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Prove that $x^3+x+1$ has a root on $[-1,1]$ using the function $g(x)=-{x^2+x\over x^2+x+1}$ and the Fixed Point Iteration Theorem.

Another question asked the same about $g(x)=-{1\over x^2+1}$ which was immediate. By the Fixed Point Iteration Theorem, $g(x)$ has a fixed point $x$ on the interval, and setting $g(x)=x$, one arrives at the desired root of the above expression. Here I am a little challenged. I arrived at the following equality: $$-x-{x+1\over x^2+x+1}=-{x^2+x\over x^2+x+1}$$. Should I define a new variable to be the LHS? How can I make sure all is preserved? I am almost certain the LHS is monotonic which guarantees injectivity. As for Surjectivity, I am not sure. I could use help. Update: Under $[-1,1]$, the LHS is sent to $[-1{2\over 3},1]$. Assuming monotone, it is also a one-to-one correspondence. It still not so assured in my view.

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    $\begingroup$ Easier might be to notice that $x^3+x+1$ is continuous and negative at $x=-1$ and positive at $x=1$. $\endgroup$ – robjohn Aug 25 '16 at 21:45
  • $\begingroup$ @robjohn I have noticed it, but I am required to practice the specif method, as I was stating "using". $\endgroup$ – Meitar Aug 25 '16 at 21:50
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    $\begingroup$ I know what fixed point iteration is, but what, exactly, is this theorem you mention? $\endgroup$ – Arthur Aug 25 '16 at 21:52
  • $\begingroup$ @Arthur I feel you. $\endgroup$ – ÍgjøgnumMeg Aug 25 '16 at 22:20
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    $\begingroup$ I do not see that the given $g$ gives a fixed point equation for the roots of the given polynomial, $x^3+x^2+x=-(x^2+x)$ is equivalent to $x^3+2x^2+2x=0$ $\endgroup$ – LutzL Aug 25 '16 at 23:06

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