题目

已知函数f(x)=x^3/(x-1)^2,试求f(x)的凹凸区间和拐点
急用

提问时间：2020-11-03

答案

f(x) = x^3/(x-1)^2
f'(x) = { (x-1)^2*3x^2-x^3*2(x-1) } / (x-1)^4
= { (x-1)*3x^2-2x^3 } / (x-1)^3
= x^2 { 3x-3-2x } / (x-1)^3
= (x^3-3x^2) / (x-1)^3
f''(x) = { (x-1)^3 * 3x(x-2) - x^2(x-3) * 3(x-1)^2 } / (x-1)^6
= { (x-1) * 3x(x-2) - x^2(x-3) * 3 } / (x-1)^4
= 3x { (x-1)(x-2) - x(x-3) } / (x-1)^4
= 3x { x^2-3x+2-x^2+3x } / (x-1)^4
= 6x / (x-1)^4
x＜0时,f''(x)＜0,为凸区间；x＞0时,f''(x)＞0,为凹区间
∴凹区间（0,+∞）,凸区间（-∞,0）,拐点x=0