# formula

(See attached file for full problem description with proper symbols)

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2. Let f(x) = x2 +1 and g(x) = {x+1, x> =3; x-1, x<3 so both f and g map R into

Find the formula for

a. (f+g)(x)

b. (f .g)(x)

c. (f o g)(x)

d. (g o f)(x)

3. Let A = {a,b,c,d} and B = {1,2,3} and let f : A  B be a function . Let g : Z  2Z, where 2Z = {0,+-2,+-4,+-6 ...}

a. Could f be one to one? Must f be one to one? Explain

b. Could f be onto? Must f be onto? Explain

c. Could g be one to one? Must g be one to one? Explain

d. Could g be onto? Must g be onto? Explain

4. Let ≡ be the relation on Z given by n ≡ m mod 5 iff 5|(n-m). Show that equivalence mod 5 is an equivalence relation on Z.

5. Let f : A  B, g : B  C so that g o f : A  C is a function from A to C and suppose that g o f is one to one.

a. Show that f is one to one

b. Show that if in addition f is onto, g is one to one

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#### Solution Preview

Please see the attached file.

2. Let f(x) = x2 +1 and g(x) = {x+1, x> =3; x-1, x<3 so both f and g map R into

Find the formula for

a. (f+g)(x)

Since (f+g)(x) is the sum of two functions f(x) and g(x), we have

b. (f .g)(x)

Since (f.g)(x) is the product of two functions f(x) and g(x), we have

c. (f o g)(x)

d. (g o f)(x)

Since only when , we have

3. Let A = {a,b,c,d} and B = {1,2,3} and let f : A B be a function . Let g : Z 2Z, where 2Z = {0,+-2,+-4,+-6 ...}

a. Could f be one to one? Must f be one to one? Explain

F cannot be one to one. Since by definition of "one-to-one", we know that if , then . So, . ...

#### Solution Summary

This solution is comprised of a detailed explanation to find the formula.