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To do that, lets take $$f(x):=x^{15} + 7x^3-5,$$ we see that $f$ is continuous.

  1. I know that we have one real solution: $$f(0)=-5 \wedge f(1)=3\overset{IVT}{\Rightarrow}\exists c\in(0,1):f(c)=0$$
  2. Now when I need to show that it's the only real root. Let's suppose that there are two solution $c,x_2\in\mathbb{R}$, then $f(c)=f(x_2)=0$. Now from Rolle's theorem we get that $\exists m$ between $c,x_2$, such that $f'(c)=0$. But $$f'(x)=15x^{14}+21x^2\geq 0\hspace{3mm}\forall{x}\in\mathbb{R}$$. If $x_2<c$, that would mean that $x_2<m<c$, knowing that $m=0$, we get $x_2<0$, which is impossible: $$f(x)<0\hspace{3mm}\forall x\in(-\infty,0)$$. Now if $x_2>c$, we get that $c<m<x_2$. Again knowing $m=0$, we get $c<0$, which is impossible, because $c\in(0,1)$. So $c$ is the only real solution.

I don't know if the second part of my proof is correct. Is that a contradiction? How should the proof go using IVT and Rolle's theorem?

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  • $\begingroup$ I know this is an analysis exercise, but it's worth noting in passing that this can also be determined by figuring out the sign of the discriminant of $f$. $\endgroup$ – Kaj Hansen Oct 30 '17 at 19:32
  • $\begingroup$ I think it's obvious without Rolle. $\endgroup$ – Michael Rozenberg Oct 30 '17 at 19:35
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HINT: we have $$f(0)=-5<0$$ and $$f(1)=3>0$$ and $$f'(x)=15x^{14}+21x^2\geq 0$$

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$g(x)=x^5+7x-5$ has a unique real root since it is an odd-degree polynomial and an increasing function. $x\mapsto x^3$ is a bijective map, hence $f(x)=g(x^3)$ has a unique real root as well.

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You can come to this conclusion more easily by Descartes rule of signs.

The atmost positive roots of f(x) is the number of sign changes in f(x) and atmost negative roots is number of sign changes in f(-x).

Let f(x) be a polynomial with real coefficients. Let s be the number of sign changes in the sequence :that is, delete the terms of the sequence that are 0 and let s be the number of pairs of consecutive terms in the remaining sequence that have opposite signs. Let p be the number of positive roots of (counted with multiplicity). Then s-p is a nonnegative even number.

The number of sign changes in f(x) is 1.That is s=1.For s-p to be a nonegative even number p=1. The number of sign changes is f(-x) is 0.

So the total number of real roots should be 1.

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