Application of mean value and Rolle's theorems - twice differentiable functions












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Let $f$ be a function from $[a,b]$ to $Bbb{R}$ that is twice-differentiable (that is, $f'$ and $f''$ exist), and assume that $f(a) = f(b) = 0$ and $f''(x) leq 0$ for every $xin (a,b)$. Show that $f(x) geq 0$ on $[a,b]$.



I think we must use the mean value/ Rolle's theorems There will be a $c$ in $(a,b)$ such that $f'(c) = frac{f(a) - f(b)}{b-a}= 0$. Where do I go from here?



Any help will be appreciated!






calculus rolles-theorem







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    1














    Let $f$ be a function from $[a,b]$ to $Bbb{R}$ that is twice-differentiable (that is, $f'$ and $f''$ exist), and assume that $f(a) = f(b) = 0$ and $f''(x) leq 0$ for every $xin (a,b)$. Show that $f(x) geq 0$ on $[a,b]$.



    I think we must use the mean value/ Rolle's theorems There will be a $c$ in $(a,b)$ such that $f'(c) = frac{f(a) - f(b)}{b-a}= 0$. Where do I go from here?



    Any help will be appreciated!






    calculus rolles-theorem







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      1












      1








      1







      Let $f$ be a function from $[a,b]$ to $Bbb{R}$ that is twice-differentiable (that is, $f'$ and $f''$ exist), and assume that $f(a) = f(b) = 0$ and $f''(x) leq 0$ for every $xin (a,b)$. Show that $f(x) geq 0$ on $[a,b]$.



      I think we must use the mean value/ Rolle's theorems There will be a $c$ in $(a,b)$ such that $f'(c) = frac{f(a) - f(b)}{b-a}= 0$. Where do I go from here?



      Any help will be appreciated!






      calculus rolles-theorem







      share|cite|improve this question















      Let $f$ be a function from $[a,b]$ to $Bbb{R}$ that is twice-differentiable (that is, $f'$ and $f''$ exist), and assume that $f(a) = f(b) = 0$ and $f''(x) leq 0$ for every $xin (a,b)$. Show that $f(x) geq 0$ on $[a,b]$.



      I think we must use the mean value/ Rolle's theorems There will be a $c$ in $(a,b)$ such that $f'(c) = frac{f(a) - f(b)}{b-a}= 0$. Where do I go from here?



      Any help will be appreciated!






      calculus rolles-theorem




      calculus rolles-theorem






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      edited Nov 29 at 18:58









      Bernard

      118k638112




      118k638112










      asked Nov 29 at 18:00









      Aishwarya Deore

      324




      324






















          3 Answers
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          $f$ is continuous in $[a,b]$ and differentiable in $(a,b), f(a)=f(b)=0$



          By Rolle's Theorem, $exists cin(a,b)$ such that $f'(c)=frac{f(b)-f(a)}{b-a}=0$



          $f''(x)leq0, forall xin(a,b)$



          $impliesforall xin(a,c), int_{x}^{c}f''(x)dx=f'(c)-f'(x)=-f'(x)leq0$



          $impliesforall xin(a,c), f'(x)geq0$



          $implies int_{a}^{x} f'(x)dx=f(x)-f(a)=f(x)geq0$



          Similarly, $forall xin(c,b), f'(x)leq0$



          $impliesint_{x}^{b} f'(x)dx=f(b)-f(x)=-f(x)leq0$



          $implies f(x)geq0$



          As far as $f(c)$ is concerned, continuity requires $f(c)geq0$






          share|cite|improve this answer





























            0














            Assume that $f(c) < 0$ for some $c in (a, b)$, and apply Rolle's theorem
            repeatedly:




            • There is an $x_1 in (a, c)$ with $f'(x_1) = frac{f(c)-f(a)}{c-a} < 0$.


            • There is an $x_2 in (c, b)$ with $f'(x_2) = frac{f(b)-f(c)}{b-c} > 0$.


            • There is an $x_3 in (x_1, x_2)$ with $f''(x_3) = frac{f'(x_2)-f'(x_1)}{x_2-x_1} > 0$.



            This is a contradiction to the assumption that $f''(x) le 0$
            for all $x in (a, b)$.




            An alternative approach: Fix any $c in (a, b)$ and define
            $$
            g(x) = (c-a)(c-b)f(x) - (x-a)(x-b)f(c) , .
            $$
            Then $g(a) = g(c) = g(b) = 0$, and repeated application of Rolle's
            theorem shows that $g''(xi) = 0$ for some $xi in (a, b)$.
            Then
            $$
            0 = g''(xi) = (c-a)(c-b)f''(xi) - 2f(c) \
            implies f(c) = frac 12 (c-a)(c-b)f''(xi) ge 0 , .
            $$




            Yet another approach: Assume that $f$ attains its minimum at $c in (a, b)$ and $f(c) < 0$. Then $f'(c) = 0$ and Taylor's formula gives
            a contradiction:
            $$
            f(b) = f(c) + frac 12 (b-c)^2 f''(xi) < 0
            $$
            for some $xi in (c, b)$.



            The real idea here is that $f$ is concave,
            and a continuous concave function on a compact interval attains its
            minimum at one of the boundary points: Therefore
            $$
            f(x) ge min (f(a), f(b)) = 0
            $$
            for $x in (a, b)$.






            share|cite|improve this answer























            • Thank you! I get the first method
              – Aishwarya Deore
              Nov 29 at 18:31



















            0














            By Rolle's theorem, there exists $cin(a,b)$ such that $f'(c)=0$.



            Now $f''(x)le 0$ in $(a,b)$ implies $f'(x)$ is non-increasing in $[a,b]$, so




            • if $ale xle c$, then $f'(x)ge f'(c)=0$, in other words $f(x)$ is non-decreasing in $[a,c]$, so that for all $xin[a,c]$, $f(x)ge f(a)=0$;

            • if $cle xle b$, then $f'(x)le f'(c)=0$, in other words $f(x)$ is non-increasing in $[c,b]$, so that for all $xin[c,b]$, $f(x)ge f(b)=0$.


            Thus, in each case, $f(x)ge 0$.






            share|cite|improve this answer





















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              3 Answers
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              0














              $f$ is continuous in $[a,b]$ and differentiable in $(a,b), f(a)=f(b)=0$



              By Rolle's Theorem, $exists cin(a,b)$ such that $f'(c)=frac{f(b)-f(a)}{b-a}=0$



              $f''(x)leq0, forall xin(a,b)$



              $impliesforall xin(a,c), int_{x}^{c}f''(x)dx=f'(c)-f'(x)=-f'(x)leq0$



              $impliesforall xin(a,c), f'(x)geq0$



              $implies int_{a}^{x} f'(x)dx=f(x)-f(a)=f(x)geq0$



              Similarly, $forall xin(c,b), f'(x)leq0$



              $impliesint_{x}^{b} f'(x)dx=f(b)-f(x)=-f(x)leq0$



              $implies f(x)geq0$



              As far as $f(c)$ is concerned, continuity requires $f(c)geq0$






              share|cite|improve this answer


























                0














                $f$ is continuous in $[a,b]$ and differentiable in $(a,b), f(a)=f(b)=0$



                By Rolle's Theorem, $exists cin(a,b)$ such that $f'(c)=frac{f(b)-f(a)}{b-a}=0$



                $f''(x)leq0, forall xin(a,b)$



                $impliesforall xin(a,c), int_{x}^{c}f''(x)dx=f'(c)-f'(x)=-f'(x)leq0$



                $impliesforall xin(a,c), f'(x)geq0$



                $implies int_{a}^{x} f'(x)dx=f(x)-f(a)=f(x)geq0$



                Similarly, $forall xin(c,b), f'(x)leq0$



                $impliesint_{x}^{b} f'(x)dx=f(b)-f(x)=-f(x)leq0$



                $implies f(x)geq0$



                As far as $f(c)$ is concerned, continuity requires $f(c)geq0$






                share|cite|improve this answer
























                  0












                  0








                  0






                  $f$ is continuous in $[a,b]$ and differentiable in $(a,b), f(a)=f(b)=0$



                  By Rolle's Theorem, $exists cin(a,b)$ such that $f'(c)=frac{f(b)-f(a)}{b-a}=0$



                  $f''(x)leq0, forall xin(a,b)$



                  $impliesforall xin(a,c), int_{x}^{c}f''(x)dx=f'(c)-f'(x)=-f'(x)leq0$



                  $impliesforall xin(a,c), f'(x)geq0$



                  $implies int_{a}^{x} f'(x)dx=f(x)-f(a)=f(x)geq0$



                  Similarly, $forall xin(c,b), f'(x)leq0$



                  $impliesint_{x}^{b} f'(x)dx=f(b)-f(x)=-f(x)leq0$



                  $implies f(x)geq0$



                  As far as $f(c)$ is concerned, continuity requires $f(c)geq0$






                  share|cite|improve this answer












                  $f$ is continuous in $[a,b]$ and differentiable in $(a,b), f(a)=f(b)=0$



                  By Rolle's Theorem, $exists cin(a,b)$ such that $f'(c)=frac{f(b)-f(a)}{b-a}=0$



                  $f''(x)leq0, forall xin(a,b)$



                  $impliesforall xin(a,c), int_{x}^{c}f''(x)dx=f'(c)-f'(x)=-f'(x)leq0$



                  $impliesforall xin(a,c), f'(x)geq0$



                  $implies int_{a}^{x} f'(x)dx=f(x)-f(a)=f(x)geq0$



                  Similarly, $forall xin(c,b), f'(x)leq0$



                  $impliesint_{x}^{b} f'(x)dx=f(b)-f(x)=-f(x)leq0$



                  $implies f(x)geq0$



                  As far as $f(c)$ is concerned, continuity requires $f(c)geq0$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 29 at 18:27









                  Shubham Johri

                  3,859716




                  3,859716























                      0














                      Assume that $f(c) < 0$ for some $c in (a, b)$, and apply Rolle's theorem
                      repeatedly:




                      • There is an $x_1 in (a, c)$ with $f'(x_1) = frac{f(c)-f(a)}{c-a} < 0$.


                      • There is an $x_2 in (c, b)$ with $f'(x_2) = frac{f(b)-f(c)}{b-c} > 0$.


                      • There is an $x_3 in (x_1, x_2)$ with $f''(x_3) = frac{f'(x_2)-f'(x_1)}{x_2-x_1} > 0$.



                      This is a contradiction to the assumption that $f''(x) le 0$
                      for all $x in (a, b)$.




                      An alternative approach: Fix any $c in (a, b)$ and define
                      $$
                      g(x) = (c-a)(c-b)f(x) - (x-a)(x-b)f(c) , .
                      $$
                      Then $g(a) = g(c) = g(b) = 0$, and repeated application of Rolle's
                      theorem shows that $g''(xi) = 0$ for some $xi in (a, b)$.
                      Then
                      $$
                      0 = g''(xi) = (c-a)(c-b)f''(xi) - 2f(c) \
                      implies f(c) = frac 12 (c-a)(c-b)f''(xi) ge 0 , .
                      $$




                      Yet another approach: Assume that $f$ attains its minimum at $c in (a, b)$ and $f(c) < 0$. Then $f'(c) = 0$ and Taylor's formula gives
                      a contradiction:
                      $$
                      f(b) = f(c) + frac 12 (b-c)^2 f''(xi) < 0
                      $$
                      for some $xi in (c, b)$.



                      The real idea here is that $f$ is concave,
                      and a continuous concave function on a compact interval attains its
                      minimum at one of the boundary points: Therefore
                      $$
                      f(x) ge min (f(a), f(b)) = 0
                      $$
                      for $x in (a, b)$.






                      share|cite|improve this answer























                      • Thank you! I get the first method
                        – Aishwarya Deore
                        Nov 29 at 18:31
















                      0














                      Assume that $f(c) < 0$ for some $c in (a, b)$, and apply Rolle's theorem
                      repeatedly:




                      • There is an $x_1 in (a, c)$ with $f'(x_1) = frac{f(c)-f(a)}{c-a} < 0$.


                      • There is an $x_2 in (c, b)$ with $f'(x_2) = frac{f(b)-f(c)}{b-c} > 0$.


                      • There is an $x_3 in (x_1, x_2)$ with $f''(x_3) = frac{f'(x_2)-f'(x_1)}{x_2-x_1} > 0$.



                      This is a contradiction to the assumption that $f''(x) le 0$
                      for all $x in (a, b)$.




                      An alternative approach: Fix any $c in (a, b)$ and define
                      $$
                      g(x) = (c-a)(c-b)f(x) - (x-a)(x-b)f(c) , .
                      $$
                      Then $g(a) = g(c) = g(b) = 0$, and repeated application of Rolle's
                      theorem shows that $g''(xi) = 0$ for some $xi in (a, b)$.
                      Then
                      $$
                      0 = g''(xi) = (c-a)(c-b)f''(xi) - 2f(c) \
                      implies f(c) = frac 12 (c-a)(c-b)f''(xi) ge 0 , .
                      $$




                      Yet another approach: Assume that $f$ attains its minimum at $c in (a, b)$ and $f(c) < 0$. Then $f'(c) = 0$ and Taylor's formula gives
                      a contradiction:
                      $$
                      f(b) = f(c) + frac 12 (b-c)^2 f''(xi) < 0
                      $$
                      for some $xi in (c, b)$.



                      The real idea here is that $f$ is concave,
                      and a continuous concave function on a compact interval attains its
                      minimum at one of the boundary points: Therefore
                      $$
                      f(x) ge min (f(a), f(b)) = 0
                      $$
                      for $x in (a, b)$.






                      share|cite|improve this answer























                      • Thank you! I get the first method
                        – Aishwarya Deore
                        Nov 29 at 18:31














                      0












                      0








                      0






                      Assume that $f(c) < 0$ for some $c in (a, b)$, and apply Rolle's theorem
                      repeatedly:




                      • There is an $x_1 in (a, c)$ with $f'(x_1) = frac{f(c)-f(a)}{c-a} < 0$.


                      • There is an $x_2 in (c, b)$ with $f'(x_2) = frac{f(b)-f(c)}{b-c} > 0$.


                      • There is an $x_3 in (x_1, x_2)$ with $f''(x_3) = frac{f'(x_2)-f'(x_1)}{x_2-x_1} > 0$.



                      This is a contradiction to the assumption that $f''(x) le 0$
                      for all $x in (a, b)$.




                      An alternative approach: Fix any $c in (a, b)$ and define
                      $$
                      g(x) = (c-a)(c-b)f(x) - (x-a)(x-b)f(c) , .
                      $$
                      Then $g(a) = g(c) = g(b) = 0$, and repeated application of Rolle's
                      theorem shows that $g''(xi) = 0$ for some $xi in (a, b)$.
                      Then
                      $$
                      0 = g''(xi) = (c-a)(c-b)f''(xi) - 2f(c) \
                      implies f(c) = frac 12 (c-a)(c-b)f''(xi) ge 0 , .
                      $$




                      Yet another approach: Assume that $f$ attains its minimum at $c in (a, b)$ and $f(c) < 0$. Then $f'(c) = 0$ and Taylor's formula gives
                      a contradiction:
                      $$
                      f(b) = f(c) + frac 12 (b-c)^2 f''(xi) < 0
                      $$
                      for some $xi in (c, b)$.



                      The real idea here is that $f$ is concave,
                      and a continuous concave function on a compact interval attains its
                      minimum at one of the boundary points: Therefore
                      $$
                      f(x) ge min (f(a), f(b)) = 0
                      $$
                      for $x in (a, b)$.






                      share|cite|improve this answer














                      Assume that $f(c) < 0$ for some $c in (a, b)$, and apply Rolle's theorem
                      repeatedly:




                      • There is an $x_1 in (a, c)$ with $f'(x_1) = frac{f(c)-f(a)}{c-a} < 0$.


                      • There is an $x_2 in (c, b)$ with $f'(x_2) = frac{f(b)-f(c)}{b-c} > 0$.


                      • There is an $x_3 in (x_1, x_2)$ with $f''(x_3) = frac{f'(x_2)-f'(x_1)}{x_2-x_1} > 0$.



                      This is a contradiction to the assumption that $f''(x) le 0$
                      for all $x in (a, b)$.




                      An alternative approach: Fix any $c in (a, b)$ and define
                      $$
                      g(x) = (c-a)(c-b)f(x) - (x-a)(x-b)f(c) , .
                      $$
                      Then $g(a) = g(c) = g(b) = 0$, and repeated application of Rolle's
                      theorem shows that $g''(xi) = 0$ for some $xi in (a, b)$.
                      Then
                      $$
                      0 = g''(xi) = (c-a)(c-b)f''(xi) - 2f(c) \
                      implies f(c) = frac 12 (c-a)(c-b)f''(xi) ge 0 , .
                      $$




                      Yet another approach: Assume that $f$ attains its minimum at $c in (a, b)$ and $f(c) < 0$. Then $f'(c) = 0$ and Taylor's formula gives
                      a contradiction:
                      $$
                      f(b) = f(c) + frac 12 (b-c)^2 f''(xi) < 0
                      $$
                      for some $xi in (c, b)$.



                      The real idea here is that $f$ is concave,
                      and a continuous concave function on a compact interval attains its
                      minimum at one of the boundary points: Therefore
                      $$
                      f(x) ge min (f(a), f(b)) = 0
                      $$
                      for $x in (a, b)$.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Nov 29 at 18:47

























                      answered Nov 29 at 18:17









                      Martin R

                      26.9k33251




                      26.9k33251












                      • Thank you! I get the first method
                        – Aishwarya Deore
                        Nov 29 at 18:31


















                      • Thank you! I get the first method
                        – Aishwarya Deore
                        Nov 29 at 18:31
















                      Thank you! I get the first method
                      – Aishwarya Deore
                      Nov 29 at 18:31




                      Thank you! I get the first method
                      – Aishwarya Deore
                      Nov 29 at 18:31











                      0














                      By Rolle's theorem, there exists $cin(a,b)$ such that $f'(c)=0$.



                      Now $f''(x)le 0$ in $(a,b)$ implies $f'(x)$ is non-increasing in $[a,b]$, so




                      • if $ale xle c$, then $f'(x)ge f'(c)=0$, in other words $f(x)$ is non-decreasing in $[a,c]$, so that for all $xin[a,c]$, $f(x)ge f(a)=0$;

                      • if $cle xle b$, then $f'(x)le f'(c)=0$, in other words $f(x)$ is non-increasing in $[c,b]$, so that for all $xin[c,b]$, $f(x)ge f(b)=0$.


                      Thus, in each case, $f(x)ge 0$.






                      share|cite|improve this answer


























                        0














                        By Rolle's theorem, there exists $cin(a,b)$ such that $f'(c)=0$.



                        Now $f''(x)le 0$ in $(a,b)$ implies $f'(x)$ is non-increasing in $[a,b]$, so




                        • if $ale xle c$, then $f'(x)ge f'(c)=0$, in other words $f(x)$ is non-decreasing in $[a,c]$, so that for all $xin[a,c]$, $f(x)ge f(a)=0$;

                        • if $cle xle b$, then $f'(x)le f'(c)=0$, in other words $f(x)$ is non-increasing in $[c,b]$, so that for all $xin[c,b]$, $f(x)ge f(b)=0$.


                        Thus, in each case, $f(x)ge 0$.






                        share|cite|improve this answer
























                          0












                          0








                          0






                          By Rolle's theorem, there exists $cin(a,b)$ such that $f'(c)=0$.



                          Now $f''(x)le 0$ in $(a,b)$ implies $f'(x)$ is non-increasing in $[a,b]$, so




                          • if $ale xle c$, then $f'(x)ge f'(c)=0$, in other words $f(x)$ is non-decreasing in $[a,c]$, so that for all $xin[a,c]$, $f(x)ge f(a)=0$;

                          • if $cle xle b$, then $f'(x)le f'(c)=0$, in other words $f(x)$ is non-increasing in $[c,b]$, so that for all $xin[c,b]$, $f(x)ge f(b)=0$.


                          Thus, in each case, $f(x)ge 0$.






                          share|cite|improve this answer












                          By Rolle's theorem, there exists $cin(a,b)$ such that $f'(c)=0$.



                          Now $f''(x)le 0$ in $(a,b)$ implies $f'(x)$ is non-increasing in $[a,b]$, so




                          • if $ale xle c$, then $f'(x)ge f'(c)=0$, in other words $f(x)$ is non-decreasing in $[a,c]$, so that for all $xin[a,c]$, $f(x)ge f(a)=0$;

                          • if $cle xle b$, then $f'(x)le f'(c)=0$, in other words $f(x)$ is non-increasing in $[c,b]$, so that for all $xin[c,b]$, $f(x)ge f(b)=0$.


                          Thus, in each case, $f(x)ge 0$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 29 at 19:13









                          Bernard

                          118k638112




                          118k638112






























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