Positive value of function and root

Question

Suppose $j,k\in\mathbb{N}$ and $1\leqslant j\leqslant k$.

The polynomial $$ f(x)=x^{2k}-\sum_{i=0}^{k-j-1}x^i $$ has exactly one positive root $x_0$ by Descartes' rule.

Do we also have that $$ f(x)>0\implies x>x_0? $$

Answer 1

Since $x_0$ is the only positive root of $f$, it is the largest one. Then, there is no sign change of $f$ on $(x_0, \infty)$ and since $\underset{x\to + \infty}{\lim} f = + \infty$. You may indeed assert that $ x > x_0 \Rightarrow f(x) > 0 $.

However $ f(x) > 0 \Rightarrow x > x_0 $ does not generally hold. Take $j=k=1$, then $f(x) = x^2$ and $f(-1)>0$ but $-1 < 0$.