9.4. It is known that the values of the quadratic trinomial $a x^{2}+b x+c$ on the interval $[-1,1]$ do not exceed 1 in absolute value. Find the maximum possible value of the sum $|a|+|b|+|c|$. Answer. 3.
Solution. Substituting the values $x=0,1,-1$ from the interval $[-1,1]$ into the polynomial $a x^{2}+b x+c$, we obtain three inequalities: $-1 \leq c \leq 1$, $-1 \leq a+b+c \leq 1$, and $-1 \leq a-b+c \leq 1$. Adding the second and third inequalities, we also get $-1 \leq a+c \leq 1$. Subtracting the second from the third (they are double and symmetric!), we have $-1 \leq b \leq 1$. Subtracting $-1 \leq c \leq 1$ from $-1 \leq a+c \leq 1$, we get $-2 \leq a \leq 2$. Due to the symmetry of the problem's conditions with respect to multiplication by -1, we can assume the coefficient $a$ is positive. If $b, c \geq 0$, then $|a|+|b|+|c|=$ $a+b+c \leq 1$ as proven. If $b<0, c \geq 0$, then $|a|+|b|+|c|=a-b+c \leq 1$, as proven. If $b \geq 0, c<0$, then $|a|+|b|+|c|=a+b-c=(a+b+c)-2 c \leq 1+2=3$. If $b, c<0$, then $|a|+|b|+|c|=a-b-c=(a-b+c)-2 c \leq 1+2=3$. Thus, the sum $|a|+|b|+|c|$ under the conditions of the problem does not exceed 3.

The value 3 is achieved, for example, on the polynomial $f(x)=2 x^{2}-1$: its minimum value is reached inside the interval at the vertex of the parabola when $x=0$, and its maximum values are reached at the ends of the interval when $x=1,-1$.

Grading criteria. Only the boundaries of the coefficients of the equation and their sums from statements 1-3 are found: 2-3 points. The estimate $|a|+|b|+|c| \leq 3$ is proven: 5 points. An example is provided and justified where this boundary is achieved: 2 points. If the example is provided without justification: minus 1 point. Any incorrect boundary: 0 points.