Question

求函数y=(x^2+7x+19)/(x+2),(x>-2)的最值

Answer 1

y'=[(2x+7)/(x+2)-(x^2+7x+19)/(x+2)^2]

=[(2x+7)(x+2)-(x^2+7x+19)]/(x+2)^2

=[2x^2+11x+14-x^2-7x-19]/(x+2)^2

={x^2+4x-5]/(x+2)^2

=(x-1)(x+5)/(x+2)^2

x>-2,x+5>0

当 x>=1,y'>0,y 递增

当 x