驗(yàn)證:y=(2x+3)/(x-1) 在x>1時(shí)為減函數(shù)

其他人氣：299 ℃時(shí)間：2020-05-19 02:43:50

優(yōu)質(zhì)解答

驗(yàn)證就免了,隨便代幾個(gè)>1的數(shù)進(jìn)去看一下,是不是x越取得大,y的值越小,就可以了.下面給出證明過(guò)程：
證：
設(shè)1f(x2)-f(x1)
=(2x2+3)/(x2-1) -(2x1+3)/(x1-1)
=[(2x2+3)(x1-1)-(2x1+3)(x2-1)]/[(x2-1)(x1-1)]
=(2x1x2-2x2+3x1-3-2x1x2+2x1-3x2+3)/[(x2-1)(x1-1)]
=(-5x2+5x1)/[(x2-1)(x1-1)]
=-5(x2-x1)/[(x2-1)(x1-1)]
x1>1x2>1x2-1>0x1-1>0
x2>x1x2-x1>0
-5<0
-5(x2-x1)/[(x2-1)(x1-1)]<0
f(x2)-f(x1)<0
f(x2)函數(shù)是減函數(shù).