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

Ответ: у max = 0 ; y min = - 1