4. Given the function
$$
f(x)=x^{3}-6 x^{2}+17 x-5 \text {, }
$$

real numbers $a, b$ satisfy $f(a)=3, f(b)=23$. Then $a+b=$ $\qquad$
4.4 .

Notice,
$$
\begin{array}{l}
f(x)=x^{3}-6 x^{2}+17 x-5 \\
=(x-2)^{3}+5(x-2)+13 \text {. } \\
\text { Let } g(y)=y^{3}+5 y .
\end{array}
$$

Then $g(y)$ is an odd function and monotonically increasing.
And $f(a)=(a-2)^{3}+5(a-2)+13=3$, so, $g(a-2)=-10$.
Also, $f(b)=(b-2)^{3}+5(b-2)+13=23$, then $g(b-2)=10$.
Thus $g(a-2)=-g(b-2)=g(2-b)$
$$
\Rightarrow a-2=2-b \Rightarrow a+b=4
$$