ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 1071

\[1)\ f(x) = 4x^{3} + 6x^{2}\]

\[D(x) = ( - \infty;\ + \infty);\]

\[ни\ четная,\ ни\ нечетная:\]

\[f( - x) = 4( - x)^{3} + 6( - x)^{2} =\]

\[= - 4x^{3} + 6x^{2};\]

\[f^{'}(x) = 4 \bullet 3x^{2} + 6 \bullet 2x =\]

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

\[Промежуток\ возрастания:\]

\[12x^{2} + 12x \geq 0\]

\[x^{2} + x \geq 0\]

\[(x + 1)x \geq 0\]

\[x \leq - 1;\text{\ \ }\ x \geq 0.\]

\[Максимум\ и\ минимум:\]

\[y( - 1) = - 4 + 6 = 2;\]

\[y(0) = 4 \bullet 0^{3} + 6 \bullet 0^{2} = 0.\]

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

\[D(x) = ( - \infty;\ + \infty);\]

\[ни\ четная,\ ни\ нечетная:\]

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

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

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

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

\[Промежуток\ возрастания:\]

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

\[x - x^{2} \geq 0\]

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

\[x(x - 1) \leq 0\]

\[0 \leq x \leq 1.\]

\[Максимум\ и\ минимум:\]

\[y(1) = 3 - 2 = 1;\]

\[y(0) = 3 \bullet 0^{2} - 2 \bullet 0^{3} = 0.\]

\[3)\ f(x) = \frac{1}{3}x^{3} - x\]

\[D(x) = ( - \infty;\ + \infty);\]

\[является\ нечетной:\]

\[f( - x) = \frac{1}{3}( - x)^{3} - ( - x) =\]

\[= - \frac{1}{3}x^{3} + x = - f(x).\]

\[f^{'}(x) = \frac{1}{3} \bullet 3x^{2} - 1 = x^{2} - 1.\]

\[Промежуток\ возрастания:\]

\[x^{2} - 1 \geq 0\]

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

\[x \leq - 1;\text{\ \ \ x} \geq 1.\]

\[Максимум\ и\ минимум:\]

\[y( - 1) = \frac{1}{3} \bullet ( - 1) + 1 = \frac{2}{3};\]

\[y(1) = \frac{1}{3} \bullet 1 - 1 = - \frac{2}{3}.\]

\[4)\ f(x) = x^{4} - \frac{1}{2}x^{2}\]

\[D(x) = ( - \infty;\ + \infty);\]

\[является\ четной:\]

\[f( - x) = ( - x)^{4} - \frac{1}{2}( - x)^{2} =\]

\[= x^{4} - \frac{1}{2}x^{2} = f(x).\]

\[f^{'}(x) = 4x^{3} - \frac{1}{2} \bullet 2x = 4x^{3} - x.\]

\[Промежуток\ возрастания:\]

\[4x^{3} - x \geq 0\]

\[x\left( 4x^{2} - 1 \right) \geq 0\]

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

\[- \frac{1}{2} \leq x \leq 0;\ \ \ x \geq \frac{1}{2}.\]

\[Максимум\ и\ минимум:\]

\[y(0) = 0^{4} - \frac{1}{2} \bullet 0^{2} = 0;\]

\[y\left( \frac{1}{2} \right) = \frac{1}{16} - \frac{1}{2} \bullet \frac{1}{4} = - \frac{1}{16}.\]