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Let $ f \in C^2([a,b], R)$ and

$$\phi(x)=x-\frac{f(x)}{f'(x)}$$

Such that

1.$ f'(x)\neq 0, \forall x \in [a,b]$

  1. $\exists \alpha \in [a,b]: f(\alpha)=0$

Show that $\phi'(x) = 0$

My attempt:

$$\phi'(x) = 1 -\frac{f'(x)f'(x) - f(x)f''(x)}{(f'(x))^2} = 0 \iff f(x)f''(x)=0$$

I don't see why this must be truth, using the hypothesis.

tinlyx
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Giiovanna
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    Choosing $f(x)=x^3+x$ over the interval $[-1,1]$ gives a counterexample since $f'(x)=3x^2+1>0$ and $f(0)=0$ but $\phi(x)=\dfrac{2x^3}{3x^2+1}$ is not constant and so $\phi'(x)$ doesn't vanish everywhere. Should it be $\phi'(\alpha)=0$? That would follow directly from your last equation. – Semiclassical Sep 11 '14 at 20:12
  • I think the professor meant $\phi'(\alpha) = 0$. I think $\phi'(x) = 0$ is too strong. – Isomorphism Sep 11 '14 at 20:14
  • Well, $\phi'(\alpha)=0$ makes sence, but I dont see why to ask to prove it. The subject is about successive aproximation method and I think it should be used to prove the convergence of the method using the pho goven by the newton's method – Giiovanna Sep 11 '14 at 20:23
  • @Semiclassical Indeed your $\phi'(x)$ does not vanish anywhere in the interval (or indeed anywhere in the reals) except $x=0$. – almagest Sep 11 '14 at 20:26

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