[tex]f(x) = \frac{x^{2}+7x }{x-9} \: \: \: \: \: \: x \in [-4;1]\\x \neq 9\\f'(x) =\frac{(x^{2}+7x)'(x-9)-(x-9)'(x^{2}+7x) }{(x-9)^{2} }=\frac{(2x+7)(x-9)-(x^{2} +7x)}{(x-9)^{2}} =\\=\frac{2x^{2} -18x+7x-63-x^{2} -7x}{(x-9)^{2}}=\frac{x^{2}-18x-63 }{(x-9)^{2}} =\frac{(x-21)(x+3)}{(x-9)^{2}} \\x_{1} =21 \: \: \: \: \: \: x_{2} =-3 \: \: \: \: \: \: x \neq 9\\++++[-3]----(9)----[21]++++\\x_{max} =-3\\x_{min} =21[/tex]

[tex]y(-4)=\frac{(-4)^{2} +7*(-4)}{-4-9}=-\frac{16-28}{13} =-\frac{-12}{13} =\frac{12}{13} \\y(-3)=\frac{(-3)^{2} +7*(-3)}{-3-9}=-\frac{9-21}{12} =-\frac{-12}{12} =1\\y(1)=\frac{1^{2} +7*1}{1-9}=-\frac{1+7}{8} =-\frac{8}{8} =-1[/tex]

Ответ: y max = 1 ; y min = - 1