Question

设x=1和x=2是函数f(x)=x5+ax3+bx+1的两个极值点.问题:(1)求a和b的值;(2)求f(x)的单调区间.

Answer 1

求导f’(x)=5*x^4+3*a*x^2+b=0，5+3*a+b=0；80+12*a+b=0；得：a=-25/3，b=20
f’(x)=5*x^4-25*x^2+20=（x^2-1)(x^2-4)=0得x1=-2，x2=-1，x3=1，x4=2；增区间（负无穷,-2]，[-1,1][2,正无穷]；减区间 [-2,-1][1,2]

Answer 2

f'(x)=5x4+3ax2+b,
f'(1)=5+3a+b=0,
f'(2)=80+12a+b=0,
a=-25/3,b=20
f(x)=x5-25/3x3+20x+1
f'(x)=5x4-25x2+20=5(x-2)(x-1)(x+1)(x+2)
剩下你懂的，写不下了