Let $g(x)=x^{10000}+x^{100}-1$. Obviously,$g(x)$ is continuous and increasing in $[0,+\infty)$, because $g(0)=-1 \ and \ g(1)=1$, so $x_0\in (0,1)$ if $g(x_0)=0$
Added:
For a function in the form of $f(x)=x^n, n\in \mathbb N$,we can see that $f(x)$ is increasing on$[0,+\infty)$.
For any $x>y, x,y\in (0,+\infty)$, we have $f(y)>0$, and $\frac{f(x)}{f(y)}=(\frac{x}{y})^n>1$, since $\frac{x}{y}>1$, therefore, we getf(x)>f(y). Also, for any$x\in(0,+\infty)$,f(x)>f(0)=0, so we can say $f(x)$ is increasing on $[0,+\infty)$.
Of course, the continuity tells you that there is at least one point $x_0$ satisfying $g(x_0)=0$, however, my conclusion is stronger: there is only one point $x_0$ satisfying $g(x_0)=0$.