Without using derivative, prove that the equation $$x^5-2x^4-3x^3-4x^2-5x-6=0$$ has unique positive real solution.
I tried, consider function $f: \mathbb{R} \rightarrow \mathbb{R}$ with $$f(x)=x^5-2x^4-3x^3-4x^2-5x-6.$$ We have $f$ is a continous function and $f(0)<0$. Another way, $$\displaystyle \lim_{x \to +\infty} f(x) = +\infty, $$ therefore there is exits a number $a >0$ such that $f(a) >0$.
Because, $f(0) \cdot f(a) <0$, therefore the equation $f(x) = 0$ has a solution $x_0 $ belongs to $(0,a)$.
Now I can not prove $x_0$ is unique solution. How can I do that without using derivative?
Ingeneral, how can I prove that the equation $$x^n = P_{n-1}(x),$$ where, $P_{n-1}(x)$ is a polynomial of $n-1$ degree and all positive real coefficients has unique positive real solution.
If $ x<0,$ always $f(x)<0$
$f(x)=(x^2-2x-3)(x^3-1)-3(x^2+x+1)$
$=((x-1)^2-4)(x^3-1)-3(x^2+x+1)$
at least x>3 we know.$f(4)=5*63-3*25>0$
In the area of $3<x<4 $ there is an unique solution.