ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 920

\[\boxed{\mathbf{920}\mathbf{.}}\]

\[1)\ y = \frac{(2 - x)^{3}}{(3 - x)^{2}}\]

\[Промежуток\ возрастания:\]

\[(3 - x)(x - 5) > 0\]

\[(x - 3)(x - 5) < 0\]

\[3 < x < 5.\]

\[Выражение\ имеет\ смысл\ при:\]

\[3 - x \neq 0\ \]

\[x \neq 3.\]

\[y(5) = \frac{(2 - 5)^{3}}{(3 - 5)^{2}} = \frac{( - 3)^{3}}{( - 2)^{2}} =\]

\[= - \frac{27}{4}.\]

\[Ответ:\ \ x = 5 - точка\ \]

\[максимума,\ \ \ \ y(5) = - \frac{27}{4}.\]

\[2)\ y = \frac{x^{3} + 2x^{2}}{(x - 1)^{2}}\]

\[= \frac{(x - 1)\left( x^{3} - 3x^{2} - 4x \right)}{(x - 1)^{4}} =\]

\[= \frac{(x - 1) \bullet x \bullet (x^{2} - 3x - 4)}{(x - 1)^{4}}.\]

\[x^{2} - 3x - 4 = (x + 1)(x - 4)\]

\[D = 3^{2} + 4 \bullet 4 = 9 + 16 = 25\]

\[x_{1} = \frac{3 - 5}{2} = - 1\ \ и\ \ \]

\[x_{2} = \frac{3 + 5}{2} = 4.\]

\[y^{'}(x) =\]

\[= \frac{(x - 1) \bullet x \bullet (x + 1)(x - 4)}{(x - 1)^{4}}\]

\[Промежуток\ возрастания:\]

\[(x + 1) \bullet x \bullet (x - 1) \bullet (x - 4) > 0\]

\[x < - 1;\ \ \ \]

\[0 < x < 1\ или\ x > 4.\]

\[Выражение\ имеет\ смысл\ при:\]

\[x - 1 \neq 0\ \]

\[x \neq 1.\]

\[y( - 1) = \frac{( - 1)^{3} + 2 \bullet ( - 1)^{2}}{( - 1 - 1)^{2}} =\]

\[= \frac{- 1 + 2}{( - 2)^{2}} = \frac{1}{4};\]

\[y(0) = \frac{0^{3} + 2 \bullet 0^{2}}{(0 - 1)^{2}} = \frac{0}{( - 1)^{2}} = 0;\]

\[y(4) = \frac{4^{3} + 2 \bullet 4^{2}}{(4 - 1)^{2}} =\]

\[= \frac{64 + 2 \bullet 16}{3^{2}} = \frac{64 + 32}{9} = \frac{96}{9} =\]

\[= \frac{32}{3}.\]

\[Ответ:\ \ x = 0 - точка\ \]

\[минимума,\ \ \ y(0) = 0;\]

\[x = 4 - точка\ минимума,\ \ \ \]

\[y(4) = 10\frac{2}{3};\]

\[x = - 1 - точка\ максимума,\ \ \ \]

\[y( - 1) = 0,25.\]

\[3)\ y = (x - 1) \bullet e^{3x}\]

\[y^{'}(x) =\]

\[= (x - 1)^{'} \bullet e^{3x} + (x - 1) \bullet \left( e^{3x} \right)^{'}\]

\[y^{'}(x) =\]

\[= 1 \bullet e^{3x} + (x - 1) \bullet 3e^{3x} =\]

\[= e^{3x} \bullet \left( 1 + 3(x - 1) \right) =\]

\[= e^{3x} \bullet (1 + 3x - 3) =\]

\[= e^{3x} \bullet (3x - 2).\]

\[Промежуток\ возрастания:\]

\[3x - 2 > 0\]

\[3x > 2\ \]

\[x > \frac{2}{3}.\]

\[y\left( \frac{2}{3} \right) = \left( \frac{2}{3} - 1 \right) \bullet e^{2} = - \frac{e^{2}}{3}.\]

\[Ответ:\ \ x = \frac{2}{3} - точка\ \]

\[минимума;\ \ \ y\left( \frac{2}{3} \right) = - \frac{e^{2}}{3}.\]

\[4)\ y = \sin x + \frac{1}{2}\sin{2x}\]

\[y^{'}(x) = \left( \sin x \right)^{'} + \frac{1}{2} \bullet \left( \sin{2x} \right)^{'};\]

\[y^{'}(x) = \cos x + \frac{1}{2} \bullet 2\cos{2x} =\]

\[= \cos x + \cos{2x} =\]

\[= \cos x + \cos^{2}x - \sin^{2}x =\]

\[= \cos x + \cos^{2}x - 1 + \cos^{2}x =\]

\[= 2\cos^{2}x + \cos x - 1.\]

\[Пусть\ y = \cos x:\]

\[2y^{2} + y - 1 > 0\]

\[D = 1^{2} + 4 \bullet 2 = 1 + 8 = 9\]

\[y_{1} = \frac{- 1 - 3}{2 \bullet 2} = - 1\ \ и\ \ \]

\[y_{2} = \frac{- 1 + 3}{2 \bullet 2} = \frac{1}{2}.\]

\[(y + 1)(y - 0,5) > 0\]

\[y < - 1\ или\ y > 0,5.\]

\[1)\ \cos x < - 1 - нет\ корней.\]

\[2)\ \cos x > \frac{1}{2}\]

\[- \frac{\pi}{3} + 2\pi n < x < \frac{\pi}{3} + 2\pi n.\]

\[y\left( - \frac{\pi}{3} + 2\pi n \right) =\]

\[= \sin\left( - \frac{\pi}{3} \right) + \frac{1}{2}\sin\left( - \frac{2\pi}{3} \right) =\]

\[= - \frac{\sqrt{3}}{2} + \frac{1}{2} \bullet \left( - \frac{\sqrt{3}}{2} \right) = - \frac{3\sqrt{3}}{4};\]

\[y\left( \frac{\pi}{3} + 2\pi n \right) = \sin\frac{\pi}{3} + \frac{1}{2}\sin\frac{2\pi}{3} =\]

\[= \frac{\sqrt{3}}{2} + \frac{1}{2} \bullet \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{4}.\]

\[Ответ:\ \ x = - \frac{\pi}{3} + 2\pi n - точка\ \]

\[минимума,\ \ \ \]

\[y\left( - \frac{\pi}{3} + 2\pi n \right) = - \frac{3\sqrt{3}}{4};\]

\[x = \frac{\pi}{3} + 2\pi n - точка\ \]

\[максимума,\ \ \ \]

\[y\left( \frac{\pi}{3} + 2\pi n \right) = \frac{3\sqrt{3}}{4}.\]

\[5)\ y = e^{\sqrt{3 - x^{2}}}\]

\[Пусть\ u = \sqrt{3 - x^{2}};\text{\ y}(u) = e^{u}:\]

\[y^{'}(x) = {\left( 3 - x^{2} \right)^{\frac{1}{2}}}^{'} \bullet \left( e^{u} \right)^{'};\]

\[y^{'}(x) = - 2x \bullet \left( 3 - x^{2} \right)^{- \frac{1}{2}} \bullet e^{u} =\]

\[= - \frac{2x}{\sqrt{3 - x^{2}}} \bullet e^{\sqrt{3 - x^{2}}}.\]

\[Промежуток\ возрастания:\]

\[- 2x > 0\ \]

\[x < 0.\]

\[Выражение\ имеет\ смысл\ при:\]

\[3 - x^{2} \geq 0\]

\[x^{2} \leq 3\]

\[- \sqrt{3} < x < \sqrt{3}.\]

\[y(0) = e^{\sqrt{3 - 0^{2}}} = e^{\sqrt{3}}.\]

\[Ответ:\ \ x = 0 - точка\ \]

\[максимума;\ \ \ y(0) = e^{\sqrt{3}}.\]

\[6)\ y = \sqrt{e^{x} - x}\]

\[Пусть\ u = e^{x} - x;y(u) = \sqrt{u}:\]

\[y^{'}(x) = \left( e^{x} - x \right)^{'} \bullet \left( \sqrt{u} \right)^{'} =\]

\[= \left( e^{x} - 1 \right) \bullet \frac{1}{2\sqrt{u}} = \frac{e^{x} - 1}{2\sqrt{e^{x} - x}}.\]

\[Промежуток\ возрастания:\]

\[e^{x} - 1 > 0\]

\[e^{x} > 1\]

\[e^{x} > e^{0}\ \]

\[x > 0.\]

\[y(0) = \sqrt{e^{0} - 0} = \sqrt{1 - 0} =\]

\[= \sqrt{1} = 1.\]

\[Ответ:\ \ x = 0 - точка\ \]

\[минимума,\ \ \ y(0) = 1.\]