Question

已知1/3≤a≤1,若函数f(x)=ax^2-2x+1在区间[1,3]上的最大值为M(a),最小值为N(a),令g(a)=M(a)-N(a) (1)求g

已知1/3≤a≤1,若函数f(x)=ax^2-2x+1在区间[1,3]上的最大值为M(a),最小值为N(a),令g(a)=M(a)-N(a)
(1)求g(a)的函数表达式.

Answer 1

1/3≤a≤1，则有1≤1/a≤3，
y=ax^-2x+1对称轴方程为X=1/a,抛物线开口向上,
1)当1≤1/a<2,即,1/2<a≤1.
f(x)max=f(3)=M(a)=9a-6+1=9a-5.
f(x)min=f(1/a)=N(a)=-1/a+1.
g(a)=9a+1/a-6

2)1/a=2时,即,a=1/2,
M(a)=9a-5=-1/2.
N(a)=-1/a+1=-1.
g(a)=1/2
3)2<1/a≤3,即,1/3≤a<1/2.
M(a)=f(1)=a-1.
N(a)=f(1/a)=-1/a+1.
g(a)=a+1/a-2

追问

为什么要用1<=1/a<2,1/a=2,2<1/a<=3呢？

Answer 2

1/3≤a≤1，则有1≤1/a≤3，
y=ax^-2x+1对称轴方程为X=1/a,抛物线开口向上,
1)当1≤1/a<2,即,1/2<a≤1.
f(x)max=f(3)=M(a)=9a-6+1=9a-5.
f(x)min=f(1/a)=N(a)=-1/a+1.
g(a)=9a+1/a-6

2)1/a=2时,即,a=1/2,
M(a)=9a-5=-1/2.
N(a)=-1/a+1=-1.
g(a)=1/2
3)2<1/a≤3,即,1/3≤a<1/2.
M(a)=f(1)=a-1.
N(a)=f(1/a)=-1/a+1.
g(a)=a+1/a-2