Question

ДАЮ 40 БАЛІВ
Знайти точки екстремуму
$f(x) = \frac{3}{4} {x}^{4} - 4 {x}^{3} + 4.5 {x}^{2} + 3$
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Answer 1

$f(x) = \frac{3}{4} {x}^{4} - 4 {x}^{3} + 4.5 {x}^{2} + 3 \\ f'(x) = \frac{3 }{4} \times 4 {x}^{4 - 1} - 4 \times 3 {x}^{3 - 1} + \frac{9}{2} \times 2 {x}^{2 - 1} = \\ 3 {x}^{3} - 12 {x}^{2} + 9x = 3x( {x}^{2} - 4x + 3) \\ f'(x) = 0 \\ x( {x}^{2} - 4x + 3) = 0 \\ x( {x}^{2} - x - 3x + 3) = 0 \\ x(x(x - 1) - 3(x - 1)) = 0 \\ x(x - 1)(x - 3) = 0 \\ x _{1} =0 \\ x_{2} = 1\\ x _{3} = 3 \\ - - - [0] + + + [1] - - - [3] + + + \\ x _{min} = 0\\ x_{max} = 1\\ x_{min} = 3$