Question

1. Решите неравенство:
а). $( frac{1}{3}) ^{ x^{2} +2x}<( frac{1}{3}) ^{ 32-2x}$
б). $log_{8}( x^{2} -4x+3) geq 1$
2. Постройте график функции $y=log _{2} frac{ x^{2} -16}{x-4}$

Answer 1

$(frac{1}{3}) ^{x^{2}+2x}<(frac{1}{3})^{32-2x}, \ x^2+2x>32-2x, \ x^2+4x-32>0, \ x^2+4x-32=0, \ x_1=-8, x_2=4, \ (x+8)(x-4)>0, \ left [ {{x<-8,} atop {x>4;}} right. \ xin(-infty;-8)cup(4;+infty);$

$log_{8}(x^{2} -4x+3) geq 1, \ left { {{x^{2} -4x+3>0,} atop {x^{2} -4x+3 geq 8;}} right. \ x^{2} -4x+3=0, \ x_1=1, x_2=3, \ x^{2} -4x-5=0, \ x_1=-1, x_2=5, \ left { {{(x-1)(x-3)>0,} atop {(x+1)(x-5)geq0;}} right. left { {{ left[ {{x<1,} atop {x>3,}} right. } atop { left [ {{x leq -1,} atop {x geq 5;}} right. }} right. left [ {{x leq -1,} atop {x geq 5;}} right. \ xin(-infty;-1]cup[5;+infty);$

$y=log _{2} frac{x^{2}-16}{x-4}, \ 1) frac{x^{2}-16}{x-4}>0, \ frac{(x+4)(x-4)}{x-4}>0, \ x-4 neq 0,x neq 4, \ (x+4)(x-4)^2>0, \ x+4>0, \ x>-4, \ D_y=(-4;4)cup(4;+infty) \ 2) y=log _{2} (x+4), \ x+4=2^y, \ x=2^y-4, \ y=2^x-4, \ E_y=R; \ 3) f(-x)=log _{2} frac{(-x)^{2}-16}{-x-4}=-log _{2} frac{x^{2}-16}{x+4}, \ f(-x) neq f(x), f(-x) neq -f(x);$
$4)y=log _{2} (x+4), \ log _{2} (x+4)gtrless0, \ x+4gtrless1, \ xgtrless-3, \ x<-3, y<0, \ x>-3, y>0;$