Question

1. Решите неравенство: а). б). 2. Постройте график функции

Answer 1

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

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

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