If A and D are sets and f A rightarrow B then for any subset

If A and D are sets and f: A rightarrow B, then for any subset of we define f(S) = {b epsilon B: b = f(a) for some a epsilon S} Similarly, for any subset T of B we define the pre-image of T as f^-1 (T) = {a epsilon A f(a) epsilon T}. Now let f: M rightarrow M be defined as f(x) = x^2. Let S_1 denote the closed interval [-2, 1], that is all x EK that satisfy - 2 lessthanorequalto x lessthanorequalto 1, and let S_2 be the open interval (-1, 2), that is all x epsilon M that satisfy - 1

Solution

Ok, so starting with part A:

S1=T1={-2,-1,0,1,2}

S2=T2={0,1}

Also, fx=x^2 is surjective: simply apply x^2 on these sets:

f ( S1US2)=f(S1)={4,1,0}

f(S1) U f(S2)={4,1,0} U {0,1}={4,0,1}

f(S1 int S2)=f(0,1)={0,1}

f(S1) int f(S2)={4,1,0} int {0,1}={0,1}

Now pre image: is same asinv image, so f-inverse would be sqrt (x) so apply sqare root:

Note: If x < 0, you would be taking the square root of a negative number, so x must be 0.

Also, f-inv({2})= since we can\'t sqare root a negative number. Also, Its not necessary that everything in Set is in the range/image of f, its just a subset, thats why f-inv is well defined, so they\'ll exist.Because x² is an even function, the preimage under f(x) = (x²) will be two, symmetric, intervals of real numbers

f-inv ( T1UT2)=f-inv(T1)={0,1,-1,2 ,-2}

f-inv(T1) U f(T2)={0,1,-1,2 ,-2}

f-inv(T1 int T2)=f-inv(T2)={0,1,-1}

f-inv(T1) int f(T2)={0,1,-1}.