作业帮高中高数f(x)=x+lnx,g(x)=f(x)+(1/2)x^2-bx 1、若g(x)存在单调递减区间,求实数b的取值范围 2、设x1,x2(x1
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高中高数f(x)=x+lnx,g(x)=f(x)+(1/2)x^2-bx 1、若g(x)存在单调递减区间,求实数b的取值范围 2、设x1,x2(x1
高中高数
f(x)=x+lnx,g(x)=f(x)+(1/2)x^2-bx 1、若g(x)存在单调递减区间,求实数b的取值范围 2、设x1,x2(x1
高中高数f(x)=x+lnx,g(x)=f(x)+(1/2)x^2-bx 1、若g(x)存在单调递减区间,求实数b的取值范围 2、设x1,x2(x1
1.g(x)=f(x)+(1/2)x^2-bx = x+lnx +(1/2)x^2-bx
g' = 1 + 1/x + x -b = x + 1/x + 1 - b
g(x)在单调递减区间上g'= 2
x + 1/x < b-1 有解
只需要x+1/x的最小值 小于 b-1 即可
2 < b-1
b>3
2.x + 1/x + 1 - b = 0
x^2+1+(1-b)x = 0
x1+x2 = b-1
x1*x2 = 1
将x1、x2由b表示
g(x1)-g(x2) = x1+lnx1 +(1/2)x1^2-bx1 - (x2+lnx2 +(1/2)x2^2-bx2)
=(1-b)(x1-x2) + ln x1/x2 +1/2(x1^2-x2^2)
=(1-b)(x1-x2) + ln x1/x2 +1/2(x1+x2)(x1-x2)
=[1-b+1/2(x1+x2)](x1-x2) + ln x1/x2
=1/2(1-b)(x1-x2) + ln x1/x2
将x1、x2代入,由b>=7/2
求最小值
1.g(x)=f(x)+(1/2)x^2-bx = x+lnx +(1/2)x^2-bx g' = 1 + 1/x + x -b = x + 1/x + 1 - b g(x)在单调递减区间上g'0(因为lnx) 又x+1/x >= 2 x + 1/x < b-1 有解 只需要x+1/x的最小值 小于 b-1 即可 2 < b-1 b>3 2.x + 1/x + 1 - b = 0 x^2+1+(1...
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1.g(x)=f(x)+(1/2)x^2-bx = x+lnx +(1/2)x^2-bx g' = 1 + 1/x + x -b = x + 1/x + 1 - b g(x)在单调递减区间上g'0(因为lnx) 又x+1/x >= 2 x + 1/x < b-1 有解 只需要x+1/x的最小值 小于 b-1 即可 2 < b-1 b>3 2.x + 1/x + 1 - b = 0 x^2+1+(1-b)x = 0 x1+x2 = b-1 x1*x2 = 1 将x1、x2由b表示 g(x1)-g(x2) = x1+lnx1 +(1/2)x1^2-bx1 - (x2+lnx2 +(1/2)x2^2-bx2) =(1-b)(x1-x2) + ln x1/x2 +1/2(x1^2-x2^2) =(1-b)(x1-x2) + ln x1/x2 +1/2(x1+x2)(x1-x2) =[1-b+1/2(x1+x2)](x1-x2) + ln x1/x2 =1/2(1-b)(x1-x2) + ln x1/x2 将x1、x2代入，由b>=7/2 求最小值
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