Question

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

Answer 1

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