Question

已知函数f(x)=|x+2|-x+3

1,写出函数y=f(x)的单调区间 2求函数y=f(x^2-3)的值域 3求不等式f(1-x^2)>f(2x)的解集

Answer 1

x<-2时，f(x)= -(x+2) - x + 3 = -2x+1, f(x)单调递减。
x>=-2时，f(x) = (x+2) - x + 3 = 5. f(x)为常数。

x^2 - 3 < -2, x^2< 1, -1<x<1时，y=f(x^2-3)=-2(x^2-3) + 1 = -2x^2 + 7, 7 >= y > 7-2=5.
x^2 - 3>=-2, x^2 >=1, x>=1或x<=-1时，y=f(x^2-3) = 5.
因此，y=f(x^2-3)的值域为[5,7].

1-x^2 < -2, x^2>3, x>3^(1/2)或x<-3^(1/2)时，f(1-x^2) = -2(1-x^2) + 1 = 2x^2 - 1,
x<-3^(1/2)时，2x < -2*3^(1/2) < -2, f(2x) = -2(2x) + 1 = -4x + 1,
f(2x) = -4x+1 < f(1-x^2) = 2x^2 - 1, 0 < 2x^2 + 4x - 2, 0 < x^2 + 2x - 1 = (x+1)^2 - 2 = [x+1-2^(1/2)][x+1+2^(1/2)], x > 2^(1/2)-1或x<-1-2^(1/2).
又，x<-3^(1/2), 因此，x < -1 - 2^(1/2).

x>3^(1/2)时，2x > 2*3^(1/2) > 0 >= -2,f(2x) = 5,
f(2x) = 5 < f(1-x^2)=2x^2 - 1, 0 < 2x^2 - 6, 0< x^2 - 3, x>3^(1/2)或，x<-3^(1/2).
又，x>3^(1/2),因此，x>3^(1/2).

-3^(1/2)<=x<-1时，3>=x^2>1, 2 >=x^2 - 1>0, -2<=1-x^2<0, f(1-x^2)=5.
2x < - 2, f(2x) = -2(2x)+ 1=1-4x,
f(2x)=1-4x<f(1-x^2)=5, -4<4x, -1<x.
又,-3^(1/2)<=x<-1, 因此,无解。

-1<=x<=3^(1/2)时，x^2 <=3, x^2 - 1 <=2, 1-x^2 >= -2, f(1-x^2) = 5.
-2 <= 2x, f(2x) = 5,
f(2x) = 5 < f(1-x^2) = 5, 无解。

综合，有，f(1-x^2)>f(2x)的解集为{x|x<-1-2^(1/2)或x>3^(1/2)}