已知函数f(x)=1/2x^4+bx^3+cx^2+dx+e分别在x=0处和x=1处取得极值

f'(x)=2x^3+3bx^2+2cx+df''(x)=6x^2+6bx+2cf'(0)=d=0,f'(1)=2+3b+2c+d=0所以b=-2c/3-2/3又f''(0)=2c≠0,f''(1)=6+6b+2c=6+6(-2c/3-2/3)+2c=-2c+2≠0所以c≠0且c≠12)函数f(x)在x=0处取得极大值,那么f''(0)0,c...