How do I know there are no more then few solutions, using, for instance Rolle? For instance, show that $x^2-2x+1=0$ has one solution only. Let us define, firstly, $f(x)=x^2-2x+1$

Formally, still, is has at least one solution, that is: $x=1$. How do I show there aren't more, using the theorem of continuity, Rolle's, etc? Because I might encounter more complicated parametric equations and it wouldn't be that easy to look and conclude immediately. I was told I should use Rolle's theorem. Okay, suppose there is one more solution, $x=c$. Then $f(1)=f(c)$ meaning there is $1<k<c$ (or $c<k<1$) such that $f'(k)=2k-2=0 \Rightarrow K=1\Rightarrow$ contradiction?? I am not quite sure. Would really appreciate your help.

How can I know there aren'y more points in which the derivative vanishes?


Between any two roots of $f(x)$ lies one root of $f^{'}(x)$ .Now $x=1$ is a root of $f(x)$

If it has another root say $c$ then $f^{'}(x)$ would vanish at some point $y\in (x,c)$ or $(c,x)$

But $f^{'}(x)$ vanishes only at $1$.hence no other root of $f(x)$

  • $\begingroup$ Which brings me to that: if I have unequivocally found few points in which $f'=0$, they are probably the only one? Assuming the function doesn't have parts where you can't derive roots... $\endgroup$ – Meitar Mar 1 '15 at 13:52
  • $\begingroup$ cant get what are you saying $\endgroup$ – Learnmore Mar 1 '15 at 13:57

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