题目

函数x/x^2-3x+2的增区间（用导数求解）

提问时间：2020-12-19

答案

y=x/(x^2-3x+2)=x/(x-1)(x-2)
y'=x'/[(x-1)(x-2)]+[x/(x-2)]*[1/(x-1)]'+[x/(x-1)]*[1/(x-2)]'
=1/[(x-1)(x-2)]-[x/(x-2)]*[1/(x-1)^2]-[x/(x-1)]*[1/(x-2)^2]
=1/[(x-1)(x-2)]-x/[(x-2)(x-1)^2]-x/[(x-1)(x-2)^2]
=(x-1)(x-2)/[(x-1)^2(x-2)^2]-x(x-2)/[(x-2)(x-1)^2]-x(x-1)/[(x-1)^2(x-2)^2]
=[(x-1)(x-2)-x(x-2)-x(x-1)]/[(x-1)^2(x-2)^2]
=[(x-1)(x-2)-x(x-2)-x(x-1)]/[(x-1)^2(x-2)^2]
=(2-x^2)/[(x-1)^2(x-2)^2]
令y'=0,则x^2=2 【分母[(x-1)^2(x-2)^2]>0】
y'>0时,增区间为：[-√2,√2]