已知函数f(x)=ax3+bx2+cx(a不等0,x属于R)为奇函数,且f(x)在x=1处取极大值2.
问题描述：
已知函数f(x)=ax3+bx2+cx(a不等0,x属于R)为奇函数,且f(x)在x=1处取极大值2.
已知函数f(x)=ax^3+bx^2+cx(a不等0,x属于R)为奇函数,且f(x)在x=1处取极大值2.(1)求函数y=f(x)的解析式 (2)记g(x)=f(x)/x+(k+1)lnx,求函数y=g(x)的单调区间.
答
.f奇,b=0,f'(x)=3ax^2+c,f'(1)=3a+c=0,f(1)=a+c=2,解得a=-1,c=3.f(x)=-x^3+3x.2.g(x)=-x^2+3+(k+1)lnx(x>0),g'(x)=-2x+(k+1)/x=-2[x^2-(k+1)/2]/x,k0,g(x)↑；x>√[(k+1)/2],g'(x)0,x>0,m>-x^2-x+3+3lnx,记为h(x),h'(x)=-2x-1+3/x=(-2x^2-x+3)/x=-(x-1)(2x+3)/x,0