Ответ:

jdjsjsj

Объяснение:

[tex]f(x) = \frac{ {x}^{2} + 3x }{x + 4} \: \: \: \: \: \: \: \: [ - 3; \: - 1] \\ x\neq - 4 \\ f'(x) = \frac{( {x}^{2} + 3x)'(x + 4) - (x + 4)'( {x}^{2} + 3x) }{(x + 4) {}^{2} } = \\ = \frac{(2x + 3)(x + 4) - ( {x}^{2} + 3x) }{(x + 4) {}^{2} } = \\ = \frac{2 {x}^{2} + 8x + 3x + 12 - {x}^{2} - 3x }{(x + 4) {}^{2} } = \\ = \frac{ {x}^{2} + 8x + 12}{(x + 4) {}^{2} } = \frac{(x + 2)(x + 6)}{(x + 4) {}^{2} } \\ x_{1} = -6 \: \: \: \: \: \: x_{2} = - 2 \: \: \: \: \: \: x\neq - 4 \\ + + + [ - 6] - - - ( - 4) - - - [ - 2] + + + \\ x_{min} = - 2[/tex]

Проверим все варианты на промежутке:

[tex]f( - 3) = \frac{( -3 ) {}^{2} + 3 \times ( - 3) }{ - 3 + 4} = \frac{9 - 9}{1} = \frac{0}{1} = 0 \\ f( - 2) = \frac{( - 2) {}^{2} + 3 \times ( - 2)}{ - 2 + 4} = \frac{4 - 6}{2} = - \frac{2}{2} = - 1 \\ f( - 1) = \frac{( - 1) {}^{2} + 3 \times ( - 1)}{ - 1 + 4} = \frac{1 - 3}{3} = - \frac{2}{3} [/tex]

Ответ: - 1