Show that $x^{15} + 7x^3-5=0$ has exactly one real solution, using IVT and Rolle's theorem.

Question

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?

Answer 1

HINT: we have $$f(0)=-5<0$$ and $$f(1)=3>0$$ and $$f'(x)=15x^{14}+21x^2\geq 0$$

Answer 2

$g(x)=x^5+7x-5$ 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.