ГДЗ по алгебре и начала математического анализа 11 класс Колягин Проверь себя II

\[\mathbf{1.}\ \]

\[f(x) = 2x^{3} + 3x^{2} - x;\text{\ x} = - 2:\]

\[f^{'}(x) = 2 \bullet 3x^{2} + 3 \bullet 2x - 1;\]

\[f^{'}( - 2) = 6 \bullet 4 + 6 \bullet ( - 2) - 1 =\]

\[= 24 - 12 - 1 = 11.\]

\[Ответ:\ \ 11.\]

\[\mathbf{2.\ }\]

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

\[f^{'}(x) = 2 \bullet \left( - \frac{1}{x^{2}} \right) + 4 \bullet \frac{1}{2\sqrt{x}} - e^{x} =\]

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

\[2)\ f(x) = (3x - 5)^{3};\]

\[f^{'}(x) = 3 \bullet 3(3x - 5)^{2} =\]

\[= 9(3x - 5)^{2}.\]

\[3)\ f(x) = 3\sin{2x} \bullet \cos x;\]

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

\[= 3 \bullet 2\cos{2x} \bullet \cos x + 3\sin{2x} \bullet \left( - \sin x \right) =\]

\[= 6\cos{2x} \bullet \cos x - 3\sin{2x} \bullet \sin x.\]

\[4)\ f(x) = \frac{x^{3}}{x^{2} + 5};\]

\[f^{'}(x) = \frac{3x^{2} \bullet \left( x^{2} + 5 \right) - x^{3} \bullet 2x}{\left( x^{2} + 5 \right)^{2}} =\]

\[= \frac{3x^{4} + 15x^{2} - 2x^{4}}{\left( x^{2} + 5 \right)^{2}} =\]

\[= \frac{x^{4} + 15x^{2}}{\left( x^{2} + 5 \right)^{2}}.\]

\[\mathbf{3.}\]

\[\ y = x^{4} - 2x^{3} + 3;\text{\ \ \ }x_{0} = \frac{1}{2}:\]

\[y^{'}(x) = 4x^{3} - 2 \bullet 3x^{2};\]

\[y^{'}\left( x_{0} \right) = 4 \bullet \frac{1}{8} - 6 \bullet \frac{1}{4} =\]

\[= 0,5 - 1,5 = - 1;\]

\[a = arctg( - 1) = - \frac{\pi}{4}.\]

\[Ответ:\ - \frac{\pi}{4}.\]

\[\mathbf{4}.\]

\[\text{\ f}(x) = \ln(3x + 1);\]

\[f^{'}(x) = 3 \bullet \frac{1}{3x + 1} > 0;\]

\[3x + 1 > 0\]

\[3x > - 1\]

\[x > - \frac{1}{3}.\]

\[Ответ:\ \ \left( - \frac{1}{3};\ + \infty \right).\]

\[\mathbf{5}.\]

\[\text{\ f}(x) = \sin{2x};\ x_{0} = - \frac{\pi}{6}:\]

\[f\left( x_{0} \right) = \sin\left( - \frac{\pi}{3} \right) =\]

\[= - \sin\frac{\pi}{3} = - \frac{\sqrt{3}}{2};\]

\[f^{'}(x) = 2\cos{2x};\]

\[f^{'}\left( x_{0} \right) = 2\cos\left( - \frac{\pi}{3} \right) =\]

\[= 2\cos\frac{\pi}{3} = 1;\]

\[y = - \frac{\sqrt{3}}{2} + 1 \bullet \left( x + \frac{\pi}{6} \right) =\]

\[= x - \frac{\sqrt{3}}{2} + \frac{\pi}{6}.\]

\[Ответ:\ \ y = x - \frac{\sqrt{3}}{2} + \frac{\pi}{6}.\]