函数f(x)=x2+bx+c,且f(3)=f(0),f(1)=0
代入得
9+3b+c=c
0=1+b+c
解得b=-3,c=2
f(x)=x^2-3x+2
F(x)=log2/3f(x)=log2/3[x^2-3x+2]
由于底数小于0,因此f(x)单增时,F(x)单减；f(x)单减时,F(x)单增
且f(x)＞0
f(x)=x^2-3x+2=(x-2)(x-1)>0
x>2或x2时,f(x)单增,F(x)单减
当x