1. (10) State whether each of these functions, where sets A and B are as stated in #1, are one-to-one, onto, both, or neither, and give a brief explanation for each answer:
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Answers:

A = {q, r, s, t} and B = {17, 18, 19, 20}

a)f is one to one because  x1,x2  A , x1≠x2 then f(x1) ≠ f(x2)

f is onto because  y  B , xA / f(x)=y

b)g is not one to one because t,r  A , t≠r and g(t) = g(r) =20

g is not onto because 18B and g(x)≠18  x  A

c)h is not one to one because t,r  A , t≠r and h(t) = h(r) =20

h is not onto because 18B and h(x)≠18  x  A

d)k is not one to one because 17,18  B , 17≠18 and k(17) = k(18) =17

k is not onto because 18B and k(x)≠18  x  B

e)l is one to one because  x1,x2  A , x1≠x2 then f(x1) ≠ f(x2)

l is onto because  y  A , xA / f(x)=y

  1. (8) Determine all bijections from A into B.
  1. A = {q, r, s} and B = {2, 3, 4}
  1. A = {1, 2, 3, 4} and B = {5, 6, 7, 8}

Answers:

a)There are 3! = 6 bijections from A into B

f1 = {(q,2),(r,3),(s,4)}, f2 = {(q,2),(r,4),(s,3)} , f3 = {(q,3),(r,2),(s,4)}

f4 = {(q,3),(r,4),(s,2)}, f5 = {(q,4),(r,2),(s,3)} , f6 = {(q,4),(r,3),(s,2)}

b) There are 4! = 24 bijections from A into B

f1 = {(1,5),(2,6),(3,7),(4,8)}, f2 = {(1,5),(2,6),(3,8),(4,7)} , f3 ={(1,5),(2,7),(3,6),(4,8)}

f4 = {(1,5),(2,7),(3,8),(4,6)}, f5 = {(1,5),(2,8),(3,6),(4,7)} , f6 ={(1,5),(2,8),(3,7),(4,6)}

f7 = {(1,6),(2,5),(3,7),(4,8)}, f8 = {(1,6),(2,5),(3,8),(4,7)} , f9 ={(1,6),(2,7),(3,5),(4,8)}

f10 = {(1,6),(2,7),(3,8),(4,5)}, f11 = {(1,6),(2,8),(3,5),(4,7)} ,f12={(1,6),(2,8),(3,7),(4,5)}

f13 = {(1,7),(2,5),(3,6),(4,8)}, f14 = {(1,7),(2,5),(3,8),(4,6)} ,f15={(1,7),(2,6),(3,5),(4,8)}

f16 = {(1,7),(2,6),(3,5),(4,8)}, f17 = {(1,7),(2,8),(3,5),(4,6)} ,f18={(1,7),(2,8),(3,6),(4,5)}

f19 = {(1,8),(2,5),(3,6),(4,7)}, f20 = {(1,8),(2,5),(3,7),(4,6)} ,f21={(1,8),(2,6),(3,5),(4,7)}

f22 = {(1,8),(2,6),(3,7),(4,5)}, f23 = {(1,8),(2,7),(3,5),(4,6)} ,f24={(1,8),(2,7),(3,6),(4,5)}

  1. (10) Which of the following functions from are one-to-one, onto, or both? Prove your answers.

Answers:

a)  x1,x2  R , If f(x1)=f(x2) then 3(x1)-4 = 3(x2) -4 then

3(x1) = 3(x2) then x1 = x2 then f is one to one

If y  R and f(x)=y then 3x-4 = y then 3x = y+4 then x = (y+4)/3 R

Then  y  R, xR / f(x) = y then f is onto

b) 3,-3  R and 3≠-3, g(3)=g(-3) = 7 then g is not one to one

If y = -3 R and f(x) = -3 then x2-2= -3 then x2 =-1 then x R

Then  x  R, f(x) ≠ -3R then g is not onto

c) h(x) is not a function from R to R because h(0) = 2/0 is undefined

d) k(x) is not a function from R to R because k(0) = ln(0) is undefined

e)  x1,x2  R , If l(x1)=l(x2) then ex1=ex2 then ln(ex1) = ln(ex2)

then: x1 = x2 then l is one to one

If y = -3 R and f(x) = -3 then ex= -3 then x R

Then  x  R, f(x) ≠ -3R then l is not onto

  1. (12) State a domain that makes the function one-to-one and explain your answer. If no such domain exists, explain why not.
  1. h(x) = x2-4x+7
  1. |

Answers:

a)Domain = D ={x/x≥3}

In this case if x1,x2  D and f(x1)=f(x2)

then x1-2 = x2-2 thenn x1-2 = x2-2 and x1 = x2 , then f is one to one.

b)Domain = D ={x/x≥1}

In this case if x1,x2  D and f(x1)=f(x2)

then (x1)2 -2 = (x2)2 -2 then (x1)2 = (x2)2 then x1 = x2 , then f is one to one.

c)Domain = D ={x/x≥3}

In this case if x1,x2  D and f(x1)=f(x2)

then (x1)2-4(x1)+7 = (x2)2-4(x2)+7 Since the vertex of f is (2,-5) f is an increasing function in the interval: [2,∞) then x1 = x2 , then f is one to one.

d)Domain = D =R

In this case if x1,x2  D and f(x1)=f(x2)

then (x1)3 +4 = (x2)3 +4 then (x1)3 = (x2)3 then x1 = x2 , then f is one to one.

e)Domain = D =R

In this case if x1,x2  D and f(x1)=f(x2)

then 3(x1) -4 = 3(x2) -4 then 3(x1) = 3(x2) then x1 = x2 , then f is one to one.

f)Domain = D ={x/x≥3}

In this case if x1,x2  D and f(x1)=f(x2)

then -3x1-6 = -3x2-6 then -(3x1-6)=-(3x2-6) then 3x1-6=3x2-6 then 3x1=3x2

then x1 = x2 , then f is one to one.

  1. (12) State a codomain that makes the function onto and explain your answer. If no such codomain exists, explain why not.
  1. |

Answers:

a)Codomain = Range of f = B ={y/y≥0}

In this case if y  B exists xR where f(x) = y then f is onto

b)Codomain = Range of g = B ={y/y≥-2}

In this case if y  B exists xR where g(x) = y then f is onto

c)Codomain = Range of h = B = {y/y ≥ 3}

In this case if y  B exists xR where h(x) = y then f is onto

d)Codomain = Range of k = B = R

In this case if y  B exists xR where h(x) = y then f is onto

e)Codomain = Range of k = B = R

In this case if y  B exists xR where f(x) = y then f is onto

f)Codomain = Range of k = B = {y/y ≤ 0}

In this case if y  B exists xR where l(x) = y then f is onto

  1. (10) Let f = {(-2, 3), (-1, 1), (0, 0), (1, -1), (2, -3)} and

let g = {(-3, 1), (-1, -2), (0, 2), (2, 2), (3, 1)}. Find:

  1. f(1)
  1. g(-1)

a)f(1) = -1

b)g(-1) = -2

c)(gf)(1)= g(f(1)) = g(-1) = -2

d)(gff)(-1)= g(f(f(-1))) = g(f(1)) = g(-1)=-2

e)(fg)(3)= f(g(3)) = f(1) = -1

  1. (8) Define q, r, and s, all functions on the integers, by , , and . Determine:

Answers:

a) qrs = q(r(s)) = q(r(5n-3)) = q(5n-3-7) = q(5n-10) = (5n-10)2 +2(5n-10)

b) rsq = r(s(q)) = r(s(n2+2n)) = r(5(n2+2n)-3) = 5(n2+2n)-3-7 =5(n2+2n)-10

c) (qsr)(8) = q(s(r(8))) = q(s(8-7)) = q(s(1)) = q(5-3)=q(2)=22+22= 8

d) (srq)(2) = s(r(q(2))) = s(r(22+22)) = s(r(8)) = s(8-7)= s(1)= 5-3 = 2

  1. (10) Consider the functions f, g (both on the reals) defined by and .
  1. Show that f is injective.
  1. Show that f is surjective.
  1. Find .
  1. Find .
  1. Find .

Answer:

a)Domain = R

In this case if x1,x2  R and f(x1)=f(x2)

then 8(x1)+5 = 8(x2)+5 then 8(x1)=8(x2) then x1 = x2 , then f is injective

b)

Let yR , if y = f(x) then y = 8x+5 then 8x = y-5 then x = (y-5)/8 R

Then  yR xR / f(x)=y then f is surjective

c)

y = 8x+5 then 8x = y-5 then x = (y-5)/8

Then f-1(x) = (x-5)/8

d) gf(x) = g(f(x)) = g(8x+5) = (8x+5)2 = 64x2+80x+25

e) fg(x) = f(g(x)) = f(x2) = 8x2+5