Решение задания 9:
Для функции $$y = \frac{1}{x^2 + 2}$$ и заданного интервала $$-3 \le x \le 3$$ с шагом 1, заполним таблицу значений.
1. x = -3: $$y = \frac{1}{(-3)^2 + 2} = \frac{1}{9 + 2} = \frac{1}{11}$$
2. x = -2: $$y = \frac{1}{(-2)^2 + 2} = \frac{1}{4 + 2} = \frac{1}{6}$$
3. x = -1: $$y = \frac{1}{(-1)^2 + 2} = \frac{1}{1 + 2} = \frac{1}{3}$$
4. x = 0: $$y = \frac{1}{(0)^2 + 2} = \frac{1}{0 + 2} = \frac{1}{2}$$
5. x = 1: $$y = \frac{1}{(1)^2 + 2} = \frac{1}{1 + 2} = \frac{1}{3}$$
6. x = 2: $$y = \frac{1}{(2)^2 + 2} = \frac{1}{4 + 2} = \frac{1}{6}$$
7. x = 3: $$y = \frac{1}{(3)^2 + 2} = \frac{1}{9 + 2} = \frac{1}{11}$$
Решение задания 10:
Для функции $$y = 3(x - 8)$$, заполним таблицу значений:
1. x = 8: $$y = 3(8 - 8) = 3(0) = 0$$
2. x = 10: $$y = 3(10 - 8) = 3(2) = 6$$
3. x = 10,2: $$y = 3(10,2 - 8) = 3(2,2) = 6,6$$
4. x = 16: $$y = 3(16 - 8) = 3(8) = 24$$
5. x = 18: $$y = 3(18 - 8) = 3(10) = 30$$
6. x = 18,1: $$y = 3(18,1 - 8) = 3(10,1) = 30,3$$
7. x = 20,4: $$y = 3(20,4 - 8) = 3(12,4) = 37,2$$