Страница 243 номер 809, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(f(x) = x^{3} - 2x\)
\(f'(x) = 3x^{2} - 2, 3x^{2} - 2 = 0, x^{2} = \dfrac{2}{3}, x = \pm\sqrt{\dfrac{2}{3}};\)

2) \(f(x) = -x^{2} + 3x + 1\)
\(f'(x) = -2x + 3, -2x + 3 = 0, x = \dfrac{3}{2} = 1{,}5;\)

3) \(f(x) = 2x^{3} + 3x^{2} - 12x - 3\)
\(f'(x) = 6x^{2} + 6x - 12, x^{2} + x - 2 = 0,\)
\(x_{1{,}2} = \dfrac{-1 \pm \sqrt{1 + 8}}{2}, x_{1} = 1, x_{2} = -2\)

4) \(f(x) = x^{3} + 2x^{2} - 7x + 1\)
\(f'(x) = 3x^{2} + 4x - 7, 3x^{2} + 4x - 7 = 0,\)
\(x_{1{,}2} = \dfrac{-4 \pm \sqrt{16 + 84}}{6}, x_{1} = 1, x_{2} = -\dfrac{7}{3};\)

5) \(f(x) = 3x^{4} - 4x^{3} - 12x^{2}\)
\(f'(x) = 12x^{3} - 12x^{2} - 24x\)
\(12x(x^{2} - x - 2) = 0,\)
\(x_{1} = 0\)
\(x_{2{,}3} = \dfrac{1 \pm \sqrt{1 + 8}}{2}\)
\(x_{2} = 2, x_{3} = -1;\)

6) \(f(x) = x^{4} + 4x^{3} - 8x^{2} - 5\)
\(f'(x) = 4x^{3} + 12x^{2} - 16x\)
\(4x(x^{2} + 3x - 4) = 0,\)
\(x_{1} = 0\)
\(x_{2{,}3} = \dfrac{-3 \pm \sqrt{9 + 16}}{2},\)
\(x_{2} = 1, x_{3} = -4.\)