2. Given $f(x)=-x^{4}+4 x^{3}-2 x^{2}-2 x+\frac{13}{9}$, $x \in[0,1]$. Provide the following conclusions:
(1) $f(x)>0$;
(2) $f(x)<0$;
(3) There exists $x_{0} \in[0,1)$, such that $f\left(x_{0}\right)=0$;
(4) There exists $x_{0} \in[0,1]$, such that $f\left(x_{0}\right)<0$.
Among them, the correct conclusion numbers are $\qquad$.
2. (1)
$$
\begin{array}{l}
f(x)=-x^{4}+4 x^{3}-2 x^{2}-2 x+\frac{13}{9} \\
=x^{3}(1-x)+3 x^{3}-3 x^{2}+(x-1)^{2}+\frac{4}{9} \\
=x^{3}(1-x)+(x-1)^{2}+\frac{3}{2} x^{3}+\frac{3}{2} x^{3}+\frac{4}{9}-3 x^{2} \\
\geqslant x^{3}(1-x)+(x-1)^{2}+3 \sqrt[3]{\frac{3}{2} x^{3} \cdot \frac{3}{2} x^{3} \cdot \frac{4}{9}}-3 x^{2} \\
=x^{3}(1-x)+(x-1)^{2} \geqslant 0 .
\end{array}
$$

The first equality holds when $x=\frac{2}{3}$, and the second equality holds when $x=1$.
Therefore, the equalities cannot hold simultaneously.
Thus, $f(x)>0$.