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] ] ]

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- October 27th 2006, 06:15 PMjenjenMore 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] ] ] - October 27th 2006, 10:31 PMCaptainBlack
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]]

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] ]

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

Quote:

iv) f[f^-1 [ [-3, -1] ] ]

**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 - October 28th 2006, 03:46 PMThePerfectHacker
- October 28th 2006, 10:19 PMCaptainBlackQuote:

Originally 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 - October 28th 2006, 11:14 PMjenjen
Thank you so much for the replies ThePerfectHacker and CaptainBlack!!!!!