Let $p(x)$ be a polynomial of degree $2n+1$ with real coefficients. then $p(x)$ has

(I) exactly $2n+1$ fixed points

(II) at least one fixed point

(III) at most one fixed point

(Iv) $n$ fixed points.

Let $p(x)=a_0+a_1 x+\dots a_{2n+1}x^{2n+1}$ for fixed points we have $p(x)=x$ i.e $a_0+(a_1-1) x+\dots a_{2n+1}x^{2n+1}=0$, so $(I)$ is true?

  • $\begingroup$ Why would $a_0+(a_1-1) x+\dots a_{2n+1}x^{2n+1}=0$ implies there are exactly $2n+1$ fixed points?? all roots of $a_0+(a_1-1) x+\dots a_{2n+1}x^{2n+1}=0$ " may not be real"... even though they are all real why would that imply they all are distinct? $\endgroup$ – user87543 Dec 28 '13 at 15:26


Make $p(x)=x$ we get $x^5+1=0\Rightarrow x=-1$ (If by fixed point you mean only real)

This tells there is only one fixed point which says :

first option is false as there is only one fixed point... (we should have $5$ for that to be true)

fourth option is false as there is only one fixed point.. (we should have $2$ for that to be true)

Take $p(x)=x^3-x$ consider $p(x)=x$ i.e., $x^3-x=x\Rightarrow x^3-2x=0\Rightarrow x(x^2-2)=0$

which has three fixed points.

So, Only thing you can say is :

Any odd degree polynomial with real coefficient will have at least one fixed point.

  • $\begingroup$ This is just a question with no "what i have done is".. So... :D $\endgroup$ – user87543 Dec 29 '13 at 13:31

HINT: Look at $p(x)=x^3+x+1$. $\quad$

  • $\begingroup$ Or even $p(x)=x+1$! $\endgroup$ – apt1002 Nov 14 '13 at 14:10

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