# More help.. on functions

• Oct 27th 2006, 06:15 PM
jenjen
More help.. on functions
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] ] ]
• Oct 27th 2006, 10:31 PM
CaptainBlack
Quote:

Originally Posted by jenjen
Hi,

Let f: R --> R be the function f(x) = (x + 1)^2. Compute the following sets:

i) f [{-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}.

Quote:

ii) f^-1 [[-1, 1]]
Is the set of all points which f maps into the closed interval [-1,1], or all
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].
Quote:

iii) f[ [-1, -1] ] union [1, 3] ]
The use of brackets here is a bit confusing, but I will assume this means:

f([-1,-1] Union [1,3]) = {f(-1)} Union f([1,3]) = {0} Union [1,16]

Quote:

iv) f[f^-1 [ [-3, -1] ] ]
As we are working in R f^-1([-3,-1]) = NullSet (because there is no real
x such that (x+1)^2 is negative), and f(NullSet)=NullSet, so:

f(f^-1 ([-3, -1])) = NullSet.

RonL
• Oct 28th 2006, 03:46 PM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlank
As we are working in R f^-1([-3,-1]) = NullSet (because there is no real
x such that (x+1)^2 is negative), and f(NullSet)=NullSet, so:

f(f^-1 ([-3, -1])) = NullSet.

I would say that such a question cannot be asked. Since the inverse image is the empty set we cannot calculate the image of that since mappings of null sets was not defined (at least not in my book).
• Oct 28th 2006, 10:19 PM
CaptainBlack
Quote:

Originally Posted by ImperfectHacker
Quote:

Originally Posted by CaptainBlack
As we are working in R f^-1([-3,-1]) = NullSet (because there is no real
x such that (x+1)^2 is negative), and f(NullSet)=NullSet, so:

f(f^-1 ([-3, -1])) = NullSet.

I would say that such a question cannot be asked. Since the inverse image is the empty set we cannot calculate the image of that since mappings of null sets was not defined (at least not in my book).

We have extended the definition of f from its domain D to the power set of
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
• Oct 28th 2006, 11:14 PM
jenjen
Thank you so much for the replies ThePerfectHacker and CaptainBlack!!!!!