2个回答
2012-01-04 · 知道合伙人教育行家
关注
展开全部
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
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
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询