Task 3. Let $\min \frac{x-1}{x^{2}+1}=a ; \max \frac{x-1}{x^{2}+1}=b$. What are the minimum and maximum of the functions a) $\frac{x^{3}-1}{x^{6}+1}$; b) $\frac{x+1}{x^{2}+1}?$
Answer: a) $\min \frac{x^{3}-1}{x^{6}+1}=a ; \max \frac{x^{3}-1}{x^{6}+1}=b$; b) $\min \frac{x+1}{x^{2}+1}=-b ; \max \frac{x+1}{x^{2}+1}=-a$.

Solution. Let $\frac{x-1}{x^{2}+1}=f(x)$.

a) We have $\frac{x^{3}-1}{x^{6}+1}=f\left(x^{3}\right)$. The value $x^{3}$ runs through all numerical values, so $f\left(x^{3}\right)$ takes the same values as $f(x)$.

b) We have $f(-x)=\frac{-x-1}{x^{2}+1}=-\frac{x+1}{x^{2}+1}$, that is, $\frac{x+1}{x^{2}+1}=-f(-x)$, so this function takes values from $-b$ to $-a$.

Grading criteria: Full solution of part a) - 7 points. Solution of part a) using derivatives without considering the behavior of the function at infinity - 5 points. Full solution of part b) - 13 points. Solution of part b) using derivatives without considering the behavior of the function at infinity - 5 points. If max and min were taken as the values of the argument at the extremal points instead of the values of the function - 0 points.