Решебник по алгебре 11 класс Никольский Параграф 4. Производная Задание 64

\[\boxed{\mathbf{64}\mathbf{.}}\]

\[\textbf{а)}\ f(x) = \sqrt{x^{2} - 3x + 10} =\]

\[= \left( x^{2} - 3x + 10 \right)^{\frac{1}{2}}\]

\[x^{2} - 3x + 10 \geq 0\]

\[D = 9 - 40 = - 31 < 0;\]

\[a = 1 > 0;\ \ x \in R.\]

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

\[= \frac{1}{2}\left( x^{2} - 3x + 10 \right)^{- \frac{1}{2}} \cdot (2x - 3) =\]

\[= \frac{1}{2} \cdot \frac{2x - 3}{\left( x^{2} - 3x + 10 \right)^{\frac{1}{2}}} =\]

\[= \frac{2x - 3}{2\sqrt{x^{2} - 3x + 10}};\ \ x \in R.\]

\[\textbf{б)}\ f(x) = \sqrt{3x^{2} + 5x + 4} =\]

\[= \left( 3x^{2} + 5x + 4 \right)^{\frac{1}{2}}\]

\[3x^{2} + 5x + 4 = 0\]

\[D = 25 - 48 = - 23 < 0\]

\[a > 0;\ \ x \in R.\]

\[f^{'}(x) = \frac{1}{2}\left( 3x^{2} + 5x + 4 \right)^{- \frac{1}{2}} \cdot\]

\[\cdot (6x + 5) = \frac{6x + 5}{2\sqrt{3x^{2} + 5x + 4}};\ \]

\[\ x \in R.\]

\[\textbf{в)}\ f(x) = \sqrt{x^{2} - 3x + 2} =\]

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

\[x^{2} - 3x + 2 = 0\]

\[D = 9 - 8 = 1\]

\[x_{1} = \frac{3 + 1}{2} = 2;\ \ x_{2} = \frac{3 - 1}{2} = 1;\]

\[(x - 1)(x - 2) \geq 0\]

\[x \leq 1;\ \ x \geq 2.\]

\[f^{'}(x) = \frac{1}{2}\left( x^{2} - 3x + 2 \right)^{- \frac{1}{2}} \cdot\]

\[\cdot (2x - 3) =\]

\[= \frac{2x - 3}{2\sqrt{x^{2} - 3x + 2}};\ \ \]

\[x \in ( - \infty;1) \cup (2; + \infty).\]

\[\textbf{г)}\ f(x) = \sqrt{4x^{2} - 5x + 1} =\]

\[= \left( 4x^{2} - 5x + 1 \right)^{\frac{1}{2}}\]

\[4x^{2} - 5x + 1 = 0\]

\[D = 25 - 16 = 9\]

\[x_{1} = \frac{5 + 3}{8} = 1;\ \ \]

\[x_{2} = \frac{5 - 3}{8} = \frac{2}{8} = \frac{1}{4};\]

\[\left( x - \frac{1}{4} \right)(x - 1) \geq 0\]

\[x \leq \frac{1}{4};\ \ \ x \geq 1.\]

\[f^{'}(x) = \frac{1}{2}\left( 4x^{2} - 5x + 1 \right)^{- \frac{1}{2}} \cdot\]

\[\cdot (8x - 5) =\]

\[= \frac{8x - 5}{2\sqrt{4x^{2} - 5x + 1}};\ \ \]

\[x \in \left( - \infty;\frac{1}{4} \right) \cup (1; + \infty).\]

\[\textbf{д)}\ f(x) = \sqrt[3]{x^{2} - 6x + 9} =\]

\[= \left( x^{2} - 6x + 9 \right)^{\frac{1}{3}};\ \ x \in R\]

\[f^{'}(x) = \frac{1}{3}\left( x^{2} - 6x + 9 \right)^{- \frac{2}{3}} \cdot\]

\[\cdot (2x - 6) =\]

\[= \frac{2x - 6}{3\sqrt[3]{\left( x^{2} - 6x + 9 \right)^{2}}} =\]

\[= \frac{2 \cdot (x - 3)}{3 \cdot \sqrt[3]{\left( (x - 3)^{2} \right)^{2}}} =\]

\[= \frac{2(x - 3)}{3\sqrt[3]{(x - 3)^{4}}} =\]

\[= \frac{2(x - 3)}{3(x - 3)\sqrt[3]{x - 3}} = \frac{2}{3\sqrt[3]{x - 3}};\ \]

\[\ x \neq 3.\]

\[\textbf{е)}\ f(x) = \sqrt[3]{x^{2} + 4x + 4} =\]

\[= \left( x^{2} + 4x + 4 \right)^{\frac{1}{3}};\ \ x \in R\]

\[f^{'}(x) = \frac{1}{3}\left( x^{2} + 4x + 4 \right)^{- \frac{2}{3}} \cdot\]

\[\cdot (2x + 4) =\]

\[= \frac{1}{3} \cdot \frac{2x + 4}{\sqrt[3]{\left( x^{2} + 4x + 4 \right)^{2}}} =\]

\[= \frac{1}{3} \cdot \frac{2x + 4}{\sqrt[3]{\left( (x + 2)^{2} \right)^{2}}} =\]

\[= \frac{2 \cdot (x + 2)}{3 \cdot (x + 2)\sqrt[3]{x + 2}} = \frac{2}{3\sqrt[3]{x + 2}};\ \]

\[\ \ x \neq - 2.\]

\[\textbf{ж)}\ f(x) = \sqrt[3]{x^{2} - 6x + 5} =\]

\[= \left( x^{2} - 6x + 5 \right)^{\frac{1}{3}};\ \ x \in R\]

\[x^{2} - 6x + 5 = 0\]

\[D_{1} = 9 - 5 = 4\]

\[x_{1} = 3 + 2 = 5;\]

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

\[f^{'}(x) = \frac{1}{3}\left( x^{2} - 6x + 5 \right)^{- \frac{2}{3}} \cdot\]

\[\cdot (2x - 6) = \frac{2(x - 3)}{3\sqrt[3]{\left( x^{2} - 6x + 5 \right)^{2}}};\ \ \]

\[x \neq 1;\ \ x \neq 5.\]

\[\textbf{з)}\ f(x) = \sqrt[3]{x^{2} + 4x + 3} =\]

\[= \left( x^{2} + 4x + 3 \right)^{\frac{1}{3}};\ \ x \in R\]

\[x^{2} + 4x + 3 = 0\]

\[D_{1} = 4 - 3 = 1\]

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

\[x_{2} = - 2 + 1 = - 1.\]

\[f^{'}(x) = \frac{1}{3}\left( x^{2} + 4x + 3 \right)^{- \frac{2}{3}} \cdot\]

\[\cdot (2x + 4) = \frac{2(x + 2)}{3\sqrt[3]{\left( x^{2} + 4x + 3 \right)^{2}}};\ \ \]

\[x \neq - 1;\ \ x \neq - 3.\ \]