已知9^x-10*3^x+9≤0求函数y=(1/4)^(x-1)-4*(1/2)^x+2的最大值与最小值

已知9^x-10*3^x+9≤0求函数y=(1/4)^(x-1)-4*(1/2)^x+2的最大值与最小值

9^x-10*3^x+9≤0 =====> (3^x)^2-10*3^x+9≤0
设3^x=t, 那么t^2-10t+9=<0

1 -1
1 -9

(t-1)(t-9)=<0
1=<t=<9

3^x>=1, x>=0,

3^x=<3^3, x=<3

y=(1/4)^(x-1)-4*(1/2)^x+2 =(1/4)^x /1/4 - 4*(1/2)^x+2

=4*[(1/2)^x]^2-4(1/2)^x+2
=4k^2-4k+2

3^x>=1, x>=0,

3^x=<3^3, x=<3

k=(1/2)^x=(1/2)^0=1, k=(1/2)^x=(1/2)^3=1/8

所以 ymax=4k^2-4k+2=4-4+2=2

ymin=4k^2-4k+2=1/16-1/2+2 =32/16 - 7/16=25/16