.
解：令t=2x+8-x^2
a=t^1/2
f(a)=(1/3)^a。定义域t>=0
2x+8-x^2>=0
x^2-2x-8<=0
x^2-2x-8=0
(x-4)(x+2)=0
x-4=0orx+2=0
x1=4,x2=-2
-2<=x<=4
t=-x^2+2x+8
=-(x^2-2x-8)
=-[(x-1)^2-1-8]
=-[(x-1)^2-9]
=-(x-1)^2+9
x=1,tmax=9,amax=9^1/2=3,fmin=(1/3)^3=1/27
x=4orx=-2,tmin=0,amin=0^1/2=0,fmax=(1/3)^0=1
答：fmax=1,fmin=1/27。
内容全部来源于网络收集，如有侵权，请联系网站删除：QQ：24596024