设原点坐标是O
1)AO段:
f(x)=y=2x
g(x)=2x*(x-1)
=2x^2-2x
=2(x^2-x+1/4)-1/2
=2(x-1/2)^2-1/2
所以该段:x=1/2时,g(x)最大值为1/2
2)AB段
f(x)=y=3-x
g(x)=(3-x)(x-1)
=3x-3-x^2+x
=-x^2+4x-3
=-(x^2-4x+4)+1
=-(x-2)^2+1
所以该段:x=2时,最大值为1
综上,g(x)最大值为1
设原点坐标是O
1)AO段:
f(x)=y=2x
g(x)=2x*(x-1)
=2x^2-2x
=2(x^2-x+1/4)-1/2
=2(x-1/2)^2-1/2
所以该段:x=1/2时,g(x)最大值为1/2
2)AB段
f(x)=y=3-x
g(x)=(3-x)(x-1)
=3x-3-x^2+x
=-x^2+4x-3
=-(x^2-4x+4)+1
=-(x-2)^2+1
所以该段:x=2时,最大值为1
综上,g(x)最大值为1