Prove or disprove: If $f$ is increasing and differentiable on $(a,b)$ then $f'(x)ge 0$ on $(a,b)$












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Here's the question again:




Prove or disprove: If $f$ is increasing and differentiable on $(a,b)$
then $f'(x)ge 0$ on $(a,b)$.




I could not find any counterexamples to it so here's my attempt at a proof:



Suppose there is a point $cin (a,b)$ such that $f'(c)<0$. Let $varepsilon = - f'(c)/2$. Then there exists a $delta > 0$ such that $frac{f(x)-f(c)}{x-c} < f'(c)/2 < 0$ for all $0<|x-c|< delta $.



Let $x_1 in (c-delta , c )$ then $x_1 -c<0$ thus we have that $frac{f(x_1)-f(c)}{x_1-c} <0$ implies $f(x_1) > f(c)$. Now, let $x_2 in (c, c+delta )$ then in a similar way, $f(x_2) < f(c)$. We note that $x_1 < x_2$ but we obtain $f(x_2)< f(c) < f(x_1)$, a contradiction since $f$ is increasing on $(a,b)$.



Is this attempt correct? Could there be any direct proofs?






real-analysis proof-verification alternative-proof monotone-functions







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$endgroup$

















    0












    $begingroup$


    Here's the question again:




    Prove or disprove: If $f$ is increasing and differentiable on $(a,b)$
    then $f'(x)ge 0$ on $(a,b)$.




    I could not find any counterexamples to it so here's my attempt at a proof:



    Suppose there is a point $cin (a,b)$ such that $f'(c)<0$. Let $varepsilon = - f'(c)/2$. Then there exists a $delta > 0$ such that $frac{f(x)-f(c)}{x-c} < f'(c)/2 < 0$ for all $0<|x-c|< delta $.



    Let $x_1 in (c-delta , c )$ then $x_1 -c<0$ thus we have that $frac{f(x_1)-f(c)}{x_1-c} <0$ implies $f(x_1) > f(c)$. Now, let $x_2 in (c, c+delta )$ then in a similar way, $f(x_2) < f(c)$. We note that $x_1 < x_2$ but we obtain $f(x_2)< f(c) < f(x_1)$, a contradiction since $f$ is increasing on $(a,b)$.



    Is this attempt correct? Could there be any direct proofs?






    real-analysis proof-verification alternative-proof monotone-functions







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Here's the question again:




      Prove or disprove: If $f$ is increasing and differentiable on $(a,b)$
      then $f'(x)ge 0$ on $(a,b)$.




      I could not find any counterexamples to it so here's my attempt at a proof:



      Suppose there is a point $cin (a,b)$ such that $f'(c)<0$. Let $varepsilon = - f'(c)/2$. Then there exists a $delta > 0$ such that $frac{f(x)-f(c)}{x-c} < f'(c)/2 < 0$ for all $0<|x-c|< delta $.



      Let $x_1 in (c-delta , c )$ then $x_1 -c<0$ thus we have that $frac{f(x_1)-f(c)}{x_1-c} <0$ implies $f(x_1) > f(c)$. Now, let $x_2 in (c, c+delta )$ then in a similar way, $f(x_2) < f(c)$. We note that $x_1 < x_2$ but we obtain $f(x_2)< f(c) < f(x_1)$, a contradiction since $f$ is increasing on $(a,b)$.



      Is this attempt correct? Could there be any direct proofs?






      real-analysis proof-verification alternative-proof monotone-functions







      share|cite|improve this question









      $endgroup$




      Here's the question again:




      Prove or disprove: If $f$ is increasing and differentiable on $(a,b)$
      then $f'(x)ge 0$ on $(a,b)$.




      I could not find any counterexamples to it so here's my attempt at a proof:



      Suppose there is a point $cin (a,b)$ such that $f'(c)<0$. Let $varepsilon = - f'(c)/2$. Then there exists a $delta > 0$ such that $frac{f(x)-f(c)}{x-c} < f'(c)/2 < 0$ for all $0<|x-c|< delta $.



      Let $x_1 in (c-delta , c )$ then $x_1 -c<0$ thus we have that $frac{f(x_1)-f(c)}{x_1-c} <0$ implies $f(x_1) > f(c)$. Now, let $x_2 in (c, c+delta )$ then in a similar way, $f(x_2) < f(c)$. We note that $x_1 < x_2$ but we obtain $f(x_2)< f(c) < f(x_1)$, a contradiction since $f$ is increasing on $(a,b)$.



      Is this attempt correct? Could there be any direct proofs?






      real-analysis proof-verification alternative-proof monotone-functions




      real-analysis proof-verification alternative-proof monotone-functions






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 24 '18 at 22:12









      Ashish KAshish K

      831613




      831613






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          Your proof is correct and almost a direct proof, let me rephrase it: Let $x<y$. Then
          $ f(y)-f(x) > 0$. Therefore
          $$ frac{f(y) - f(x)}{y-x} > 0.$$
          Sending $yto x$ yields
          $$ f'(x) ge 0.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Since $f$ is increasing on $(a,b)$, should not it be $f(y)-f(x) ge 0$ albeit it makes no difference?
            $endgroup$
            – Ashish K
            Nov 24 '18 at 22:20










          • $begingroup$
            @AshishK It depends on what you define as increasing. I'm always confused, since its "$>$" for increasing sequences...
            $endgroup$
            – Calvin Khor
            Nov 24 '18 at 22:22



















          3












          $begingroup$

          Your proof is correct. A direct proof goes like this:



          For any $cin (a,b)$, since $f$ is differentiable at $c$, we have
          $$lim_{xto c}frac{f(x)-f(c)}{x-c}=f'(c).$$
          The expression inside the limit is always nonnegative since $f$ is increasing. Thus, the limit has to be nonnegative.






          share|cite|improve this answer









          $endgroup$













            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            Your proof is correct and almost a direct proof, let me rephrase it: Let $x<y$. Then
            $ f(y)-f(x) > 0$. Therefore
            $$ frac{f(y) - f(x)}{y-x} > 0.$$
            Sending $yto x$ yields
            $$ f'(x) ge 0.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Since $f$ is increasing on $(a,b)$, should not it be $f(y)-f(x) ge 0$ albeit it makes no difference?
              $endgroup$
              – Ashish K
              Nov 24 '18 at 22:20










            • $begingroup$
              @AshishK It depends on what you define as increasing. I'm always confused, since its "$>$" for increasing sequences...
              $endgroup$
              – Calvin Khor
              Nov 24 '18 at 22:22
















            3












            $begingroup$

            Your proof is correct and almost a direct proof, let me rephrase it: Let $x<y$. Then
            $ f(y)-f(x) > 0$. Therefore
            $$ frac{f(y) - f(x)}{y-x} > 0.$$
            Sending $yto x$ yields
            $$ f'(x) ge 0.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Since $f$ is increasing on $(a,b)$, should not it be $f(y)-f(x) ge 0$ albeit it makes no difference?
              $endgroup$
              – Ashish K
              Nov 24 '18 at 22:20










            • $begingroup$
              @AshishK It depends on what you define as increasing. I'm always confused, since its "$>$" for increasing sequences...
              $endgroup$
              – Calvin Khor
              Nov 24 '18 at 22:22














            3












            3








            3





            $begingroup$

            Your proof is correct and almost a direct proof, let me rephrase it: Let $x<y$. Then
            $ f(y)-f(x) > 0$. Therefore
            $$ frac{f(y) - f(x)}{y-x} > 0.$$
            Sending $yto x$ yields
            $$ f'(x) ge 0.$$






            share|cite|improve this answer









            $endgroup$



            Your proof is correct and almost a direct proof, let me rephrase it: Let $x<y$. Then
            $ f(y)-f(x) > 0$. Therefore
            $$ frac{f(y) - f(x)}{y-x} > 0.$$
            Sending $yto x$ yields
            $$ f'(x) ge 0.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 24 '18 at 22:16









            Calvin KhorCalvin Khor

            11.2k21438




            11.2k21438












            • $begingroup$
              Since $f$ is increasing on $(a,b)$, should not it be $f(y)-f(x) ge 0$ albeit it makes no difference?
              $endgroup$
              – Ashish K
              Nov 24 '18 at 22:20










            • $begingroup$
              @AshishK It depends on what you define as increasing. I'm always confused, since its "$>$" for increasing sequences...
              $endgroup$
              – Calvin Khor
              Nov 24 '18 at 22:22


















            • $begingroup$
              Since $f$ is increasing on $(a,b)$, should not it be $f(y)-f(x) ge 0$ albeit it makes no difference?
              $endgroup$
              – Ashish K
              Nov 24 '18 at 22:20










            • $begingroup$
              @AshishK It depends on what you define as increasing. I'm always confused, since its "$>$" for increasing sequences...
              $endgroup$
              – Calvin Khor
              Nov 24 '18 at 22:22
















            $begingroup$
            Since $f$ is increasing on $(a,b)$, should not it be $f(y)-f(x) ge 0$ albeit it makes no difference?
            $endgroup$
            – Ashish K
            Nov 24 '18 at 22:20




            $begingroup$
            Since $f$ is increasing on $(a,b)$, should not it be $f(y)-f(x) ge 0$ albeit it makes no difference?
            $endgroup$
            – Ashish K
            Nov 24 '18 at 22:20












            $begingroup$
            @AshishK It depends on what you define as increasing. I'm always confused, since its "$>$" for increasing sequences...
            $endgroup$
            – Calvin Khor
            Nov 24 '18 at 22:22




            $begingroup$
            @AshishK It depends on what you define as increasing. I'm always confused, since its "$>$" for increasing sequences...
            $endgroup$
            – Calvin Khor
            Nov 24 '18 at 22:22











            3












            $begingroup$

            Your proof is correct. A direct proof goes like this:



            For any $cin (a,b)$, since $f$ is differentiable at $c$, we have
            $$lim_{xto c}frac{f(x)-f(c)}{x-c}=f'(c).$$
            The expression inside the limit is always nonnegative since $f$ is increasing. Thus, the limit has to be nonnegative.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              Your proof is correct. A direct proof goes like this:



              For any $cin (a,b)$, since $f$ is differentiable at $c$, we have
              $$lim_{xto c}frac{f(x)-f(c)}{x-c}=f'(c).$$
              The expression inside the limit is always nonnegative since $f$ is increasing. Thus, the limit has to be nonnegative.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                Your proof is correct. A direct proof goes like this:



                For any $cin (a,b)$, since $f$ is differentiable at $c$, we have
                $$lim_{xto c}frac{f(x)-f(c)}{x-c}=f'(c).$$
                The expression inside the limit is always nonnegative since $f$ is increasing. Thus, the limit has to be nonnegative.






                share|cite|improve this answer









                $endgroup$



                Your proof is correct. A direct proof goes like this:



                For any $cin (a,b)$, since $f$ is differentiable at $c$, we have
                $$lim_{xto c}frac{f(x)-f(c)}{x-c}=f'(c).$$
                The expression inside the limit is always nonnegative since $f$ is increasing. Thus, the limit has to be nonnegative.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 24 '18 at 22:16









                Eclipse SunEclipse Sun

                6,9841437




                6,9841437






























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