兩條f6函數問題.....

1
log(2/3)(x^2-3x+2)
=log(x^2-3x+2)/log(2/3)
=-a[log(x-2)(x-1)] where a>0
So for 1<=x<=2 f(x) undefined
dy/dx
=-a(2x-3)/(x-2)(x-1)
if x>2, then dy/dx<0 and so y strictly decreasing
if x<1,then dy/dx>0 and so y strict;y increasing
2
對(1)
由第一題只需考慮f(1),f(2)
f(1)=log2/log(1/2)
f(2)=log3/log(1/2)
故最大值為f(1)=-1﹐最小值為f(3)=log6/log(1/2)
(2)
因log6>0﹐故用同法可得x在[1,3]單調上升
最大值f(3)=2﹐最小值f(1)=1