Вопрос:

Fill in the table with the values of the functions for the given inputs.

Ответ:

Решение:

Заполним таблицу, подставляя значения \( x \) в выражения для \( y = f(x) \) и вычисляя значения функции.

\( y=f(x) \)\( f(2) \)\( f(0) \)\( f(-1) \)
1)\( \frac{3}{x+2} \)\( \frac{3}{2+2} = \frac{3}{4} \)\( \frac{3}{0+2} = \frac{3}{2} \)\( \frac{3}{-1+2} = 3 \)
2)\( \frac{4}{x-3} \)\( \frac{4}{2-3} = -4 \)\( \frac{4}{0-3} = -\frac{4}{3} \)\( \frac{4}{-1-3} = -1 \)
3)\( \frac{6}{x-5} \)\( \frac{6}{2-5} = -2 \)\( \frac{6}{0-5} = -\frac{6}{5} \)\( \frac{6}{-1-5} = -1 \)
4)\( \frac{-12}{x+4} \)\( \frac{-12}{2+4} = -2 \)\( \frac{-12}{0+4} = -3 \)\( \frac{-12}{-1+4} = -4 \)
5)\( \frac{-9}{1-2x} \)\( \frac{-9}{1-2(2)} = \frac{-9}{1-4} = 3 \)\( \frac{-9}{1-2(0)} = -9 \)\( \frac{-9}{1-2(-1)} = \frac{-9}{1+2} = -3 \)
6)\( y=x^2 \)\( 2^2 = 4 \)\( 0^2 = 0 \)\( (-1)^2 = 1 \)
7)\( y=x^2+3 \)\( 2^2+3 = 7 \)\( 0^2+3 = 3 \)\( (-1)^2+3 = 4 \)
8)\( y=30-x^2 \)\( 30-2^2 = 26 \)\( 30-0^2 = 30 \)\( 30-(-1)^2 = 29 \)
9)\( y=25-x^2 \)\( 25-2^2 = 21 \)\( 25-0^2 = 25 \)\( 25-(-1)^2 = 24 \)
10)\( y=-x^2 \)\( -(2^2) = -4 \)\( -(0^2) = 0 \)\( -(-1)^2 = -1 \)
11)\( y=-x^2+4 \)\( -(2^2)+4 = 0 \)\( -(0^2)+4 = 4 \)\( -(-1)^2+4 = 3 \)
12)\( y=-x^2-9 \)\( -(2^2)-9 = -13 \)\( -(0^2)-9 = -9 \)\( -(-1)^2-9 = -10 \)
13)\( y=\sqrt{x+1} \)\( \sqrt{2+1} = \sqrt{3} \)\( \sqrt{0+1} = 1 \)\( \sqrt{-1+1} = 0 \)
14)\( y=-\sqrt{x+2} \)\( -\sqrt{2+2} = -2 \)\( -\sqrt{0+2} = -\sqrt{2} \)\( -\sqrt{-1+2} = -1 \)

Ответ: Таблица заполнена согласно расчетам.

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