Question

设函数f（x）=｛3x+4,x<0:x2-4x+6，x>=0}若互不相等的实数X1,X2,X3满足f(X1)=f(X2)=f(X3),

设函数f（x）=｛3x+4,x<0:x2-4x+6，x>=0}若互不相等的实数X1,X2,X3满足f(X1)=f(X2)=f(X3),则X1+X2+X3的取值范围是

Answer 1

解：函数f（x）=
x2−6x+6， x≥0
   3x+4，  x＜0
的图象，如图，

不妨设x1＜x2＜x3，则x2，x3关于直线x=3对称，故x2+x3=6，
且x1满足-
7
3
＜x1＜0；
则x1+x2+x3的取值范围是：-
7
3
+6＜x1+x2+x3＜0+6；
即x1+x2+x3∈（
11
3
，6）

Answer 2

x<0时，f（x）=3x+4<4
x>=0时，f（x）=x2-4x+6=(x-2)^2+2>=2
若互不相等的实数X1,X2,X3满足f(X1)=f(X2)=f(X3),=m
则2<m<4(等号舍去）
3x+4=m,x1=(m-4)/3
(x-2)^2+2=m,x2=2+根号（m-2),x3=2-根号（m-2),
X1+X2+X3=8/3+m/3
则10/3<X1+X2+X3<4

Answer 3

f（x）=3x+4 ,x<0
= x^2-4x+6 ，x>=0
without loss of generality,
Assume : x1< x2< x3

case 1: x1,x2,x3 <0
=> x1=x2=x3 ( rejected)

case 2: x1,x2<0 x3 >= 0
x1=x2 (rejected)

case 3: x1,<0 x2, x3 >= 0
for x<0
-无穷 <f(x1)< 4

for x>=0
f(x) = x^2-4x+6
= (x-2)^2 + 2
2<= f(x2)=f(x3) < +无穷

f(x1)=f(x2)=f(x3)
2<= f(x1)=f(x2)=f(x3) <4
2<= 3x1+4 <4 and 2<= x2^2-4x2+6<4
-2/3 <=x1 < 0 and x2^2-4x2+4 >=0 and x2^2-4x2+2 <0
-2/3 <=x1 < 0 and (x2-2)^2>0 and 2-√2< x2 < 2+√2
-2/3+ 2(2-√2)<x1+x2+x3 < 0+ 2(2+√2)
10/3- 2√2 < x1+x2+x3 < 2(2+√2)

case 4: x1,x2,x3>0 (rejected)

ie
10/3- 2√2 < x1+x2+x3 < 2(2+√2)

Answer 4

此题 考的就是画图
要使f(x1)=f(x2)=f(x3)
即使
2<f(x1)=f(x2)=f(x3)<4
假定x1<x2<x3
x2+x3=2*2=4
2<3x1+4<4
-2/3<x1<0
10/3<x1+x2+x3<4