ГДЗ по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 15

\[\boxed{\mathbf{15}\mathbf{.}}\]

\[\textbf{а)}\ y = \sqrt{|x|};\ \ \lbrack - 1;1\rbrack\]

\[y^{'} = \frac{1}{2\sqrt{|x|}} \cdot \frac{x}{|x|} = \frac{x}{2\sqrt{|x|^{3}}};\]

\[\frac{x}{2\sqrt{|x|^{3}}} = 0\]

\[x \neq 0.\]

\[При\ x = 0\ производная\ f^{'}(x)\]

\[\ не\ существует.\]

\[0 \in \lbrack - 1;1\rbrack.\]

\[y(0) = 0;\]

\[y( - 1) = 1;\]

\[y(1) = 1.\]

\[\max{f(x)} = 1;\]

\[\min{f(x)} = 0.\]

\[\textbf{б)}\ y = \sqrt{|x|};\ \ ( - 1;1)\]

\[y^{'} = \frac{1}{2\sqrt{|x|}} \cdot \frac{x}{|x|} = \frac{x}{2\sqrt{|x|^{3}}};\]

\[\frac{x}{2\sqrt{|x|^{3}}} = 0\]

\[x \neq 0.\]

\[При\ x = 0\ производная\ f^{'}(x)\]

\[не\ существует.\]

\[0 \in ( - 1;1).\]

\[y(0) = 0.\]

\[\max{f(x)} = 1 = нет;\]

\[\min{f(x)} = 0.\]

\[\textbf{в)}\ y = \sqrt{|x|};\ \ \lbrack - 1;1)\]

\[y^{'} = \frac{1}{2\sqrt{|x|}} \cdot \frac{x}{|x|} = \frac{x}{2\sqrt{|x|^{3}}};\]

\[\frac{x}{2\sqrt{|x|^{3}}} = 0\]

\[x \neq 0.\]

\[При\ x = 0\ производная\ f^{'}(x)\]

\[\ не\ существует.\]

\[0 \in \lbrack - 1;1).\]

\[y(0) = 0;\]

\[y( - 1) = 1.\]

\[\max{f(x)} = 1;\]

\[\min{f(x)} = 0.\]

\[\textbf{г)}\ y = \sqrt{|x|};\ \ ( - 1;1\rbrack\]

\[y^{'} = \frac{1}{2\sqrt{|x|}} \cdot \frac{x}{|x|} = \frac{x}{2\sqrt{|x|^{3}}};\]

\[\frac{x}{2\sqrt{|x|^{3}}} = 0\]

\[x \neq 0.\]

\[При\ x = 0\ производная\ f^{'}(x)\ \]

\[не\ существует.\]

\[0 \in ( - 1;1\rbrack.\]

\[y(0) = 0;\]

\[y(1) = 1.\]

\[\max{f(x)} = 1;\]

\[\min{f(x)} = 0.\]