已知f(x)=(x²+1)分之(2ax-a²+1)(a≠0),求f(x)的单调区间及极值

2个回答

  • f(x) = (2ax-a^2+1)/(x^2+1)

    f'(x) = { (x^2+1)*2a - (2ax-a^2+1)*2x } / (x^2+1)^2

    = -2{ax^2+(1-a^2)x-a} / (x^2+1)^2

    = -2a{x^2+(1/a-a)x-1} / (x^2+1)^2

    = -2a(x+1/a)(x-a) / (x^2+1)^2

    (一)当a<0时,f'(x) = -2a(x+1/a)(x-a) / (x^2+1)^2

    x<a,或x>-1/a时,f'(x)>0;a<x<-1/a时,f'(x)<0

    单调增区间(-∞,a),(-1/a,+∞)

    单调减区间(a,-1/a)

    极大值f(a) = (2a*a-a^2+1)/(a^2+1) = 1

    极小值f(-1/a) = (2a*(-1/a)-a^2+1)/((-1/a)^2+1) = -a^2

    (二),当a>0时,f'(x) = -2a(x+1/a)(x-a) / (x^2+1)^2

    x<a,或x>-1/a时,f'(x)<0;-1/a<x<a时,f'(x)>0

    单调减区间(-∞,-1/a),(a,+∞)

    单调增区间(-1/a,a)

    极小值f(-1/a) = (2a*(-1/a)-a^2+1)/((-1/a)^2+1) = -a^2

    极大值f(a) = (2a*a-a^2+1)/(a^2+1) = 1