Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 780

\[\boxed{\mathbf{780}\mathbf{.}}\]

\[1)\ f(x) = 3x + 2\]

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

\[= \frac{3(x + h) + 2 - 3x - 2}{h} =\]

\[= \frac{3x + 3h - 3x}{h} = \frac{3h}{h} = 3\]

\[f^{'}(x) = \lim_{h \rightarrow 0}3 = 3.\]

\[Ответ:\ \ 3.\]

\[2)\ f(x) = 5x + 7\]

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

\[= \frac{5(x + h) + 7 - 5x - 7}{h} =\]

\[= \frac{5x + 5h - 5x}{h} = \frac{5h}{h} = 5\]

\[f^{'}(x) = \lim_{h \rightarrow 0}5 = 5.\]

\[Ответ:\ \ 5.\]

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

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

\[= \frac{3(x + h)^{2} - 5(x + h) - 3x^{2} + 5x}{h} =\]

\[= \frac{3x^{2} + 6xh + 3h^{2} - 5x - 5h - 3x^{2} + 5x}{h} =\]

\[= \frac{6xh + 3h^{2} - 5h}{h} = 6x + 3h - 5\]

\[f^{'}(x) = \lim_{h \rightarrow 0}(6x + 3h - 5) =\]

\[= 6x + 3 \bullet 0 - 5 = 6x - 5.\]

\[Ответ:\ \ 6x - 5.\]

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

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

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

\[= \frac{- 3x^{2} - 6xh - 3h^{2} + 3x^{2}}{h} =\]

\[= \frac{- 6xh - 3h^{2}}{h} = - 6x - 3h\]

\[f^{'}(x) = \lim_{h \rightarrow 0}( - 6x - 3h) =\]

\[= - 6x - 3 \bullet 0 = - 6x.\]

\[Ответ:\ - 6x.\]