1. It is known that $x+\frac{1}{x} \leqslant 4$. Find the range of the function

$$
f(x)=x^{3}+\frac{1}{x^{3}}
$$
Solution. Let $g(x)=x+\frac{1}{x}$. If $x>0$, then $g(x) \geqslant 2$. Together with the condition of the problem, we get that $2 \leqslant g(x) \leqslant 4$ for $x>0$. Now

$$
f(x)=\left(x+\frac{1}{x}\right)\left(x^{2}-1+\frac{1}{x^{2}}\right)=g(x)\left(\left(x+\frac{1}{x}\right)^{2}-2-1\right)=g(x)\left(g^{2}(x)-3\right)
$$

If $g(x) \geqslant 2$, then $g(x)\left(g^{2}(x)-3\right) \geqslant 2\left(2^{2}-3\right)=2$.

If $g(x) \leqslant 4$, then $g(x)\left(g^{2}(x)-3\right) \leqslant 4\left(4^{2}-3\right)=52$.

Thus, if $x>0$, then $f(x) \in[2 ; 52]$.

Note that both boundaries are achieved. The lower one at $x=1$, the upper one at such $x$ for which $g(x)=4$. It is found from the equation $x+1 / x=4$ and equals $2 \pm \sqrt{3}$.

Now let $x<0$. The functions $g(x)$ and $f(x)$ are odd, so if $x<0$, then $g(x) \leqslant-2, f(x) \leqslant-2$.

For all $M<0$, from the inequality $x<\sqrt[3]{M}$ it follows that $f(x)=x^{3}+1 / x^{3}<x^{3}<M$. This means that the function $f(x)$ is not bounded from below, i.e., any negative number less than -2 belongs to its range.

Answer. $f(x) \in[-\infty ;-2] \cup[2 ; 52]$.