[tex]f(x) = \frac{2}{5} {x}^{5} - 3 \frac{1}{3} {x}^{3} + 8x \\ f'(x) = 5 \times \frac{2}{5} {x}^{5 - 1} - 3 \times \frac{10}{3} {x}^{3 - 1} + 8 {x}^{1 - 1} = \\ = 2 {x}^{4} - 10 {x}^{2} + 8 \\ f'(x) = 0 \\ 2 {x}^{4} - 10 {x}^{2} + 8 = 0 \\ {x}^{4} - 5 {x}^{2} + 4 = 0 \\ {x}^{2} = a \: , \: a \geqslant 0[/tex]
По теореме Виета:
[tex]{x}^{2} + bx + c = 0 \\ x _{1} + x_{2} = - b \\ x _{1} x_{2} = c[/tex]
[tex] {a}^{2} - 5a + 4 = 0 \\ a_{1} + a_{2} =5 \\ a_{1} a_{2} = 4\\ a_{1} = 1 \\ a_{2} =4 [/tex]
[tex] {x}^{2} = 1 \\ {x}^{2} = 4 \\ x_{1} = - 1\\ x_{2} = 1 \\ x_{3} = - 2 \\ x_{4} = 2[/tex]
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[tex]f(x) = \frac{2}{5} {x}^{5} - 3 \frac{1}{3} {x}^{3} + 8x \\ f'(x) = 5 \times \frac{2}{5} {x}^{5 - 1} - 3 \times \frac{10}{3} {x}^{3 - 1} + 8 {x}^{1 - 1} = \\ = 2 {x}^{4} - 10 {x}^{2} + 8 \\ f'(x) = 0 \\ 2 {x}^{4} - 10 {x}^{2} + 8 = 0 \\ {x}^{4} - 5 {x}^{2} + 4 = 0 \\ {x}^{2} = a \: , \: a \geqslant 0[/tex]
По теореме Виета:
[tex]{x}^{2} + bx + c = 0 \\ x _{1} + x_{2} = - b \\ x _{1} x_{2} = c[/tex]
[tex] {a}^{2} - 5a + 4 = 0 \\ a_{1} + a_{2} =5 \\ a_{1} a_{2} = 4\\ a_{1} = 1 \\ a_{2} =4 [/tex]
[tex] {x}^{2} = 1 \\ {x}^{2} = 4 \\ x_{1} = - 1\\ x_{2} = 1 \\ x_{3} = - 2 \\ x_{4} = 2[/tex]