math question, find the domain and range

a.
in order to find the domain, you need to find out all the value of x
so that the function value f(x) is undefined
for x < 2,
f(x)=-x/2
it is not difficult to see that the function is well-defined for x<2
for x >=2,
f(x)=√(x +2)
the trouble here is the square root,
consider x+2<0
x<-2
however, f(x)=√(x +2) only for """x>=2"""
so for all x>=2
f(x) is well-defined
so the domain of x is all real number

b.
you need to find out the max and min value of f(x)
so that you can find the range of f(x)
firstly, you can find out is f(x) goes to -infinite and +infinite
possible?
obviously, f(x) = -x/2, x<2
as x goes to -infinite, f(x) goes to -infinite too
also,
f(x) = √(x +2), x>=2
as x goes to +infinite, f(x) goes to +infinite too
so the range is also all real number.


if the function cannot goes to -infinite or +infinite
you can try to work out from the domain of x
by putting some value of x
to see the trends of the function value
is it increasing or decreasing or something else.

2008-05-30 20:09:32 補充：
sorry, some mistakes are made
f(x) = -x/2, x<2
as x goes to -infinite, f(x) goes to -infinite too
it is wrong because there is a "-" side so f(x) goes to infinite ...

2008-05-30 20:09:36 補充：
but the range of f(x) is still all real number
f(x) = √(x +2), x greater or equal to 2
square root can be take on positive value or negative value
so as x goes to +infinite
f(x) goes to both -infinite and +infinite

a)
x --> f(x) : R --> R

b)
for -infinity<x<2, infinity>f(x)>-1

for infinity>x>=2, infinity>f(x)>=2

range of f(x) is between infinity and -1