求y=(3x+2)/(x-2) (1

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求y=(3x+2)/(x-2) (1

求y=(3x+2)/(x-2) (1
求y=(3x+2)/(x-2) (1

求y=(3x+2)/(x-2) (1
y=(3x+2)/(x-2)
=(3x-6+8)/(x-2)
=3+8/(x-2)
由函数图像可知,f(x)=8/(x-2)在[1,2]上是单调递减的
所以f(1)>f(2)
f(2)是无穷小
f(1)=-8
所以,y的值域为 (负无穷,-5]

y=(3x+2)/(x-2)=3+8/(x-2)
∵x∈[1,2]函数单调递减
∴y∈(-∝,-5]

y=3+8/(x-2);而1<=x<=2
故y<=3-8=-5

y=(3x+2)/(x-2)
=3+8/(x-2)
(x-2)(y-3)=8
画图像
当1≤x≤2时,y≤-5

y=(3x-6+6+2)/(x-2)=3+8/(x-2),因为x属于【1,2】,故值域为(负无穷,-5]