      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?










      share|cite|improve this question














      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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Mar 6 '16 at 15:23









      Jason

      1,3091025




      1,3091025






















          2 Answers
          2






          active

          oldest

          votes


















          2














          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, $f(x_0)+f''(x_0) = 0$.






          share|cite|improve this answer



















          • 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





















          0














          Try to explore the extrema of
          $$
          g(x)=f(x)^2+f'(x)^2implies g'(x)=2f'(x)(f(x)+f''(x)).
          $$






          share|cite|improve this answer





















          • What about another, more specific, hint?
            – Jason
            Mar 6 '16 at 16:30











          Your Answer





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          2 Answers
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          2 Answers
          2






          active

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          active

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          active

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          2














          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, $f(x_0)+f''(x_0) = 0$.






          share|cite|improve this answer



















          • 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














          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, $f(x_0)+f''(x_0) = 0$.






          share|cite|improve this answer



















          • 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








          2






          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, $f(x_0)+f''(x_0) = 0$.






          share|cite|improve this answer














          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, $f(x_0)+f''(x_0) = 0$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 6 '16 at 19:23

























          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
















          • 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













          0














          Try to explore the extrema of
          $$
          g(x)=f(x)^2+f'(x)^2implies g'(x)=2f'(x)(f(x)+f''(x)).
          $$






          share|cite|improve this answer





















          • What about another, more specific, hint?
            – Jason
            Mar 6 '16 at 16:30
















          0














          Try to explore the extrema of
          $$
          g(x)=f(x)^2+f'(x)^2implies g'(x)=2f'(x)(f(x)+f''(x)).
          $$






          share|cite|improve this answer





















          • What about another, more specific, hint?
            – Jason
            Mar 6 '16 at 16:30














          0












          0








          0






          Try to explore the extrema of
          $$
          g(x)=f(x)^2+f'(x)^2implies g'(x)=2f'(x)(f(x)+f''(x)).
          $$






          share|cite|improve this answer












          Try to explore the extrema of
          $$
          g(x)=f(x)^2+f'(x)^2implies g'(x)=2f'(x)(f(x)+f''(x)).
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • 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


















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