Given $f(x)=x^3+x^2-x+2=0$ calculate $3$ iterations at$x_0=-2.4$

Question

Given $f(x)=x^3+x^2-x+2=0$

1. Find an interval[a,b] such that there exists $1$ and only root

2. Find a function $\phi(x)=x$

3. Find an interval where it is possible to use the simple iteration method and calculate $3$ iterations at initial value $x_0=-2.4$

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What I tried:

1. Using intermediate value theorem we have $f(-1)=3$ and $f(-3)=-13$ so there is at least $1$ root, I showed there this root is the only one using rolls theorem.. Assume by contradiction there are $2$ different roots $x_1,x_2$ such that $f(x_1)=f(x_2)=0$ the function is continuous and differentiable so there is a root $x_1 <c <x_2$ such that $f'(c)=0$ but $f'(x)=3x^2+2x-1=(x+1)(3x-1)=0$ so the roots are $c_1=-1$ and $c_2=\frac{1}{3}$ and there is only one of them in $[-3,-1]$

2. I added $x$ to both sides and got $x=x^3+x^2+2$ and then $\phi(x)=x^3+x^2+2$ we know that it converges if $max|\phi'(x)|<1$ so I solved $3x^2+2x<1$ and got $x \in (-1,\frac{1}{3})$but I don't know how to pick the interval from here is it $(-1,\frac{1}{3})$ or something else

3. When I first use $x_0=-2.4$ on $\phi(x)$ the function seems to diverge rather than converge..

$\phi(-2.4)=(-2.4)^3+(-2.4)^2-(-2.4)+2=-3.664$

$\phi(-3.664)=(-3.664)^3+(-3.664)^2-(-3.664)+2=-30.09992$

$\phi(-30.09992)=(-30.09992)^3+(-30.09992)^2-(-30.09992)+2=-26392.78612$

I am obviously doing something wrong but I can't figure it out

thanks for any help and tips

Answer 1

I think $\phi(x)=-(x^2-x+2)^{1/3}$ works.

We have $$\phi'(x)=\frac{-2x+1}{3(x^2-x+2)^{2/3}}$$ and $$\phi''(x)=\frac{2(x^2-x-5)}{9(x^2-x+2)^{5/3}}$$ So, we can say that for $-3\lt x\lt \frac{1-\sqrt{21}}{2}\approx -1.79$, $\phi''(x)\gt 0$ holds.

Since $$|\phi'(-3)|=\frac{14^{1/3}}{6}\lt 1$$ $$\bigg|\phi'\bigg(\frac{1-\sqrt{21}}{2}\bigg)\bigg|=\frac{7^{1/3}}{\sqrt{21}}\lt 1$$ we can say that, for $-3\lt x\lt \frac{1-\sqrt{21}}{2}\approx -1.79$, $|\phi'(x)|\lt 1$ holds.

Answer 2

Recall that, for convergence, we must have

$$|\phi'(x)|\le k \lt 1~~~\mbox{for all}~ x\in(a,b)$$

Does your choice of $\phi(x)$ satisfy that for the interval $(a, b) = (-3, -1)$ you found?

There are many ways to select candidate $\phi(x)$ functions and not all of them will work.

For example, we could have chosen

• $x = \phi_1(x) = x^3 + x^2 + 2$

• $x = \phi_2(x) = \pm \sqrt{x^3-x+2}$

• $x = \phi_3(x) = -\dfrac{2}{x^2+x-1}$

• $x = \phi_4(x) = (-x^2 + x -2)^{1/3}$

• $x = \phi_5(x) = x - \dfrac{x^3+x^2-x+2}{3x^2+2x-1}$

Try these and see what you get as iterates.

Do you recognize the last one? Which of these converge?