几道求函数值域的题 急1 f(x)=(2x^2-x+1)/(x-1) (x>1)2 f(x)={-x^2-2x (-2

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几道求函数值域的题 急1      f(x)=(2x^2-x+1)/(x-1)         (x>1)2      f(x)={-x^2-2x  (-2

几道求函数值域的题 急1 f(x)=(2x^2-x+1)/(x-1) (x>1)2 f(x)={-x^2-2x (-2
几道求函数值域的题 急
1 f(x)=(2x^2-x+1)/(x-1) (x>1)
2 f(x)={-x^2-2x (-2

几道求函数值域的题 急1 f(x)=(2x^2-x+1)/(x-1) (x>1)2 f(x)={-x^2-2x (-2
1.f(x)=(2x^2-x+1)/(x-1) (x>1)
解析:∵f(x)=(2x^2-x+1)/(x-1),其定义域为x>1
令F’(x)=[(4x-1)(x-1)-(2x^2-x+1)]/(x-1)^2=(2x^2-4x)/(x-1)^2=0
X1=0(舍),x2=2
∵2x^2-4x为开口向上的抛物线,当x渐增取过x=2时,F’(x)由负变正
∴f(x)在x=2处取最小值f(2)=7
∴函数f(x)的值域为[7,+∞)
2.f(x)={-x^2-2x (-2x=-1/2
∴f(x)在x=-1/2处取极大值
∵1