Application of Derivatives (Rolle's Theorem?)

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Let there be a cubic equation $f(x)=ax³+bx²+cx+d=0$, and the coefficients of the equation be related by $-a+b-c+d=3$ and $8a+4b+2c+d=6$. How can I show that the quadratic equation $3ax²+2bx+c-1$ has a root in $(-1,2)$?



I believe that if we apply Rolle's Theorem for the cubic and we show that $f(-1)=f(2)$ (already the cubic is continuous in $[-1,2]$ and differentiable in $(-1,2)$), we can prove that $f'(x)$ has a root in (-1,2). But the given quadratic isn't exactly equal to f'(x).



Can anybody help?




real-analysis derivatives polynomials cubic-equations rolles-theorem




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    up vote
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    down vote

    favorite












    Let there be a cubic equation $f(x)=ax³+bx²+cx+d=0$, and the coefficients of the equation be related by $-a+b-c+d=3$ and $8a+4b+2c+d=6$. How can I show that the quadratic equation $3ax²+2bx+c-1$ has a root in $(-1,2)$?



    I believe that if we apply Rolle's Theorem for the cubic and we show that $f(-1)=f(2)$ (already the cubic is continuous in $[-1,2]$ and differentiable in $(-1,2)$), we can prove that $f'(x)$ has a root in (-1,2). But the given quadratic isn't exactly equal to f'(x).



    Can anybody help?




    real-analysis derivatives polynomials cubic-equations rolles-theorem




    share|cite|improve this question























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Let there be a cubic equation $f(x)=ax³+bx²+cx+d=0$, and the coefficients of the equation be related by $-a+b-c+d=3$ and $8a+4b+2c+d=6$. How can I show that the quadratic equation $3ax²+2bx+c-1$ has a root in $(-1,2)$?



      I believe that if we apply Rolle's Theorem for the cubic and we show that $f(-1)=f(2)$ (already the cubic is continuous in $[-1,2]$ and differentiable in $(-1,2)$), we can prove that $f'(x)$ has a root in (-1,2). But the given quadratic isn't exactly equal to f'(x).



      Can anybody help?




      real-analysis derivatives polynomials cubic-equations rolles-theorem




      share|cite|improve this question













      Let there be a cubic equation $f(x)=ax³+bx²+cx+d=0$, and the coefficients of the equation be related by $-a+b-c+d=3$ and $8a+4b+2c+d=6$. How can I show that the quadratic equation $3ax²+2bx+c-1$ has a root in $(-1,2)$?



      I believe that if we apply Rolle's Theorem for the cubic and we show that $f(-1)=f(2)$ (already the cubic is continuous in $[-1,2]$ and differentiable in $(-1,2)$), we can prove that $f'(x)$ has a root in (-1,2). But the given quadratic isn't exactly equal to f'(x).



      Can anybody help?





      real-analysis derivatives polynomials cubic-equations rolles-theorem





      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited Aug 1 at 14:05









      Michael Rozenberg

      87.4k1577179




      87.4k1577179









      asked Aug 1 at 13:46









      Arka Seth

      73




      73




















          2 Answers
          2






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          up vote
          3
          down vote



          accepted










          Take $g(x)=f(x)-x$ and use Rolle.



          Because $g(-1)=g(2)=4.$






          share|cite|improve this answer





















          • Thanks a lot! It was really silly of me to have not considered/spotted this.
            – Arka Seth
            Aug 1 at 14:05










          • You are welcome!
            – Michael Rozenberg
            Aug 1 at 14:05

















          up vote
          2
          down vote













          By Lagrange's theorem, for some $xin(-1, 2)$:
          $$f(2)-f(-1)=(2-(-1))f'(x) implies f'(x)=1$$






          share|cite|improve this answer

















          • 1




            Thanks for the help!
            – Arka Seth
            Aug 1 at 14:06










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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          3
          down vote



          accepted










          Take $g(x)=f(x)-x$ and use Rolle.



          Because $g(-1)=g(2)=4.$






          share|cite|improve this answer





















          • Thanks a lot! It was really silly of me to have not considered/spotted this.
            – Arka Seth
            Aug 1 at 14:05










          • You are welcome!
            – Michael Rozenberg
            Aug 1 at 14:05














          up vote
          3
          down vote



          accepted










          Take $g(x)=f(x)-x$ and use Rolle.



          Because $g(-1)=g(2)=4.$






          share|cite|improve this answer





















          • Thanks a lot! It was really silly of me to have not considered/spotted this.
            – Arka Seth
            Aug 1 at 14:05










          • You are welcome!
            – Michael Rozenberg
            Aug 1 at 14:05












          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          Take $g(x)=f(x)-x$ and use Rolle.



          Because $g(-1)=g(2)=4.$






          share|cite|improve this answer













          Take $g(x)=f(x)-x$ and use Rolle.



          Because $g(-1)=g(2)=4.$







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Aug 1 at 13:56









          Michael Rozenberg

          87.4k1577179




          87.4k1577179











          • Thanks a lot! It was really silly of me to have not considered/spotted this.
            – Arka Seth
            Aug 1 at 14:05










          • You are welcome!
            – Michael Rozenberg
            Aug 1 at 14:05
















          • Thanks a lot! It was really silly of me to have not considered/spotted this.
            – Arka Seth
            Aug 1 at 14:05










          • You are welcome!
            – Michael Rozenberg
            Aug 1 at 14:05















          Thanks a lot! It was really silly of me to have not considered/spotted this.
          – Arka Seth
          Aug 1 at 14:05




          Thanks a lot! It was really silly of me to have not considered/spotted this.
          – Arka Seth
          Aug 1 at 14:05












          You are welcome!
          – Michael Rozenberg
          Aug 1 at 14:05




          You are welcome!
          – Michael Rozenberg
          Aug 1 at 14:05










          up vote
          2
          down vote













          By Lagrange's theorem, for some $xin(-1, 2)$:
          $$f(2)-f(-1)=(2-(-1))f'(x) implies f'(x)=1$$






          share|cite|improve this answer

















          • 1




            Thanks for the help!
            – Arka Seth
            Aug 1 at 14:06














          up vote
          2
          down vote













          By Lagrange's theorem, for some $xin(-1, 2)$:
          $$f(2)-f(-1)=(2-(-1))f'(x) implies f'(x)=1$$






          share|cite|improve this answer

















          • 1




            Thanks for the help!
            – Arka Seth
            Aug 1 at 14:06












          up vote
          2
          down vote










          up vote
          2
          down vote









          By Lagrange's theorem, for some $xin(-1, 2)$:
          $$f(2)-f(-1)=(2-(-1))f'(x) implies f'(x)=1$$






          share|cite|improve this answer













          By Lagrange's theorem, for some $xin(-1, 2)$:
          $$f(2)-f(-1)=(2-(-1))f'(x) implies f'(x)=1$$







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Aug 1 at 13:57









          Rumpelstiltskin

          1,378315




          1,378315







          • 1




            Thanks for the help!
            – Arka Seth
            Aug 1 at 14:06












          • 1




            Thanks for the help!
            – Arka Seth
            Aug 1 at 14:06







          1




          1




          Thanks for the help!
          – Arka Seth
          Aug 1 at 14:06




          Thanks for the help!
          – Arka Seth
          Aug 1 at 14:06












           

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