9.1.1. (12 points) Find the minimum value of the function $f(x)=\operatorname{tg}^{2} x+3 \operatorname{tg} x+6 \operatorname{ctg} x+4 \operatorname{ctg}^{2} x-1$ on the interval $\left(0 ; \frac{\pi}{2}\right)$.
Answer: $3+6 \sqrt{2} \approx 11.49$.

Solution. For $x>0$, we have $t=\operatorname{tg} x+2 \operatorname{ctg} x \geqslant 2 \sqrt{2}$ (equality is achieved at $x=\operatorname{arctg} \sqrt{2}$), and $f(x)=(\operatorname{tg} x+2 \operatorname{ctg} x)^{2}-3(\operatorname{tg} x+2 \operatorname{ctg} x)-5$. Therefore, we need to find the minimum value of the function $g(t)=t^{2}-3 t-5$ for $t \geqslant 2 \sqrt{2}$. The function $g(t)$ is increasing for $t \geqslant \frac{3}{2}$, and since $2 \sqrt{2} \geqslant \frac{3}{2}$, the desired minimum value is $g(2 \sqrt{2})=3+6 \sqrt{2} \approx 11.49$.