Answer Part B only. Show proof with explanation step by step. Given a function f

Question

Answer Part B only. Show proof with explanation step by step.

Given a function f : D rightarrow R and a subset B R. let f-1(B) be the set of all points from the domain D that get mapped into B: that is, f-1(B) = {x D : f(x) B}. This set is called the preimage of B. Let f(x) = x2. If A is the closed interval [0,4] and B is the closed interval [-1, 1], find f-1(A) and f-l(B). Does f-1(A B) = f-l(A) f-l(B) in this case? Does f-l(A B) = f-1(A) f-1(B)? (b) The good behavior of preimages demonstrated in (a) is completely general. Show that for an arbitrary function g : R rightarrow R, it is always true that g-l(A B) = g-1(A) g-l(B) and g-l(A B) = g-1 (A) g-l(B) for all sets A, B R.

Explanation / Answer

a) ( inter stands for intersection)

Let

g^-1 ( A inter B) = x

hence

g(x) belongs to A inter B

hence

g(x) individually belongs to both A and B

Hence all such points which belong to both A and B are present in (A inter B).

Now the preimages of A and those of B have only these x points in common.

hence

g^-1(A inter B) = g^-1(A) inter g^-1(B)

b) ( U stands for union)

Let

g^-1 ( A U B) = x

hence

g(x) belongs to A U B

hence

g(x) individually belongs to either A or B or both

Hence all such points which belong to either A or B or both are present in (A U B).

Now the preimages of A and those of B have only these x points in common.

hence

g^-1(A U B) = g^-1(A) U g^-1(B)

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