Question

Срочно помогите по математике.

#1
Пусть f(x) = 7-x^2
найдите:
a) f(0)
б) f(1)-f(-1)
в) f(1-√2)+f(1+√2)

#2
Найдите область определения функции:
а) y=√2x-18
б) y=2/√x^2+x-20
в) у=log2(-4+2x)

#3 Постройте график функции y=|x^2+2|x|-8| и укажите:
А) область определения
Б) множество значений
В) Нули функции
Г) наибольшее и наименьшее значение

Answer 1

Ответ:

$\#1.\ a)\ 7,\ b)\ 0, \ c)\ 8 \\\#2.\ a)\ x\ c\ [9; +\infty),\ b)\ x\ c\ (-\infty;\ -5)U(4;\ +\infty),\ c)\ x\ c\ (2;\ +\infty)\\\#3. a) x\ c\ R,\ x\ c\ (-\infty;\ +\infty),\ b)y\ c\ [0;\ +\infty),\\.\ \ \ \ c)y = 0, \ x = \{ -2;\ 2;\},\ d)y_{min} = 0, \ y_{max} = +\infty$

Пошаговое объяснение:

$\#1. f(x)=7-x^2\\a) f(0) = 7 - 0^2 = 7\\b) f(1)-f(-1) = (7 - 1^2) - (7 - (-1)^2) = 6-6=0\\c)f(1-\sqrt(2)) + f(1+\sqrt2) = (7 - (1-\sqrt2)^2) +(7 - (1+\sqrt2)^2) =\\=7-(1-2\sqrt2+2) + 7 - (1+2\sqrt2+2) = 14-1-1-2-2+2\sqrt2-2\sqrt2 = 8\\\\\#2\\a)y = \sqrt{2x-18}\\2x-18 \geq 0\\2x \geq 18\\x\geq 9\\x\ c\ [\ 9;\ +\infty)\\b) y = \frac{2}{\sqrt{x^2+x-20}} \\\left \{ {{\sqrt{x^2+x-20} \neq 0} \atop {x^2+x-20\geq 0}} \right. \\x^2+x-20> 0\\D = 1-4*1*(-20)=81=9^2\\x_1 = \frac{-1+9}{2} =4$

$x_2 = \frac{-1-9}{2} =-5\\(x-4)(x+5)>0\\\\|\ \ \ \ + \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\\----o-------o---->x\\|\ \ \ \ \ \ \ -5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4\\x\ c\ (-\infty;\ -5)U(4;\ +\infty)\\c) y =log_2(-4+2x)\\-4+2x > 0\\2x > 4\\x > 2\\x\ c\ (2; \ +\infty)$

$\#3.\ y = |x^2 + 2|x| - 8|\\\left \{ {{y =|x^2 + 2x - 8|\ \ \ \ \ \ x \ \geq\ 0} \atop {y = |x^2 - 2x - 8|\ \ \ \ \ \ x \ <\ 0}} \right. \\\\x^2 + 2x - 8 \geq 0\\---------\\D = 4+32 = 6^2\\x_1 = \frac{-2 + 6}{2} = 2\\x_2 = \frac{-2 - 6}{2} = -4\\---------\\(x-2)(x+4) \geq 0\\\\|\ \ \ \ + \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\\----.-------.---->x\\|\ \ \ \ \ \ \ -4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\\\\x\ c\ (-\infty;\ -4]U[2;\ +\infty)$

$x^2 - 2x - 8 \geq 0\\---------\\D = 4+32 = 6^2\\x_1 = \frac{2 + 6}{2} = 4\\x_2 = \frac{2 - 6}{2} = -2\\---------\\(x-4)(x+2) \geq 0\\\\|\ \ \ \ + \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\\----.-------.---->x\\|\ \ \ \ \ \ \ -2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4\\\\x\ c\ (-\infty;\ -2]U[4;\ +\infty)$

$y = x^2 + 2x - 8,\ \ \ \ \ x\ c\ [2;\ +\infty)\\y = -x^2-2x+8, \ \ \ x\ c\ [0;\ 2)\\y = -x^2+2x+8,\ \ \ x\ c\ (-2;\ 0)\\y = x^2-2x-8, \ \ \ \ \ x\ c\ (-\infty;\ -2]$

$a) x\ c\ R,\ x\ c\ (-\infty;\ +\infty)\\b)y\ c\ [0;\ +\infty)\\c)y = 0, \ x = \{ -2;\ 2;\}\\d)y_{min} = 0, \ y_{max} = +\infty$