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求下列函数的值域(1)y=x/x2+2(x<0)(2)y=2x2-x+1/2x-1
1个回答
展开全部
y=x/x2+2
1/y=x+2/x≥2√2 (x>0)
1/y=x+2/x≤-2√2 (x<0)
-2√2≤y≤2√2
(2)y=2x2-x+1/2x-1=2x+1/[2x-1]=2x-1+1/[2x-1]+1≥2+1=3
1/y=x+2/x≥2√2 (x>0)
1/y=x+2/x≤-2√2 (x<0)
-2√2≤y≤2√2
(2)y=2x2-x+1/2x-1=2x+1/[2x-1]=2x-1+1/[2x-1]+1≥2+1=3
追问
题目搞错了......y=x/(x^2+2),
y=(2x^2-x+1)/(2x-1),(x<1/2)
追答
y=x/x2+2
1/y=x+2/x≥2√2 (x>0)
1/y=x+2/x≤-2√2 (x<0)
-√2/4≤y≤√2/4 (x=√2最大值,可验证)
(2)y=2x2-x+1/2x-1=2x+1/[2x-1]=2x-1+1/[2x-1]+1>2+1=3
y=2x2-x+1/2x-1>3 (x=1/2可取等号)
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