[tex]f(x) = \frac{ {x}^{2} + 5 }{2 - x} \\ f'(x) = \frac{( {x}^{2} + 5)'(2 - x) - (2 - x) '( {x}^{2} + 5)}{(2 - x) {}^{2} } = \\ \frac{2x(2 - x) - ( - 1)(x {}^{2} + 5) }{(2 - {x}^{2}) } = \\ \frac{4x - 2 {x}^{2} + {x}^{2} + 5}{(2 - x) {}^{2} } = \frac{ - {x}^{2} + 4x + 5 }{(2 - x) {}^{2} } = \\ \frac{ - (x + 1)(x - 5)}{(2 - x) {}^{2} } \\ f'(x) = 0 \\ \frac{ - (x + 1)(x - 5)}{(2 - x) {}^{2} } = 0 \\ x _{1} = - 1 \\ x_{2} = 5[/tex]

Ответ:

[tex]x_{min} = - 1 \\ x_{max} = 5[/tex]