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求函数y=(x^2+7x+19)/(x+2),(x>-2)的最值
人气:327 ℃ 时间:2019-10-19 09:13:34
解答
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
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