Question

знайдіть найбільше і найменшою значення функції на проміжку: f(x)=x^3-4x^2+4x+3, [1;3]
помогите пожалуйста ​

Answer 1

$f(x) = {x}^{3} - 4 {x}^{2} + 4x + 3 \\ f'(x) = 3 {x}^{2} - 2 \times 4x + 4 = 3 {x}^{2} - 8x + 4 \\ 3 {x}^{2} - 8x + 4 = 0 \\ a = 3 \\ b = - 8\\ c = 4 \\ D = {b}^{2} - 4ac = ( - 8) {}^{2} - 4 \times 3 \times 4 = 64 - 48 = 16 \\ x_{1} = \frac{8 - 4}{2 \times 3} = \frac{4}{6} = \frac{2}{3} \\ x_{2} = \frac{8 + 4}{2 \times 3} = \frac{12}{6} = 2 \\ + + + + [ \frac{2}{3} ] - - - - [2] + + + + \\ x_{max} = \frac{2}{3} \\ x_{min} = 2 \\ \\ f(1) = {1}^{3} - 4 \times {1}^{2} + 4 \times 1 + 3 = 1 - 4 + 4 + 3 = \\ = 4 \\ \\ f( \frac{2}{3} ) = ( \frac{2}{3} ) {}^{3} - 4 \times (\frac{2}{3} ) {}^{2} + 4 \times \frac{2}{3} + 3 = \frac{8}{27} - 4 \times \frac{4}{9} + \frac{8}{3} + 3 = \\ = \frac{8}{27} - \frac{16}{9} + \frac{8}{3} + 3 = \frac{8}{27} - \frac{48}{27} + \frac{72}{27} + \frac{81}{27} = \frac{113}{27} = 4 \frac{5}{27} \\ \\ f(2) = {2}^{3} - 4 \times {2}^{2} + 4 \times 2 + 3 = \\ = 8 - 4 \times 4 + 8 + 3 = 19 - 16 = 3 \\ \\ f(3) = {3}^{3} - 4 \times {3}^{2} + 4 \times 3 + 3 = \\ = 27 - 4 \times 9 + 12 + 3 = 42 - 36 = 6$

Ответ: у min = 3 ; y max = 6