Hi,
Let f: R --> R be the function f(x) = (x + 1)^2. Compute the following sets:
i) f [{-1}]
ii) f^-1 [[-1, 1]]
iii) f[ [-1, -1] ] union [1, 3] ]
iv) f[f^-1 [ [-3, -1] ] ]
Is the set of points which are the image of all the points in {-1} under f.
As {-1} contains but one point f({-1}) = {f(-1)} = {0}.
Is the set of all points which f maps into the closed interval [-1,1], or allii) f^-1 [[-1, 1]]
points x such that |f(x)|<=1. As the domain of f is R and f(x)>0 for
all x in R this is the set of all x in R such that:
(x+1)^2 <=1
or:
-1<x+1<=1,
or:
-2<=x<=0
which is the closed interval [-2,0].
The use of brackets here is a bit confusing, but I will assume this means:iii) f[ [-1, -1] ] union [1, 3] ]
f([-1,-1] Union [1,3]) = {f(-1)} Union f([1,3]) = {0} Union [1,16]
As we are working in R f^-1([-3,-1]) = NullSet (because there is no realiv) f[f^-1 [ [-3, -1] ] ]
x such that (x+1)^2 is negative), and f(NullSet)=NullSet, so:
f(f^-1 ([-3, -1])) = NullSet.
RonL
We have extended the definition of f from its domain D to the power set ofOriginally Posted by ImperfectHacker
its domain by defining:
f(A): {all b in the Range of f, such that there exists a in A such that f(a)=b}.
Thus if A is the null set (which is in the power set of D), there is no b in the
range of f which is the image of an a in A. Therefore f(NullSet)=NullSet.
RonL