Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 938

\[\boxed{\mathbf{938}\mathbf{.}}\]

\[1)\ f(x) = x^{4} - 8x^{2} + 5;\ \ \lbrack - 3;\ 2\rbrack\]

\[f^{'}(x) = \left( x^{4} \right)^{'} - 8\left( x^{2} \right)^{'} + (5)^{'} =\]

\[= 4x^{3} - 8 \bullet 2x + 0 = 4x^{3} - 16x.\]

\[Точки\ экстремума:\]

\[4x^{3} - 16x = 0\]

\[4x \bullet \left( x^{2} - 4 \right) = 0\]

\[(x + 2) \bullet 4x \bullet (x - 2) = 0\]

\[x_{1} = - 2;\ \ \ x_{2} = 0;\text{\ \ }x_{3} = 2.\]

\[f( - 3) =\]

\[= ( - 3)^{4} - 8 \bullet ( - 3)^{2} + 5 =\]

\[= 81 - 72 + 5 = 14;\]

\[f( \pm 2) =\]

\[= ( \pm 2)^{4} - 8 \bullet ( \pm 2)^{2} + 5 =\]

\[= 16 - 32 + 5 = - 11;\]

\[f(0) = 0^{4} - 8 \bullet 0^{2} + 5 = 5.\]

\[Ответ:\ \ y_{\min} = - 11;\ \ y_{\max} = 14.\]

\[2)\ f(x) = x + \frac{1}{x};\ \ \lbrack - 2;\ - 0,5\rbrack\]

\[f^{'}(x) = (x)^{'} + \left( \frac{1}{x} \right)^{'} = 1 - \frac{1}{x^{2}}.\]

\[Точки\ экстремума:\]

\[1 - \frac{1}{x^{2}} = 0\]

\[x^{2} - 1 = 0\]

\[x^{2} = 1\]

\[x = \pm 1.\]

\[y( - 2) = - 2 + \frac{1}{- 2} = - 2 - 0,5 =\]

\[= - 2,5;\]

\[y( - 1) = - 1 + \frac{1}{- 1} = - 1 - 1 =\]

\[= - 2;\]

\[y( - 0,5) = - 0,5 + \frac{1}{- 0,5} =\]

\[= - 0,5 - 2 = - 2,5.\]

\[Ответ:\ \ y_{\min} = - 2,5;\ \ \]

\[y_{\max} = - 2.\]

\[3)\ f(x) = \sin x + \cos x;\ \left\lbrack \pi;\ \frac{3\pi}{2} \right\rbrack\]

\[f^{'}(x) = \left( \sin x \right)^{'} + \left( \cos x \right)^{'} =\]

\[= \cos x - \sin x.\]

\[Точки\ экстремума:\]

\[\cos x - \sin x = 0\ \ \ \ \ |\ :\cos x\]

\[1 - tg\ x = 0\]

\[tg\ x = 1\]

\[x = arctg\ 1 + \pi n\]

\[x = \frac{\pi}{4} + \pi n.\]

\[y(\pi) = \sin\pi + \cos\pi = 0 - 1 =\]

\[= - 1;\]

\[y\left( \frac{5\pi}{4} \right) = \sin\frac{5\pi}{4} + \cos\frac{5\pi}{4} =\]

\[= - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = - \sqrt{2};\]

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

\[= - 1 + 0 = - 1.\]

\[Ответ:\ \ y_{\min} = - \sqrt{2};\ \ y_{\max} = - 1.\]