Найдите наименьшее значение функции y=loq4(3/x^2+4x+12) на отрезке [-6;0]

[latex] y'= \frac{1}{\frac{3}{x^2+4x+12}*ln4 } = \frac{x^2+4x+12}{3ln4} \\ y(-6)=\frac{36-24+12}{3ln4}= \frac{24}{3ln4} \\ y(-5)=\frac{25-20+12}{3ln4}=\frac{17}{3ln4} \\ y(-4)=\frac{16-16+12}{3ln4}=\frac{12}{3ln4} \\ y(-3)=\frac{9-12+12}{3ln4}=\frac{9}{3ln4} \\ y(-2)=\frac{4-8+12}{3ln4}=\frac{8}{3ln4} \\ y(-1)=\frac{1-4+12}{3ln4}=\frac{9}{3ln4} \\ y(0)=\frac{0-0+12}{3ln4}=\frac{12}{3ln4} \\ ------------\\ x^2+4x+12=0\\ d=16-4*1*12=d<0 \\ Answer: min =\frac{8}{3ln4} [/latex]