ГДЗ по Алгебре 7 Класс Номер 22.3 Мордкович — Подробные Ответы

Дана функция у = f(х), где f(х) = -\(\frac{1}{3}\) х + 2. Найдите: а) f(0), f(-3), f(\(\frac{1}{3}\)); б) f(-x), -f(x), f(3x); \(в) f(x^2), f(2x^2), f(-2x^2)\); г) f(3), f(-6), f(-\(\frac{1}{3}\)); д) f(-2x), -f(-2x), f(\(\frac{1}{3}\) x); \(е) f(-x^2), f(-3x^2), f(27x^2)\).

а)
\( f(0) = -\frac{1}{3} \cdot 0 + 2 = 2 \)

\( f(-3) = -\frac{1}{3} \cdot (-3) + 2 = 1 + 2 = 3 \)

\( f(\frac{1}{3}) = -\frac{1}{3} \cdot \frac{1}{3} + 2 = -\frac{1}{9} + 2 = \frac{17}{9} \)

б)
\( f(-x) = -\frac{1}{3} \cdot (-x) + 2 = \frac{1}{3}x + 2 \)

\( -f(x) = — (-\frac{1}{3}x + 2) = \frac{1}{3}x — 2 \)

\( f(3x) = -\frac{1}{3} \cdot (3x) + 2 = -x + 2 \)

в)
\( f(x^2) = -\frac{1}{3} \cdot (x^2) + 2 = -\frac{1}{3}x^2 + 2 \)

\( f(2x^2) = -\frac{1}{3} \cdot (2x^2) + 2 = -\frac{2}{3}x^2 + 2 \)

\( f(-2x^2) = -\frac{1}{3} \cdot (-2x^2) + 2 = \frac{2}{3}x^2 + 2 \)

г)
\( f(3) = -\frac{1}{3} \cdot 3 + 2 = -1 + 2 = 1 \)

\( f(-6) = -\frac{1}{3} \cdot (-6) + 2 = 2 + 2 = 4 \)

\( f(-\frac{1}{3}) = -\frac{1}{3} \cdot (-\frac{1}{3}) + 2 = \frac{1}{9} + 2 = \frac{19}{9} \)

д)
\( f(-2x) = -\frac{1}{3} \cdot (-2x) + 2 = \frac{2}{3}x + 2 \)

\( -f(-2x) = -(\frac{2}{3}x + 2) = -\frac{2}{3}x — 2 \)

\( f(\frac{1}{3}x) = -\frac{1}{3} \cdot (\frac{1}{3}x) + 2 = -\frac{1}{9}x + 2 \)

е)
\( f(-x^2) = -\frac{1}{3} \cdot (-x^2) + 2 = \frac{1}{3}x^2 + 2 \)

\( f(-3x^2) = -\frac{1}{3} \cdot (-3x^2) + 2 = x^2 + 2 \)

\( f(27x^2) = -\frac{1}{3} \cdot (27x^2) + 2 = -9x^2 + 2 \)

а) Значения функции \(f(x)\)

Функция задана как \(f(x) = -\frac{1}{3}x + 2\). Подставляем значения:

1. \(f(0)\):
\[
f(0) = -\frac{1}{3} \cdot 0 + 2 = 2
\]

2. \(f(-3)\):
\[
f(-3) = -\frac{1}{3} \cdot (-3) + 2 = 1 + 2 = 3
\]

3. \(f\left(\frac{1}{3}\right)\):
\[
f\left(\frac{1}{3}\right) = -\frac{1}{3} \cdot \frac{1}{3} + 2 = -\frac{1}{9} + 2 = \frac{17}{9}
\]

б) Преобразования функции

1. \(f(-x)\):
\[
f(-x) = -\frac{1}{3} \cdot (-x) + 2 = \frac{1}{3}x + 2
\]

2. \(-f(x)\):
\[
-f(x) = -\left(-\frac{1}{3}x + 2\right) = \frac{1}{3}x — 2
\]

3. \(f(3x)\):
\[
f(3x) = -\frac{1}{3} \cdot (3x) + 2 = -x + 2
\]

в) Функция от квадратов

1. \(f(x^2)\):
\[
f(x^2) = -\frac{1}{3} \cdot (x^2) + 2 = -\frac{1}{3}x^2 + 2
\]

2. \(f(2x^2)\):
\[
f(2x^2) = -\frac{1}{3} \cdot (2x^2) + 2 = -\frac{2}{3}x^2 + 2
\]

3. \(f(-2x^2)\):
\[
f(-2x^2) = -\frac{1}{3} \cdot (-2x^2) + 2 = \frac{2}{3}x^2 + 2
\]

г) Дополнительные значения функции

1. \(f(3)\):
\[
f(3) = -\frac{1}{3} \cdot 3 + 2 = -1 + 2 = 1
\]

2. \(f(-6)\):
\[
f(-6) = -\frac{1}{3} \cdot (-6) + 2 = 2 + 2 = 4
\]

3. \(f\left(-\frac{1}{3}\right)\):
\[
f\left(-\frac{1}{3}\right) = -\frac{1}{3} \cdot \left(-\frac{1}{3}\right) + 2 = \frac{1}{9} + 2 = \frac{19}{9}
\]

д) Преобразования с учетом множителей

1. \(f(-2x)\):
\[
f(-2x) = -\frac{1}{3} \cdot (-2x) + 2 = \frac{2}{3}x + 2
\]

2. \(-f(-2x)\):
\[
-f(-2x) = -\left(\frac{2}{3}x + 2\right) = -\frac{2}{3}x — 2
\]

3. \(f\left(\frac{1}{3}x\right)\):
\[
f\left(\frac{1}{3}x\right) = -\frac{1}{3} \cdot \left(\frac{1}{3}x\right) + 2 = -\frac{1}{9}x + 2
\]

е) Функция от отрицательных квадратов

1. \(f(-x^2)\):
\[
f(-x^2) = -\frac{1}{3} \cdot (-x^2) + 2 = \frac{1}{3}x^2 + 2
\]

2. \(f(-3x^2)\):
\[
f(-3x^2) = -\frac{1}{3} \cdot (-3x^2) + 2 = x^2 + 2
\]

3. \(f(27x^2)\):
\[
f(27x^2) = -\frac{1}{3} \cdot (27x^2) + 2 = -9x^2 + 2
\]