Proof $y=x^3/(x^2+1)$ is injective












0












$begingroup$


i need to proof that $$f(x)=frac{x^3}{x^2+1}$$ is injective



I tried $f(x_1)=f(x_2)$ and I also tried with Rolle's theorem.






functions







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$endgroup$

















    0












    $begingroup$


    i need to proof that $$f(x)=frac{x^3}{x^2+1}$$ is injective



    I tried $f(x_1)=f(x_2)$ and I also tried with Rolle's theorem.






    functions







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      i need to proof that $$f(x)=frac{x^3}{x^2+1}$$ is injective



      I tried $f(x_1)=f(x_2)$ and I also tried with Rolle's theorem.






      functions







      share|cite|improve this question











      $endgroup$




      i need to proof that $$f(x)=frac{x^3}{x^2+1}$$ is injective



      I tried $f(x_1)=f(x_2)$ and I also tried with Rolle's theorem.






      functions




      functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 16 '18 at 14:05









      egreg

      184k1486205




      184k1486205










      asked Dec 16 '18 at 14:00









      shayshay

      103




      103






















          4 Answers
          4






          active

          oldest

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          1












          $begingroup$

          Assume $exists a,b: f(a)=f(b)$, then begin{align*}frac{a^3}{a^2+1}=frac{b^3}{b^2+1}&iff a^3b^2+a^3=b^3a^2+b^3 \ &iff a^3b^2-a^2b^3+a^3-b^3=0 \&iff a^2b^2(a-b)+(a-b)(a^2-ab+b^2)=0 \ &iff (a-b)left[a^2b^2+a^2+ab+b^2right]=0\&iff a=bvee a^2b^2+a^2+ab+b^2=0end{align*}



          Edit: based on @egreg's comment: $$a^2b^2+a^2+ab+b^2=(ab)^2+left(a+frac{b}{2}right)^2+frac{3b^2}{4}ge 0$$ and equality holds if and only if $a=b=0$.



          Either way, we found that $a=b$, hence the function is injective.






          share|cite|improve this answer











          $endgroup$









          • 2




            $begingroup$
            $a^2b^2+a^2+ab+b^2=(ab)^2+(a+b/2)^2+3b^2/4$. A sum of squares is nonnegative and is zero if and only if each term is zero.
            $endgroup$
            – egreg
            Dec 16 '18 at 14:46










          • $begingroup$
            @egreg, thanks, I have looked after the factorization, but didn't manage to compute it by myself.
            $endgroup$
            – Galc127
            Dec 16 '18 at 14:48



















          3












          $begingroup$

          Hint: Prove that, for each real $x$, $f'(x)geqslant0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I did it, but still dont know how it's can help me
            $endgroup$
            – shay
            Dec 16 '18 at 14:08






          • 1




            $begingroup$
            SInce $f'(x)$ is always non-negative, $f$ is invreasing. Now, suppose that $f(x)=f(y)$ for some $x$ and some $y$ such that $x<y$. Then $f$ is constant in $[x,y]$ and therefore $f'$ is $0$ there. But $f'$ only has one single zero.
            $endgroup$
            – José Carlos Santos
            Dec 16 '18 at 14:10



















          2












          $begingroup$

          Here is an approach if you want to avoid derivatives. For $x neq 0$, we have
          $$frac{x^3}{x^2 + 1} = frac{1}{frac{1}{x} + frac{1}{x^3}}.$$



          The function $1/x + 1/x^3$ for $x > 0$ is the sum of two strictly decreasing functions, hence itself strictly decreasing and therefore one to one with range $(0, infty)$. Hence $1/(1/x + 1/x^3)$ is strictly increasing and one to one with range $(0, infty)$. For $x < 0$, notice $1/x + 1/x^3$ is an odd function, and therefore a similar argument shows $1/(1/x + 1/x^3)$ is one to one with range $(-infty, 0)$.



          In fact, this shows that if you add back the point $x = 0$, the function is a bijection since the function is zero here.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            Suppose $f(a)=f(b)$: then
            $$
            a^3(b^2+1)=b^3(a^2+1)
            $$
            This implies
            $$
            a^3b^2-a^2b^3+a^3-b^3=0
            $$
            and so
            $$
            a^2b^2(a-b)+(a-b)(a^2+ab+b^2)=0
            $$
            Conclude.




            If you know that $a^2+ab+b^2le0$ if and only if $a=b=0$, you're done.




            With calculus,
            $$
            f'(x)=frac{3x^2(x^2+1)-2x^4}{(x^2+1)^2}
            $$
            The derivative only vanishes at $0$, so Rolle's theorem would only apply to an interval of the form $[a,b]$, with $a<0$ and $b>0$. Can $f(a)=f(b)$ in this case?






            share|cite|improve this answer









            $endgroup$













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              4 Answers
              4






              active

              oldest

              votes








              4 Answers
              4






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              Assume $exists a,b: f(a)=f(b)$, then begin{align*}frac{a^3}{a^2+1}=frac{b^3}{b^2+1}&iff a^3b^2+a^3=b^3a^2+b^3 \ &iff a^3b^2-a^2b^3+a^3-b^3=0 \&iff a^2b^2(a-b)+(a-b)(a^2-ab+b^2)=0 \ &iff (a-b)left[a^2b^2+a^2+ab+b^2right]=0\&iff a=bvee a^2b^2+a^2+ab+b^2=0end{align*}



              Edit: based on @egreg's comment: $$a^2b^2+a^2+ab+b^2=(ab)^2+left(a+frac{b}{2}right)^2+frac{3b^2}{4}ge 0$$ and equality holds if and only if $a=b=0$.



              Either way, we found that $a=b$, hence the function is injective.






              share|cite|improve this answer











              $endgroup$









              • 2




                $begingroup$
                $a^2b^2+a^2+ab+b^2=(ab)^2+(a+b/2)^2+3b^2/4$. A sum of squares is nonnegative and is zero if and only if each term is zero.
                $endgroup$
                – egreg
                Dec 16 '18 at 14:46










              • $begingroup$
                @egreg, thanks, I have looked after the factorization, but didn't manage to compute it by myself.
                $endgroup$
                – Galc127
                Dec 16 '18 at 14:48
















              1












              $begingroup$

              Assume $exists a,b: f(a)=f(b)$, then begin{align*}frac{a^3}{a^2+1}=frac{b^3}{b^2+1}&iff a^3b^2+a^3=b^3a^2+b^3 \ &iff a^3b^2-a^2b^3+a^3-b^3=0 \&iff a^2b^2(a-b)+(a-b)(a^2-ab+b^2)=0 \ &iff (a-b)left[a^2b^2+a^2+ab+b^2right]=0\&iff a=bvee a^2b^2+a^2+ab+b^2=0end{align*}



              Edit: based on @egreg's comment: $$a^2b^2+a^2+ab+b^2=(ab)^2+left(a+frac{b}{2}right)^2+frac{3b^2}{4}ge 0$$ and equality holds if and only if $a=b=0$.



              Either way, we found that $a=b$, hence the function is injective.






              share|cite|improve this answer











              $endgroup$









              • 2




                $begingroup$
                $a^2b^2+a^2+ab+b^2=(ab)^2+(a+b/2)^2+3b^2/4$. A sum of squares is nonnegative and is zero if and only if each term is zero.
                $endgroup$
                – egreg
                Dec 16 '18 at 14:46










              • $begingroup$
                @egreg, thanks, I have looked after the factorization, but didn't manage to compute it by myself.
                $endgroup$
                – Galc127
                Dec 16 '18 at 14:48














              1












              1








              1





              $begingroup$

              Assume $exists a,b: f(a)=f(b)$, then begin{align*}frac{a^3}{a^2+1}=frac{b^3}{b^2+1}&iff a^3b^2+a^3=b^3a^2+b^3 \ &iff a^3b^2-a^2b^3+a^3-b^3=0 \&iff a^2b^2(a-b)+(a-b)(a^2-ab+b^2)=0 \ &iff (a-b)left[a^2b^2+a^2+ab+b^2right]=0\&iff a=bvee a^2b^2+a^2+ab+b^2=0end{align*}



              Edit: based on @egreg's comment: $$a^2b^2+a^2+ab+b^2=(ab)^2+left(a+frac{b}{2}right)^2+frac{3b^2}{4}ge 0$$ and equality holds if and only if $a=b=0$.



              Either way, we found that $a=b$, hence the function is injective.






              share|cite|improve this answer











              $endgroup$



              Assume $exists a,b: f(a)=f(b)$, then begin{align*}frac{a^3}{a^2+1}=frac{b^3}{b^2+1}&iff a^3b^2+a^3=b^3a^2+b^3 \ &iff a^3b^2-a^2b^3+a^3-b^3=0 \&iff a^2b^2(a-b)+(a-b)(a^2-ab+b^2)=0 \ &iff (a-b)left[a^2b^2+a^2+ab+b^2right]=0\&iff a=bvee a^2b^2+a^2+ab+b^2=0end{align*}



              Edit: based on @egreg's comment: $$a^2b^2+a^2+ab+b^2=(ab)^2+left(a+frac{b}{2}right)^2+frac{3b^2}{4}ge 0$$ and equality holds if and only if $a=b=0$.



              Either way, we found that $a=b$, hence the function is injective.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Dec 16 '18 at 14:50

























              answered Dec 16 '18 at 14:16









              Galc127Galc127

              3,3121334




              3,3121334








              • 2




                $begingroup$
                $a^2b^2+a^2+ab+b^2=(ab)^2+(a+b/2)^2+3b^2/4$. A sum of squares is nonnegative and is zero if and only if each term is zero.
                $endgroup$
                – egreg
                Dec 16 '18 at 14:46










              • $begingroup$
                @egreg, thanks, I have looked after the factorization, but didn't manage to compute it by myself.
                $endgroup$
                – Galc127
                Dec 16 '18 at 14:48














              • 2




                $begingroup$
                $a^2b^2+a^2+ab+b^2=(ab)^2+(a+b/2)^2+3b^2/4$. A sum of squares is nonnegative and is zero if and only if each term is zero.
                $endgroup$
                – egreg
                Dec 16 '18 at 14:46










              • $begingroup$
                @egreg, thanks, I have looked after the factorization, but didn't manage to compute it by myself.
                $endgroup$
                – Galc127
                Dec 16 '18 at 14:48








              2




              2




              $begingroup$
              $a^2b^2+a^2+ab+b^2=(ab)^2+(a+b/2)^2+3b^2/4$. A sum of squares is nonnegative and is zero if and only if each term is zero.
              $endgroup$
              – egreg
              Dec 16 '18 at 14:46




              $begingroup$
              $a^2b^2+a^2+ab+b^2=(ab)^2+(a+b/2)^2+3b^2/4$. A sum of squares is nonnegative and is zero if and only if each term is zero.
              $endgroup$
              – egreg
              Dec 16 '18 at 14:46












              $begingroup$
              @egreg, thanks, I have looked after the factorization, but didn't manage to compute it by myself.
              $endgroup$
              – Galc127
              Dec 16 '18 at 14:48




              $begingroup$
              @egreg, thanks, I have looked after the factorization, but didn't manage to compute it by myself.
              $endgroup$
              – Galc127
              Dec 16 '18 at 14:48











              3












              $begingroup$

              Hint: Prove that, for each real $x$, $f'(x)geqslant0$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                I did it, but still dont know how it's can help me
                $endgroup$
                – shay
                Dec 16 '18 at 14:08






              • 1




                $begingroup$
                SInce $f'(x)$ is always non-negative, $f$ is invreasing. Now, suppose that $f(x)=f(y)$ for some $x$ and some $y$ such that $x<y$. Then $f$ is constant in $[x,y]$ and therefore $f'$ is $0$ there. But $f'$ only has one single zero.
                $endgroup$
                – José Carlos Santos
                Dec 16 '18 at 14:10
















              3












              $begingroup$

              Hint: Prove that, for each real $x$, $f'(x)geqslant0$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                I did it, but still dont know how it's can help me
                $endgroup$
                – shay
                Dec 16 '18 at 14:08






              • 1




                $begingroup$
                SInce $f'(x)$ is always non-negative, $f$ is invreasing. Now, suppose that $f(x)=f(y)$ for some $x$ and some $y$ such that $x<y$. Then $f$ is constant in $[x,y]$ and therefore $f'$ is $0$ there. But $f'$ only has one single zero.
                $endgroup$
                – José Carlos Santos
                Dec 16 '18 at 14:10














              3












              3








              3





              $begingroup$

              Hint: Prove that, for each real $x$, $f'(x)geqslant0$.






              share|cite|improve this answer









              $endgroup$



              Hint: Prove that, for each real $x$, $f'(x)geqslant0$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 16 '18 at 14:05









              José Carlos SantosJosé Carlos Santos

              167k22132235




              167k22132235












              • $begingroup$
                I did it, but still dont know how it's can help me
                $endgroup$
                – shay
                Dec 16 '18 at 14:08






              • 1




                $begingroup$
                SInce $f'(x)$ is always non-negative, $f$ is invreasing. Now, suppose that $f(x)=f(y)$ for some $x$ and some $y$ such that $x<y$. Then $f$ is constant in $[x,y]$ and therefore $f'$ is $0$ there. But $f'$ only has one single zero.
                $endgroup$
                – José Carlos Santos
                Dec 16 '18 at 14:10


















              • $begingroup$
                I did it, but still dont know how it's can help me
                $endgroup$
                – shay
                Dec 16 '18 at 14:08






              • 1




                $begingroup$
                SInce $f'(x)$ is always non-negative, $f$ is invreasing. Now, suppose that $f(x)=f(y)$ for some $x$ and some $y$ such that $x<y$. Then $f$ is constant in $[x,y]$ and therefore $f'$ is $0$ there. But $f'$ only has one single zero.
                $endgroup$
                – José Carlos Santos
                Dec 16 '18 at 14:10
















              $begingroup$
              I did it, but still dont know how it's can help me
              $endgroup$
              – shay
              Dec 16 '18 at 14:08




              $begingroup$
              I did it, but still dont know how it's can help me
              $endgroup$
              – shay
              Dec 16 '18 at 14:08




              1




              1




              $begingroup$
              SInce $f'(x)$ is always non-negative, $f$ is invreasing. Now, suppose that $f(x)=f(y)$ for some $x$ and some $y$ such that $x<y$. Then $f$ is constant in $[x,y]$ and therefore $f'$ is $0$ there. But $f'$ only has one single zero.
              $endgroup$
              – José Carlos Santos
              Dec 16 '18 at 14:10




              $begingroup$
              SInce $f'(x)$ is always non-negative, $f$ is invreasing. Now, suppose that $f(x)=f(y)$ for some $x$ and some $y$ such that $x<y$. Then $f$ is constant in $[x,y]$ and therefore $f'$ is $0$ there. But $f'$ only has one single zero.
              $endgroup$
              – José Carlos Santos
              Dec 16 '18 at 14:10











              2












              $begingroup$

              Here is an approach if you want to avoid derivatives. For $x neq 0$, we have
              $$frac{x^3}{x^2 + 1} = frac{1}{frac{1}{x} + frac{1}{x^3}}.$$



              The function $1/x + 1/x^3$ for $x > 0$ is the sum of two strictly decreasing functions, hence itself strictly decreasing and therefore one to one with range $(0, infty)$. Hence $1/(1/x + 1/x^3)$ is strictly increasing and one to one with range $(0, infty)$. For $x < 0$, notice $1/x + 1/x^3$ is an odd function, and therefore a similar argument shows $1/(1/x + 1/x^3)$ is one to one with range $(-infty, 0)$.



              In fact, this shows that if you add back the point $x = 0$, the function is a bijection since the function is zero here.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Here is an approach if you want to avoid derivatives. For $x neq 0$, we have
                $$frac{x^3}{x^2 + 1} = frac{1}{frac{1}{x} + frac{1}{x^3}}.$$



                The function $1/x + 1/x^3$ for $x > 0$ is the sum of two strictly decreasing functions, hence itself strictly decreasing and therefore one to one with range $(0, infty)$. Hence $1/(1/x + 1/x^3)$ is strictly increasing and one to one with range $(0, infty)$. For $x < 0$, notice $1/x + 1/x^3$ is an odd function, and therefore a similar argument shows $1/(1/x + 1/x^3)$ is one to one with range $(-infty, 0)$.



                In fact, this shows that if you add back the point $x = 0$, the function is a bijection since the function is zero here.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Here is an approach if you want to avoid derivatives. For $x neq 0$, we have
                  $$frac{x^3}{x^2 + 1} = frac{1}{frac{1}{x} + frac{1}{x^3}}.$$



                  The function $1/x + 1/x^3$ for $x > 0$ is the sum of two strictly decreasing functions, hence itself strictly decreasing and therefore one to one with range $(0, infty)$. Hence $1/(1/x + 1/x^3)$ is strictly increasing and one to one with range $(0, infty)$. For $x < 0$, notice $1/x + 1/x^3$ is an odd function, and therefore a similar argument shows $1/(1/x + 1/x^3)$ is one to one with range $(-infty, 0)$.



                  In fact, this shows that if you add back the point $x = 0$, the function is a bijection since the function is zero here.






                  share|cite|improve this answer









                  $endgroup$



                  Here is an approach if you want to avoid derivatives. For $x neq 0$, we have
                  $$frac{x^3}{x^2 + 1} = frac{1}{frac{1}{x} + frac{1}{x^3}}.$$



                  The function $1/x + 1/x^3$ for $x > 0$ is the sum of two strictly decreasing functions, hence itself strictly decreasing and therefore one to one with range $(0, infty)$. Hence $1/(1/x + 1/x^3)$ is strictly increasing and one to one with range $(0, infty)$. For $x < 0$, notice $1/x + 1/x^3$ is an odd function, and therefore a similar argument shows $1/(1/x + 1/x^3)$ is one to one with range $(-infty, 0)$.



                  In fact, this shows that if you add back the point $x = 0$, the function is a bijection since the function is zero here.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 16 '18 at 14:17









                  RileyRiley

                  1825




                  1825























                      1












                      $begingroup$

                      Suppose $f(a)=f(b)$: then
                      $$
                      a^3(b^2+1)=b^3(a^2+1)
                      $$
                      This implies
                      $$
                      a^3b^2-a^2b^3+a^3-b^3=0
                      $$
                      and so
                      $$
                      a^2b^2(a-b)+(a-b)(a^2+ab+b^2)=0
                      $$
                      Conclude.




                      If you know that $a^2+ab+b^2le0$ if and only if $a=b=0$, you're done.




                      With calculus,
                      $$
                      f'(x)=frac{3x^2(x^2+1)-2x^4}{(x^2+1)^2}
                      $$
                      The derivative only vanishes at $0$, so Rolle's theorem would only apply to an interval of the form $[a,b]$, with $a<0$ and $b>0$. Can $f(a)=f(b)$ in this case?






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        Suppose $f(a)=f(b)$: then
                        $$
                        a^3(b^2+1)=b^3(a^2+1)
                        $$
                        This implies
                        $$
                        a^3b^2-a^2b^3+a^3-b^3=0
                        $$
                        and so
                        $$
                        a^2b^2(a-b)+(a-b)(a^2+ab+b^2)=0
                        $$
                        Conclude.




                        If you know that $a^2+ab+b^2le0$ if and only if $a=b=0$, you're done.




                        With calculus,
                        $$
                        f'(x)=frac{3x^2(x^2+1)-2x^4}{(x^2+1)^2}
                        $$
                        The derivative only vanishes at $0$, so Rolle's theorem would only apply to an interval of the form $[a,b]$, with $a<0$ and $b>0$. Can $f(a)=f(b)$ in this case?






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Suppose $f(a)=f(b)$: then
                          $$
                          a^3(b^2+1)=b^3(a^2+1)
                          $$
                          This implies
                          $$
                          a^3b^2-a^2b^3+a^3-b^3=0
                          $$
                          and so
                          $$
                          a^2b^2(a-b)+(a-b)(a^2+ab+b^2)=0
                          $$
                          Conclude.




                          If you know that $a^2+ab+b^2le0$ if and only if $a=b=0$, you're done.




                          With calculus,
                          $$
                          f'(x)=frac{3x^2(x^2+1)-2x^4}{(x^2+1)^2}
                          $$
                          The derivative only vanishes at $0$, so Rolle's theorem would only apply to an interval of the form $[a,b]$, with $a<0$ and $b>0$. Can $f(a)=f(b)$ in this case?






                          share|cite|improve this answer









                          $endgroup$



                          Suppose $f(a)=f(b)$: then
                          $$
                          a^3(b^2+1)=b^3(a^2+1)
                          $$
                          This implies
                          $$
                          a^3b^2-a^2b^3+a^3-b^3=0
                          $$
                          and so
                          $$
                          a^2b^2(a-b)+(a-b)(a^2+ab+b^2)=0
                          $$
                          Conclude.




                          If you know that $a^2+ab+b^2le0$ if and only if $a=b=0$, you're done.




                          With calculus,
                          $$
                          f'(x)=frac{3x^2(x^2+1)-2x^4}{(x^2+1)^2}
                          $$
                          The derivative only vanishes at $0$, so Rolle's theorem would only apply to an interval of the form $[a,b]$, with $a<0$ and $b>0$. Can $f(a)=f(b)$ in this case?







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                          answered Dec 16 '18 at 14:09









                          egregegreg

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