6.1. Let $f(x)=x^{2}+p x+q$. It is known that the inequality $|f(x)|>\frac{1}{2}$ has no solutions on the interval $[1 ; 3]$. Find $\underbrace{f(f(\ldots f}_{2017}\left(\frac{3+\sqrt{7}}{2}\right)) \ldots)$. Round your answer to the nearest hundredth if necessary.
Answer: 0.18 (exact value: $\frac{3-\sqrt{7}}{2}$).

Solution. The condition of the problem means that for all $x \in [1; 3]$, the inequality $|f(x)| \leqslant \frac{1}{2}$ holds. By making the substitution $t = x + 2$, we obtain the inequality $\left|t^{2} + (p+4)t + (2p+q+4)\right| \leqslant \frac{1}{2}$, which holds for all $t \in [-1; 1]$. Consider the functions $g(t) = \left|t^{2} - (at + b)\right|$, $G(a, b) = \max_{t \in [-1; 1]} |g(t)|$, and the points $A = \left(-1, \frac{3}{2}\right)$, $B = \left(-1, \frac{1}{2}\right)$, $C = \left(0, \frac{1}{2}\right)$, $D = \left(0, -\frac{1}{2}\right)$, $E = \left(1, \frac{3}{2}\right)$, $F = \left(1, \frac{1}{2}\right)$ on the coordinate plane $Oty$ (see figure).

If the line $y = at + b$ does not intersect at least one of the segments $AB$, $CD$, $EF$, then $G(a, b) \geqslant \max(g(-1), g(0), g(1)) > \frac{1}{2}$. The only line that intersects all three segments is the line $y = \frac{1}{2}$. Therefore, for any $a$ and $b$, we have $G(a, b) \geqslant \frac{1}{2}$, and $G(a, b) = \frac{1}{2}$ only when $a = 0$, $b = \frac{1}{2}$. Consequently, in our case, $p + 4 = 0$, $2p + q + 4 = \frac{1}{2}$, from which we get $p = -4$, $q = \frac{7}{2}$. Thus, $f(x) = x^{2} - 4x + \frac{7}{2} = (x - 2)^{2} - \frac{1}{2}$, and we find $f\left(\frac{3+\sqrt{7}}{2}\right) = \frac{3-\sqrt{7}}{2}$, $f\left(f\left(\frac{3+\sqrt{7}}{2}\right)\right) = f\left(\frac{3-\sqrt{7}}{2}\right) = \frac{3+\sqrt{7}}{2}$, $\ldots$, $\underbrace{f(f(\ldots f}_{2017}\left(\frac{3+\sqrt{7}}{2}\right)) \ldots) = \frac{3-\sqrt{7}}{2} \approx 0.18$.

![](https://cdn.mathpix.com/cropped/2024_05_06_2ec34274f48265893886g-4.jpg?height=708&width=806&top_left_y=588&top_left_x=1119)