Prove that $f(x) =0$ for at least one positive $x$

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Problem
Let $f$ be a polynomial of degree $n$ , such that first and last coefficients have opposite signs. Prove that $f(x) =0$ for at least one positive $x$.




I have an Inkling that i have to apply Bolzano's theorem .



Any hint or suggestion ?




calculus real-analysis




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  • 4




    Take the coefficient of the highest power to be 1 without loss of generality. Then $f(0) <0$ and $f(x) to infty$ as $x toinfty$. Hence there is some $x>0$ such that $f(x) >0$. The intermediate value theorem shows that there is a root in $(0,x)$.
    – copper.hat
    Jul 16 at 6:12










  • Please check whether i have understood it right or not.
    – blue boy
    Jul 16 at 6:37










  • Let Co<0 and Cn> 0 . f(0)>0 and f(x) > 0 for some x. Therefore f(c) =0 for some c between 0 and x.
    – blue boy
    Jul 16 at 6:39










  • I don't know you understood it, but copper's explanation is very clear.
    – Takahiro Waki
    Jul 16 at 11:29














up vote
2
down vote

favorite













Problem
Let $f$ be a polynomial of degree $n$ , such that first and last coefficients have opposite signs. Prove that $f(x) =0$ for at least one positive $x$.




I have an Inkling that i have to apply Bolzano's theorem .



Any hint or suggestion ?




calculus real-analysis




share|cite|improve this question

















  • 4




    Take the coefficient of the highest power to be 1 without loss of generality. Then $f(0) <0$ and $f(x) to infty$ as $x toinfty$. Hence there is some $x>0$ such that $f(x) >0$. The intermediate value theorem shows that there is a root in $(0,x)$.
    – copper.hat
    Jul 16 at 6:12










  • Please check whether i have understood it right or not.
    – blue boy
    Jul 16 at 6:37










  • Let Co<0 and Cn> 0 . f(0)>0 and f(x) > 0 for some x. Therefore f(c) =0 for some c between 0 and x.
    – blue boy
    Jul 16 at 6:39










  • I don't know you understood it, but copper's explanation is very clear.
    – Takahiro Waki
    Jul 16 at 11:29












up vote
2
down vote

favorite









up vote
2
down vote

favorite












Problem
Let $f$ be a polynomial of degree $n$ , such that first and last coefficients have opposite signs. Prove that $f(x) =0$ for at least one positive $x$.




I have an Inkling that i have to apply Bolzano's theorem .



Any hint or suggestion ?




calculus real-analysis




share|cite|improve this question














Problem
Let $f$ be a polynomial of degree $n$ , such that first and last coefficients have opposite signs. Prove that $f(x) =0$ for at least one positive $x$.




I have an Inkling that i have to apply Bolzano's theorem .



Any hint or suggestion ?





calculus real-analysis





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 16 at 6:11









Legoman

4,65421033




4,65421033









asked Jul 16 at 6:08









blue boy

562211




562211







  • 4




    Take the coefficient of the highest power to be 1 without loss of generality. Then $f(0) <0$ and $f(x) to infty$ as $x toinfty$. Hence there is some $x>0$ such that $f(x) >0$. The intermediate value theorem shows that there is a root in $(0,x)$.
    – copper.hat
    Jul 16 at 6:12










  • Please check whether i have understood it right or not.
    – blue boy
    Jul 16 at 6:37










  • Let Co<0 and Cn> 0 . f(0)>0 and f(x) > 0 for some x. Therefore f(c) =0 for some c between 0 and x.
    – blue boy
    Jul 16 at 6:39










  • I don't know you understood it, but copper's explanation is very clear.
    – Takahiro Waki
    Jul 16 at 11:29












  • 4




    Take the coefficient of the highest power to be 1 without loss of generality. Then $f(0) <0$ and $f(x) to infty$ as $x toinfty$. Hence there is some $x>0$ such that $f(x) >0$. The intermediate value theorem shows that there is a root in $(0,x)$.
    – copper.hat
    Jul 16 at 6:12










  • Please check whether i have understood it right or not.
    – blue boy
    Jul 16 at 6:37










  • Let Co<0 and Cn> 0 . f(0)>0 and f(x) > 0 for some x. Therefore f(c) =0 for some c between 0 and x.
    – blue boy
    Jul 16 at 6:39










  • I don't know you understood it, but copper's explanation is very clear.
    – Takahiro Waki
    Jul 16 at 11:29







4




4




Take the coefficient of the highest power to be 1 without loss of generality. Then $f(0) <0$ and $f(x) to infty$ as $x toinfty$. Hence there is some $x>0$ such that $f(x) >0$. The intermediate value theorem shows that there is a root in $(0,x)$.
– copper.hat
Jul 16 at 6:12




Take the coefficient of the highest power to be 1 without loss of generality. Then $f(0) <0$ and $f(x) to infty$ as $x toinfty$. Hence there is some $x>0$ such that $f(x) >0$. The intermediate value theorem shows that there is a root in $(0,x)$.
– copper.hat
Jul 16 at 6:12












Please check whether i have understood it right or not.
– blue boy
Jul 16 at 6:37




Please check whether i have understood it right or not.
– blue boy
Jul 16 at 6:37












Let Co<0 and Cn> 0 . f(0)>0 and f(x) > 0 for some x. Therefore f(c) =0 for some c between 0 and x.
– blue boy
Jul 16 at 6:39




Let Co<0 and Cn> 0 . f(0)>0 and f(x) > 0 for some x. Therefore f(c) =0 for some c between 0 and x.
– blue boy
Jul 16 at 6:39












I don't know you understood it, but copper's explanation is very clear.
– Takahiro Waki
Jul 16 at 11:29




I don't know you understood it, but copper's explanation is very clear.
– Takahiro Waki
Jul 16 at 11:29










2 Answers
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Hint: If $f(x)=a_0+a_1x+ldots +a_n x^n$, compute $f(0)$ and $f(x_0)$ for $x_0>nmax_kleft|fraca_ka_nright|$






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    Use Descartes' rule of sign and fundamental theorem of algebra. The existence of such $x$ will be guaranteed. Link: https://www.mathplanet.com/education/algebra-2/polynomial-functions/descartes-rule-of-sign






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      2 Answers
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      2 Answers
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      up vote
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      down vote













      Hint: If $f(x)=a_0+a_1x+ldots +a_n x^n$, compute $f(0)$ and $f(x_0)$ for $x_0>nmax_kleft|fraca_ka_nright|$






      share|cite|improve this answer

























        up vote
        2
        down vote













        Hint: If $f(x)=a_0+a_1x+ldots +a_n x^n$, compute $f(0)$ and $f(x_0)$ for $x_0>nmax_kleft|fraca_ka_nright|$






        share|cite|improve this answer























          up vote
          2
          down vote










          up vote
          2
          down vote









          Hint: If $f(x)=a_0+a_1x+ldots +a_n x^n$, compute $f(0)$ and $f(x_0)$ for $x_0>nmax_kleft|fraca_ka_nright|$






          share|cite|improve this answer













          Hint: If $f(x)=a_0+a_1x+ldots +a_n x^n$, compute $f(0)$ and $f(x_0)$ for $x_0>nmax_kleft|fraca_ka_nright|$







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 16 at 6:12









          Hagen von Eitzen

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          265k20258477




















              up vote
              0
              down vote













              Use Descartes' rule of sign and fundamental theorem of algebra. The existence of such $x$ will be guaranteed. Link: https://www.mathplanet.com/education/algebra-2/polynomial-functions/descartes-rule-of-sign






              share|cite|improve this answer

























                up vote
                0
                down vote













                Use Descartes' rule of sign and fundamental theorem of algebra. The existence of such $x$ will be guaranteed. Link: https://www.mathplanet.com/education/algebra-2/polynomial-functions/descartes-rule-of-sign






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Use Descartes' rule of sign and fundamental theorem of algebra. The existence of such $x$ will be guaranteed. Link: https://www.mathplanet.com/education/algebra-2/polynomial-functions/descartes-rule-of-sign






                  share|cite|improve this answer













                  Use Descartes' rule of sign and fundamental theorem of algebra. The existence of such $x$ will be guaranteed. Link: https://www.mathplanet.com/education/algebra-2/polynomial-functions/descartes-rule-of-sign







                  share|cite|improve this answer













                  share|cite|improve this answer



                  share|cite|improve this answer











                  answered Jul 16 at 6:49









                  Sujit Bhattacharyya

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