5. Given that $f(x)$ is a linear function, and $f(f[\underbrace{f \cdots f(x)]} \geqslant 1024 x+1023$, find $f(x)$.
5. Let $f(x)=a x+b(a \neq 0)$, and denote $j\{f[\underbrace{f \cdots f(x)}_{n \approx x}]\}=f^{(n)}(x)$, then
$$
\begin{array}{c}
f^{(2)}(x)=f[f(x)]=a(a x+b)+b=a^{2} x+b(a+1), \\
f^{(3)}(x)=f\{f[f(x)]\}=a\left[a^{2} x+b(a+1)\right]+b=a^{3} x+b\left(a^{2}+a+1\right),
\end{array}
$$

By analogy, we have: $f^{(10)}(x)=a^{10} x+b\left(a^{9}+a^{8}+\cdots+a+1\right)=a^{10} x+\frac{b\left(1-a^{10}\right)}{1-a}$,
Given the condition: $a^{10} x+\frac{b\left(1-a^{30}\right)}{1-a} \geqslant 1024 x+1023$ for $x \in \mathbf{R}$. Then
$$
a^{10}=1024 \text { and } \frac{b\left(1-a^{30}\right)}{1-a} \geqslant 1023,
$$

Thus, $a=2, b \geqslant 1$ or $a=-2, b \leqslant-3$,
Therefore, $f(x)=2 x+b(b \geqslant 1)$ or $f(x)=-2 x+b(b \leqslant-3)$.