Question

已知函数f(x)=ax³+bx²+cx(a≠0,x∈R）为奇函数,且f(x)在x=1处取得极大值2

Answer 1

（1）f(x)=ax³+bx²+cx(a≠0,x∈R）为奇函数,

∴b=0,f'(x)=3ax^2+c

f(x)在x=1处取得极大值2,

∴f(1)=a+c=2,

f'(1)=3a+c=0,

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=(k+1-2x^2)/x,

k0,g(x)↑；x>√[(k+1)/2]时g'(x)0),

h'(x)=-2x-1+3/x=-2(x-1)(x+3/2)/x,

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