Real Root of a Polynomial on a closed interval












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      can you give me some tips to solve this question? enter image description here






      real-analysis analysis







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      asked Nov 20 '18 at 21:13









      bebe

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          I think the simplest way of looking at this would be to construct a second polynomial $Q(x)$ that is, essentially, the antiderivative of $P(x)$ (with a constant term of $C=0$). (Depending on where you are in your coursework, you don't have to say it's the antiderivative to $P$, just say "let $Q(x)$ be ..." It's equally valid.)



          Thus, $Q'(x) = P(x)$. From there, apply the mean value theorem to $Q$ on the interval $[0,1]$.



          That should be a sufficient nudge forward.






          share|cite|improve this answer





























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            Hint: Consider the polynomial
            begin{align}
            f(x) = C_0 x+frac{C_1}{2}x^2+ldots+frac{C_n}{n+1}x^{n+1}
            end{align}
            and use Rolle's theorem.






            share|cite|improve this answer





















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              2 Answers
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              active

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              2 Answers
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              active

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              1














              I think the simplest way of looking at this would be to construct a second polynomial $Q(x)$ that is, essentially, the antiderivative of $P(x)$ (with a constant term of $C=0$). (Depending on where you are in your coursework, you don't have to say it's the antiderivative to $P$, just say "let $Q(x)$ be ..." It's equally valid.)



              Thus, $Q'(x) = P(x)$. From there, apply the mean value theorem to $Q$ on the interval $[0,1]$.



              That should be a sufficient nudge forward.






              share|cite|improve this answer


























                1














                I think the simplest way of looking at this would be to construct a second polynomial $Q(x)$ that is, essentially, the antiderivative of $P(x)$ (with a constant term of $C=0$). (Depending on where you are in your coursework, you don't have to say it's the antiderivative to $P$, just say "let $Q(x)$ be ..." It's equally valid.)



                Thus, $Q'(x) = P(x)$. From there, apply the mean value theorem to $Q$ on the interval $[0,1]$.



                That should be a sufficient nudge forward.






                share|cite|improve this answer
























                  1












                  1








                  1






                  I think the simplest way of looking at this would be to construct a second polynomial $Q(x)$ that is, essentially, the antiderivative of $P(x)$ (with a constant term of $C=0$). (Depending on where you are in your coursework, you don't have to say it's the antiderivative to $P$, just say "let $Q(x)$ be ..." It's equally valid.)



                  Thus, $Q'(x) = P(x)$. From there, apply the mean value theorem to $Q$ on the interval $[0,1]$.



                  That should be a sufficient nudge forward.






                  share|cite|improve this answer












                  I think the simplest way of looking at this would be to construct a second polynomial $Q(x)$ that is, essentially, the antiderivative of $P(x)$ (with a constant term of $C=0$). (Depending on where you are in your coursework, you don't have to say it's the antiderivative to $P$, just say "let $Q(x)$ be ..." It's equally valid.)



                  Thus, $Q'(x) = P(x)$. From there, apply the mean value theorem to $Q$ on the interval $[0,1]$.



                  That should be a sufficient nudge forward.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 20 '18 at 21:19









                  Eevee Trainer

                  4,7201634




                  4,7201634























                      1














                      Hint: Consider the polynomial
                      begin{align}
                      f(x) = C_0 x+frac{C_1}{2}x^2+ldots+frac{C_n}{n+1}x^{n+1}
                      end{align}
                      and use Rolle's theorem.






                      share|cite|improve this answer


























                        1














                        Hint: Consider the polynomial
                        begin{align}
                        f(x) = C_0 x+frac{C_1}{2}x^2+ldots+frac{C_n}{n+1}x^{n+1}
                        end{align}
                        and use Rolle's theorem.






                        share|cite|improve this answer
























                          1












                          1








                          1






                          Hint: Consider the polynomial
                          begin{align}
                          f(x) = C_0 x+frac{C_1}{2}x^2+ldots+frac{C_n}{n+1}x^{n+1}
                          end{align}
                          and use Rolle's theorem.






                          share|cite|improve this answer












                          Hint: Consider the polynomial
                          begin{align}
                          f(x) = C_0 x+frac{C_1}{2}x^2+ldots+frac{C_n}{n+1}x^{n+1}
                          end{align}
                          and use Rolle's theorem.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 20 '18 at 21:19









                          Jacky Chong

                          17.8k21128




                          17.8k21128






























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