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If I have some cubic equation $f(x)$, and I need to find how many solutions $f(x)$ has. $f'(x)$ has two zeros, does it state that $f(x)$ has $3$ solutions by Rolle's theorem?

$$f(x)= x^3+2x^2-7x+1$$

TheHolyJoker
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Veronika Kovaleva
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    Well, $p(x)=x(x+1)(x-1)+3$ has only one root but $p'(x)$ has two roots. – lulu Jan 15 '20 at 16:09
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    Rolle's Theorem only tells you for any pair of zeroes, there is (at least) one point in between where $f'=0$. – Paul Jan 15 '20 at 16:12
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    You should check whether the extrema, i.e. the points for which $f'(x)=0$ but $f''(x) \neq 0$, lie above or below the $x$-axis. $f(x)$ is not odd by the way. – bsbb4 Jan 15 '20 at 16:12
  • For what domain of x? – twentyeightknots Jan 15 '20 at 16:15
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    You can check that $f(-\infty)<0$, $f(0)=1>0$, $f(1)=-3<0$, $f(+\infty)>0$ and $f(-\infty)<0$. Therefore, the intermediate value theorem tells you the existence of a root between each consecutive pair of those points. – MoonLightSyzygy Jan 15 '20 at 16:17
  • @Paul I can't use the theorem backwards? So if I have 2 point where f'=0 , then I have at least 3 point where f=0? – Veronika Kovaleva Jan 15 '20 at 16:17
  • @VeronikaKovaleva Unfortunately, not. Consider $f(x) = 5 + \sin x$, this has an infinite amount of points with $f'(x)=0$ but no roots at all... – gt6989b Jan 15 '20 at 16:23

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If $f'(x )=0$ for $x=\alpha , \beta$ and $$f(\alpha) \cdot f(\beta) \le 0$$ Then $f(x)=0$ has three real roots.

Gabrielek
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mathsdiscussion.com
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  • If the two values are roots, then the function product of them must be zero. Where did the $<=$ come from? – NoChance Jul 16 '23 at 10:03