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I have to show that the function: $f(x) = x^3 + x^2 - 1$ has a unique positive root. It’s for numerical methods, would anyone have any tips? I’ve been trying to use fixed point methods but I’m not getting anywhere.

  • You can study the growth intervals of $f$ – Tito Eliatron Nov 13 '20 at 18:53
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    In this case, Decartes' Rule of Signs does it. – GEdgar Nov 13 '20 at 18:53
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    $f(0)=-1<0$ and $f$ has the leading coefficient positive so it eventually becomes positive - thus has at least one zero as per Intermediate Value Theorem. If it had two different zeros, then its derivative $f'(x)=3x^2+2x$ would have a zero "between" those two zeros as per Rolle's theorem; however, $3x^2+2x>0$ for $x>0$. –  Nov 13 '20 at 18:53

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Note that the number of sign changes between nonzero coefficients is $1$. Thus, by Descartes' Rule of Signs, we have that the number of positive real roots is either $1$, or an even number less than $1$. However, since we cannot have a negative number of roots, there must be $\boxed{1\text{ positive root.}}$

Joshua Wang
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You have $$f'(x)=3x^2+2x=x(3x+2)>0$$ if $x>0$ so that $f$ is increasing on $[0,\infty[$. Moreover $f(0)=-1<0$ and $f(1)=1>0$. By the intermediate value theorem there is a root on $[0,1]$ and by monotonicity it is the only positive root.

Olivier Moschetta
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