Решебник по алгебре 11 класс Никольский Параграф 4. Производная Задание 5

\[\boxed{\mathbf{5}\mathbf{.}}\]

\[f(x) = x^{2};\ \ x_{1} = x;\ \ x_{2} = x + \mathrm{\Delta}x\]

\[\textbf{а)}\ f(x) = x^{2};\]

\[f(x + \mathrm{\Delta}x) = (x + \mathrm{\Delta}x)^{2};\]

\[\mathrm{\Delta}f = f(x + \mathrm{\Delta}x) - f(x) =\]

\[= (x + \mathrm{\Delta}x)^{2} - x^{2} = x^{2} +\]

\[+ 2x \cdot \mathrm{\Delta}x + (\mathrm{\Delta}x)^{2} - x^{2} =\]

\[= 2x \cdot \mathrm{\Delta}x + (\mathrm{\Delta}x)^{2} =\]

\[= (2x + \mathrm{\Delta}x) \cdot \mathrm{\Delta}x.\]

\[\textbf{б)}\ tg\ \beta = \frac{\mathrm{\Delta}f(x)}{\mathrm{\Delta}x} =\]

\[= \frac{(2x + \mathrm{\Delta}x) \cdot \mathrm{\Delta}x}{\mathrm{\Delta}x} = 2x + \mathrm{\Delta}x.\]

\[\textbf{в)}\ f^{'}(x) = \lim_{x \rightarrow 0}\frac{\mathrm{\Delta}f}{\mathrm{\Delta}x};\ \]

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

\[= \lim_{x \rightarrow 0}\frac{(2x + \mathrm{\Delta}x) \cdot \mathrm{\Delta}x}{\mathrm{\Delta}x} =\]

\[= \lim_{x \rightarrow 0}(2x + \mathrm{\Delta}x) = 2x.\]

\[\textbf{г)}\ tg\ a = 2x;\]

\[x = 0:\]

\[tg\ a = 2x = 2 \cdot 0 = 0.\]

\[x = 1:\]

\[tg\ a = 2x = 2 \cdot 1 = 2.\]

\[x = - 1:\]

\[tg\ a = 2x = 2 \cdot ( - 1) = - 2.\]

\[x = 2:\]

\[tg\ a = 2x = 2 \cdot 2 = 4.\]

\[x = - 2:\]

\[tg\ a = 2x = 2 \cdot ( - 2) = - 4.\]