Решебник по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 6

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\[\textbf{а)}\ y = 2x^{3} - 3x^{2}\]

\[y^{'}(x) = 2 \cdot 3x^{2} - 3 \cdot 2x =\]

\[= 6x^{2} - 6x;\]

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

\[6x(x - 1) = 0\]

\[x = 0 \in \lbrack - 3;3\rbrack;\ \ \]

\[x = 1 \in \lbrack - 3;3\rbrack.\]

\[Ответ:критические\ точки\ \]

\[x = 0;x = 1.\]

\[\textbf{б)}\ y = 5x^{3} - 15x\]

\[y^{'}(x) = 5 \cdot 3x^{2} - 15 =\]

\[= 15x^{2} - 15;\]

\[15x^{2} - 15 = 0\]

\[15 \cdot \left( x^{2} - 1 \right) = 0\]

\[x^{2} = 1\]

\[x = \pm 1.\]

\[x = - 1 \in \lbrack - 2;2\rbrack;\]

\[x = 1 \in \lbrack - 2;2\rbrack.\]

\[Ответ:критические\ точки\]

\[\ x = \pm 1.\]

\[\textbf{в)}\ y = 3x^{4} + x^{3} + 7\]

\[y^{'}(x) = 3 \cdot 4x^{3} + 3x^{2} + 0 =\]

\[= 12x^{3} + 3x^{2};\]

\[12x^{3} + 3x^{2} = 0\]

\[12x^{2}\left( x + \frac{1}{4} \right) = 0\]

\[x = 0;\ \ x = - \frac{1}{4} = - 0,25.\]

\[x = 0 \in \lbrack - 3;2\rbrack;\]

\[x = - 0,25 \in \lbrack - 3;2\rbrack.\]

\[Ответ:критические\ точки\ \]

\[x = 0;x = - 0,25.\]

\[\textbf{г)}\ y = x^{4} - 4x^{2}\]

\[y^{'}(x) = 4x^{3} - 4 \cdot 2x =\]

\[= 4x^{3} - 8x;\]

\[4x^{3} - 8x = 0\]

\[4x\left( x^{2} - 2 \right) = 0\]

\[x = 0;\ \ x = \pm \sqrt{2}.\]

\[x = 0\ \in \left\{ - 4;4 \right\rbrack;\]

\[x = - \sqrt{2} \in \lbrack - 4;4\rbrack;\]

\[x = \sqrt{2} \in \lbrack - 4;4\rbrack.\]

\[Ответ:критические\ точки\ \]

\[x = 0;x = \pm \sqrt{2}.\]