11.4. Given the function $f(x)=\left(1-x^{3}\right)^{-1 / 3}$. Find $f(f(f \ldots f(2018) \ldots))$ (the function $f$ is applied 2019 times)
Solution: Let's carry out equivalent transformations (for $x \neq 1$ and $x \neq 0$):

$$
\begin{gathered}
f(f(x))=\left(1-f(x)^{3}\right)^{-1 / 3}=\left(1-\left(1-x^{3}\right)^{-1}\right)^{-1 / 3}=\left(1-\frac{1}{1-x^{3}}\right)^{-1 / 3}= \\
=\left(\frac{-x^{3}}{1-x^{3}}\right)^{-1 / 3}=\frac{-x^{-1}}{\left(1-x^{3}\right)^{-1 / 3}}=-\frac{\left(1-x^{3}\right)^{1 / 3}}{x} \\
f(f(f(x)))=f\left(-\frac{\left(1-x^{3}\right)^{1 / 3}}{x}\right)=\left(1+\frac{1-x^{3}}{x^{3}}\right)^{-1 / 3}=\left(\frac{1}{x^{3}}\right)^{-1 / 3}=x
\end{gathered}
$$

Thus, applying the function three times does not change the value of the variable. Since 2019 is divisible by three, applying the function 2019 times will also not change this value.

Answer: 2018.

Recommendations for checking:

| is in the work | points |
| :--- | :--- |
| Correct and fully justified answer | 7 points |
| Periodicity of the operation of applying the function with a step of 3 is found; no conclusion is drawn from this, or an incorrect conclusion is drawn | 5 points |
| Only $f(f(x))$ is correctly found | 2 points |
| Answer without justification and/or incorrect answer | 0 points |