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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.Originally Posted by jenjen
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
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 ofI 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).
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
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