Question

设函数f(x)=｛e^(x-1),x<1,x^(1/3),x≥1,则使得f（x）≤2成立的x的取值

设函数f(x)=｛e^(x-1),x<1,x^(1/3),x≥1,则使得f（x）≤2成立的x的取值范为

Answer 1

设X1大于X2大于等于2 f（x1）=x1^2+a/x1 f（x2）=x2^2+a/x2
因为在x区间[2,正无穷）上为增函数所以f（x1)-f（x2）大于0
x1^2+a/x1 -(x2^2+a/x2)大于0
(x1+x2)(x1-x2)+a(x2-x1)/x1x2大于0
(x1-x2)((x1+x2)x1x2-a/x1x2)大于0
因为x1-x2大于0 x1x2大于0所以(x1+x2)x1x2-a大于0
(x1+x2)x1x2最小等于(2+2)*2*2=16但x1大于x2所以这大于16
所以a的取值范围为小于16

解：f(x)=x^2+a/x
f'(x)=2x-a/x^2
若f(x)在[2,+∞)上为增函数，则：
f'(x)=2x-a/x^2≥0
a≤2x^3≤16
a∈(-∞,16]

Answer 2

当x<1时
f(x)=e^(x-1)<=2
∵e^(x-1)>0
∴两边取对数ln得
lne^(x-1)<=ln2
x-1<=ln2
x<=1+ln2

∴x<=1+ln2 1

当x>=1时
x^(1/3)<=2
x<=2^3=8
∴1<=x<=8 2
结合1式2式得
1<=x<=1+ln2

Answer 3

e^(x-1)<=2解得
x<=ln2+1
同时x<1
故x<1
x^(1/3)<=2解得
x<=8
同时x>=1
故1=<x<=8
故x取值为x<1和1=<x<=8