Question

已知函数f(x)=1/3ax^3-bx^2+(2-b)x+1在x=x1处取得极大值,在x=x2处取得极小值,且0<x1<1<x2<2

若z=a+2b,求z 的取值范围

Answer 1

f(x)=1/3ax^3-bx^2+(2-b)x+1
f'(x)=ax²-2bx+2-b
取得最值时须f'(x)=0
根据题意0<x1<1<x2<2
可知f'(x)=a(x-x1)(x-x2)
f'(x)/a=x²-2bx/a+(2-b)/a
因0<x1<1<x2<2
可知
f'(0)/a=x1*x2>0,f'(2)/a=(2-x1)(2-x2)>0,
f'(1)/a=(1-x1)(1-x2)<0
=>(2-b)/a>0,4-4b/a+(2-b)/a>0,(a-2b+2-b)/a<0,
=>(2-b)/a>0,(4a-5b+2)/a>0,(a-3b+2)/a<0
待定系数法设m(-b)+n(4a-5b)=(a-3b)
则4n=1,-5n-m=-3
=>n=1/4,m=7/4
=>(2-b)7/4a+(4a-5b+2)/4a-(a-3b+2)/a>0
=>(14-7b+4a-5b+2-4a+12b-8)/a>0
=>8/a>0
即a>0
由a>0,(2-b)/a>0,(4a-5b+2)/a>0,(a-3b+2)/a<0
=>2-b>0,4a-5b+2>0,a-3b+2<0
通过图像法解
z=a+2b
设a+2b=k(4a-5b)+m(a-3b)
得1=4k+m,2=-5k-3m
=>k=5/7,m=-13/7
则a+2b=5(4a-5b+2)/7-13(a-3b+2)/7+16/7>16/7

Answer 2

a+2b=1
特殊值吧