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Please help me how to proceed in this question is it correct if I use iteration formula and perform iteration....

taniya kapoor
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    Step 1: Read how to ask a good question. Step 2: Ask a good question. That includes formatting of formulas using mathjax, and (important) include what you have done so far. – P. Siehr Nov 09 '17 at 12:13
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    In any case, the problem has a typo. As stated, it's not true. To correct it, one way is to change the definition of $f$ to $$f(x)=3x^3 + 4x^2 + 4x + 2$$ Another possible correction is to leave $f$ as it is, but correct the iteration to $$x_{k+1} = -x_k - 2x_k^2 - {\small{\frac{3}{2}}}x_k^3 - {\small{\frac{1}{2}}}$$ – quasi Nov 09 '17 at 12:58
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    The version with $...+1$ factorizes as $(3x+1)(x^2+x+1)$ and has a complex pair of roots with real part $-0.5$ which is not very optimistic for the convergence from the given initial point. The version with $...+2$ obviously has not that factorizaton and roots close to $-0.74487$ and $ -0.294\pm 0.899i$. – Lutz Lehmann Nov 09 '17 at 19:45
  • Please type your questions (using MathJax) instead of posting links to pictures. – José Carlos Santos Nov 13 '17 at 16:42

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You have to compute the derivative of the iteration formula and find a bound on the interval around $-0.5$ where you can bound it by $1$. Then prove that that interval is mapped into itself and you have shown the convergence to the root.

For instance with $x_{n+1}=g(x_n)$, $$g(x)=-\frac13\left(1+\frac1{x^2+x+1}\right)$$ you get that $[-1,0]$ is mapped to $$ -\frac13\left(1+\frac1{([-\frac12,\frac12])^2+\frac34}\right)=\left[-\tfrac79,-\tfrac23\right] $$ and its derivative $$ g'(x)=\frac23\frac{x+\frac12}{\left((x+\frac12)^2+\frac23\right)^2} $$ has values smaller $\frac34$ so that the iteration of that formula is contractive.

Lutz Lehmann
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