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МатематикаРешебник по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 30
Задание 30
\[\boxed{\mathbf{30.}}\]
\[\textbf{а)}\ f(x) = \frac{8}{\sqrt{x}} = 8x^{- 2};\ \ x_{0} = 4\]
\[f^{'}(x) = 8 \cdot \left( - \frac{1}{2} \right)x^{- \frac{3}{2}} = - \frac{4}{\sqrt{x^{3}}};\]
\[y_{0} = f(4) = \frac{8}{\sqrt{4}} = \frac{8}{2} = 4;\]
\[k = f^{'}(4) = - \frac{4}{\sqrt{4^{3}}} = - \frac{4}{\sqrt{2^{6}}} =\]
\[= - \frac{4}{2^{3}} = - \frac{4}{8} = - \frac{1}{2};\]
\[y - y_{0} = k\left( x - x_{0} \right)\]
\[y - 4 = - \frac{1}{2}(x - 4)\]
\[y - 4 = - 0,5x + 2\]
\[y = - 0,5x + 6.\]
\[Уравнение\ касательной:\]
\[y = - 0,5x + 6.\]
\[\textbf{б)}\ f(x) = \frac{- 4}{\sqrt{x}} = - 4x^{- 2};\ \ x_{0} = 2\]
\[f^{'}(x) = - 4 \cdot \left( - \frac{1}{2} \right)x^{- \frac{3}{2}} = \frac{2}{\sqrt{x^{3}}};\]
\[y_{0} = f(2) = \frac{- 4}{\sqrt{2}} = - 2\sqrt{2};\]
\[k = f^{'}(2) = \frac{2}{\sqrt{2^{3}}} = \frac{2}{2\sqrt{2}} =\]
\[= \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2};\]
\[y - y_{0} = k\left( x - x_{0} \right)\]
\[y + 2\sqrt{2} = \frac{\sqrt{2}}{2}(x - 2)\]
\[y + 2\sqrt{2} = \frac{\sqrt{2}}{2}x - \sqrt{2}\]
\[y = \frac{\sqrt{2}}{2}x - 3\sqrt{2}.\]
\[Уравнение\ касательной:\]
\[y = \frac{\sqrt{2}}{2}x - 3\sqrt{2}.\]
\[\textbf{в)}\ f(x) = \sin\frac{\text{πx}}{2} + \ln(2 - x);\ \ \]
\[x_{0} = 1\]
\[f^{'}(x) = \frac{\pi}{2}\cos\frac{\text{πx}}{2} - \frac{1}{2 - x};\]
\[y_{0} = f(1) = \sin\frac{\pi}{2} + \ln(2 - 1) =\]
\[= 1 + 0 = 1;\]
\[k = f^{'}(1) = \frac{\pi}{2} \cdot \cos\frac{\pi}{2} - \frac{1}{2 - 1} =\]
\[= \frac{\pi}{2} \cdot 0 - 1 = - 1;\]
\[y - y_{0} = k\left( x - x_{0} \right)\]
\[y - 1 = - 1 \cdot (x - 1)\]
\[y - 1 = - x + 1\]
\[y = - x + 2.\]
\[Уравнение\ касательной:\]
\[y = - x + 2.\]
\[\textbf{г)}\ f(x) = \cos\text{πx} - e^{1 - x};\ \ x_{0} = 1\]
\[f^{'}(x) = - \pi\sin\text{πx} + e^{1 - x};\]
\[y_{0} = f(1) = \cos\pi - e^{0} =\]
\[= - 1 - 1 = - 2;\]
\[k = f^{'}(1) = - \pi\sin\pi + e^{0} =\]
\[= 0 + 1 = 1;\]
\[y - y_{0} = k\left( x - x_{0} \right)\]
\[y + 2 = 1(x - 1)\]
\[y + 2 = x - 1\]
\[y = x - 3.\]
\[Уравнение\ касательной:\]
\[y = x - 3.\]
