ГДЗ по алгебре 11 класс Никольский Параграф 4. Производная Задание 8

\[\boxed{\mathbf{8}\mathbf{.}}\]

\[\textbf{а)}\ f(x) = 3x + 8;\ \ \ X = R\]

\[\mathrm{\Delta}f = \left( 3 \cdot (x + \mathrm{\Delta}x) + 8 \right) -\]

\[- (3x + 8) = 3\mathrm{\Delta}x;\]

\[\frac{\mathrm{\Delta}f}{\mathrm{\Delta}x} = \frac{3\mathrm{\Delta}x}{\mathrm{\Delta}x} = 3;\]

\[f^{'}(x) = 3.\]

\[f^{'}(0) = f^{'}(1) = f^{'}( - 1) =\]

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

\[= f^{'}( - 3) = 3.\]

\[f^{'}(x) = 0 \rightarrow 3 \neq 0 - нет\ \]

\[решений.\]

\[f^{'}(x) = 1 \rightarrow 3 \neq 1 - нет\ \]

\[решений.\]

\[f^{'}(x) = 3 \rightarrow 3 = 3 - x \in R.\]

\[\textbf{б)}\ f(x) = 8x - 11;\ \ X = R\]

\[\mathrm{\Delta}f = \left( 8 \cdot (x + \mathrm{\Delta}x) - 1 \right) -\]

\[- (8x - 11) = 8\mathrm{\Delta}x;\]

\[\frac{\mathrm{\Delta}f}{\mathrm{\Delta}x} = \frac{8\mathrm{\Delta}x}{\mathrm{\Delta}x} = 8;\]

\[f^{'}(x) = 8.\]

\[f^{'}(0) = f^{'}(1) = f^{'}( - 1) = f^{'}(2) =\]

\[= f^{'}( - 2) = f^{'}(3) = f^{'}( - 3) = 8.\]

\[f^{'}(x) = 0 \rightarrow 8 \neq 0 - нет\ решений.\]

\[f^{'}(x) = 1 \rightarrow 8 \neq 1 - нет\ решений.\]

\[f^{'}(x) = 3 \rightarrow 8 \neq 3 - нет\ решений.\]

\[\textbf{в)}\ f(x) = kx + b;\ \ X = R\]

\[\mathrm{\Delta}f = \left( k \cdot (x + \mathrm{\Delta}x) + k \right) -\]

\[- (kx + b) = k\mathrm{\Delta}x;\]

\[\frac{\mathrm{\Delta}f}{\mathrm{\Delta}x} = \frac{k\mathrm{\Delta}x}{\mathrm{\Delta}x} = k;\]

\[f^{'}(x) = 8.\]

\[f^{'}(0) = f^{'}(1) = f^{'}( - 1) =\]

\[= f^{'}(2) = f^{'}( - 2) =\]

\[= f^{'}(3) = f^{'}( - 3) = k.\]

\[f^{'}(x) = 0 \rightarrow k \neq 0 - нет\]

\[\ решений.\]

\[f^{'}(x) = 1 \rightarrow k = 1\ при\ x \in R;\ \]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ k \neq 1 - нет\]

\[\ решений.\]

\[f^{'}(x) = 3 \rightarrow k = 3\ при\ x\ \in R;\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ k \neq 3 - нет\ \]

\[решений.\]

\[\textbf{г)}\ f(x) = x^{2} - x + 5;\ \ \ X = R\]

\[\mathrm{\Delta}f = \left( (x + \mathrm{\Delta}x)^{2} - (x + \mathrm{\Delta}x) + 5 \right) -\]

\[- \left( x^{2} - x - 5 \right) = (2x - 1)\mathrm{\Delta}x;\]

\[\frac{\mathrm{\Delta}f}{\mathrm{\Delta}x} = \frac{(2x - 1)\mathrm{\Delta}x}{\mathrm{\Delta}x} = 2x - 1;\]

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

\[f^{'}(0) = - 1;\]

\[f^{'}(1) = 1;\]

\[f^{'}( - 1) = - 3;\]

\[f^{'}(2) = 3;\]

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

\[f^{'}(3) = 5;\]

\[f^{'}( - 3) = - 7.\]

\[2x - 1 = 0 \rightarrow x = 0,5;\]

\[2x - 1 = 1 \rightarrow x = 1;\]

\[2x - 1 = 3 \rightarrow x = 2.\]

\[\textbf{д)}\ f(x) = x^{2} + 3x - 1;\ \ X = C\]

\[\mathrm{\Delta}f = \left( (x + \mathrm{\Delta}x)^{2} + 3(x + \mathrm{\Delta}x) - 1 \right) -\]

\[- \left( x^{2} + 3x - 1 \right) = (2x + 3)\mathrm{\Delta}x;\]

\[\frac{\mathrm{\Delta}f}{\mathrm{\Delta}x} = \frac{(2x + 3)\mathrm{\Delta}x}{\mathrm{\Delta}x} = 2x + 3;\]

\[f^{'}(x) = 2x + 3.\]

\[f^{'}(0) = 3;\]

\[f^{'}(1) = 5;\]

\[f^{'}( - 1) = 1;\]

\[f^{'}(2) = 7;\]

\[f^{'}( - 2) = - 1;\]

\[f^{'}(3) = 9;\]

\[f^{'}( - 3) = - 3.\]

\[2x + 3 = 0 \rightarrow x = - 1,5;\]

\[2x + 3 = 1 \rightarrow x = - 1;\]

\[2x + 3 = 3 \rightarrow x = 0.\]

\[\textbf{е)}\ f(x) = ax^{2} + bx + c;\ \ X = R\]

\[\mathrm{\Delta}f = \left( (x + \mathrm{\Delta}x)^{2} + b(x + \mathrm{\Delta}x) + c \right) -\]

\[- \left( ax^{2} + bx + c \right) =\]

\[= (2ax + b + a + \mathrm{\Delta}x)\mathrm{\Delta}x;\]

\[\frac{\mathrm{\Delta}f}{\mathrm{\Delta}x} = \frac{(2ax + b + a + \mathrm{\Delta}x)\mathrm{\Delta}x}{\mathrm{\Delta}x} =\]

\[= 2ax + b + a + \mathrm{\Delta}x;\]

\[f^{'}(x) = 2ax + b.\]

\[f^{'}(0) = b;\]

\[f^{'}(1) = 2a + b;\]

\[f^{'}( - 1) = - 2a + b;\]

\[f^{'}(2) = 4a + b;\]

\[f^{'}( - 2) = - 4a + b;\]

\[f^{'}(3) = 6a + b;\]

\[f^{'}( - 3) = - 6a + b.\]

\[2ax + b = 0 \rightarrow x = - \frac{b}{2a};\]

\[2ax + b = 1 \rightarrow x = \frac{1 - b}{2a};\]

\[2ax + b = 3 \rightarrow x = \frac{3 - b}{2a}.\]