Question

Найдите точку минимума функции y=log5(x^2-30x+249)+8

Answer 1

$y=log_5(x^2-30x+249)+8. \ x^2-30x+249 textgreater 0, \ D_{/4}=(-15)^2-249=225-249=-24 textless 0, \ a=1 textgreater 0, D textless 0, x^2-30x+249 textgreater 0 forall xin R, \ D_y=R. \ y'=frac{1}{(x^2-30x+249)ln5}cdot(2x-30+0)+0=frac{2x-30}{(x^2-30x+249)ln5}. \ left [ {{2x-30=0,} atop {x^2-30x+249=0,}} right. left [ {{x=15,} atop {xin varnothing;}} right. \ x=15, \ begin{array}{c|ccc}x&(-infty;15)&15&(15;+infty)\y'&-&0&+\y&searrow&min&nearrowend{array} \ x_{min}=15, y_{min}=log_5(15^2-30cdot15+249)+8=log_524+8approx10.$