ГДЗ по алгебре 11 класс Никольский Параграф 3. Обратные функции Задание 3

\[\boxed{\mathbf{3}\mathbf{.}}\]

\[\textbf{а)}\ y = 3x + 1\]

\[x = 3y + 1\]

\[3y = x - 1\]

\[y = \frac{1}{3}x - \frac{1}{3}.\]

\[\textbf{б)}\ y = 2x - 8\]

\[x = 2y - 8\]

\[2y = x + 8\]

\[y = \frac{1}{2} + 4.\]

\[\textbf{в)}\ y = x^{2};\ \ x \in \lbrack 0;3\rbrack\]

\[0 \leq x \leq 3\]

\[0 \leq x^{2} \leq 9\]

\[0 \leq y \leq 9\]

\[y \in \lbrack 0;9\rbrack.\]

\[x = y^{2}\]

\[y = \sqrt{x};\ \ x \in \lbrack 0;9\rbrack.\]

\[\textbf{г)}\ y = - x^{2};\ \ x \in \lbrack 0;3\rbrack\]

\[0 \leq x \leq 3\]

\[0 \leq x^{2} \leq 9\]

\[- 9 \leq - x^{2} \leq 0\]

\[- 9 \leq y \leq 0\]

\[y \in \lbrack - 9;0\rbrack.\]

\[x = - y^{2}\]

\[y^{2} = - x\]

\[y = \sqrt{- x};\ \ x \in \lbrack - 9;0\rbrack.\]

\[\textbf{д)}\ y = 8x^{3}\]

\[x = 8y^{3}\]

\[y^{3} = \frac{x}{8}\]

\[y = \sqrt[3]{\frac{x}{8}} = \frac{1}{2}\sqrt[3]{x}.\]

\[\textbf{е)}\ y = 0,5\sqrt{x};\ x \in \lbrack 0;25\rbrack\]

\[0 \leq x \leq 25\]

\[0 \leq \sqrt{x} \leq \sqrt{25}\]

\[0 \leq \sqrt{x} \leq 5\]

\[0,5 \cdot 0 \leq 0,5\sqrt{x} \leq 0,5 \cdot 5\]

\[0 \leq 0,5\sqrt{x} \leq 2,5\]

\[0 \leq y \leq 2,5\]

\[y \in \lbrack 0;2,5\rbrack\text{.\ \ }\]

\[x = 0,5\sqrt{y}\]

\[\sqrt{y} = 2x\]

\[y = (2x)^{2} = 4x^{2};\ \ x \in \lbrack 0;2,5\rbrack.\]

\[\textbf{ж)}\ y = 3^{x}\]

\[x = 3^{y}\]

\[y = \log_{3}x.\]

\[\textbf{з)}\ y = \log_{5}x;\ \ x \in (0;25)\]

\[0 < x < 25\]

\[\lim_{x \rightarrow 0}{\log_{5}x} < \log_{5}x < \log_{5}25\]

\[- \infty < y < 2\]

\[y \in ( - \infty;2).\]

\[x = \log_{5}y\]

\[y = 5^{x};\ \ x \in ( - \infty;2).\]