Help me with these problems. I will allocate the max number of points (3000) 11.
ID: 2970946 • Letter: H
Question
Help me with these problems. I will allocate the max number of points (3000)
11. Determine if the following relation C on the set of real numbers is reflexive, symmetric, or transitive:
xCy iff |x
Which of the following relations are functions? For those relations that are functions, give the domain and two sets that could be a codomain. Identify the domain, range, and another possible codomain for each of the following mappings. Assuming that the domain of each of the following functions is the largest possible subset of R, find the domain and range of Show that the following relations are not functions on R. Explain why the functions f(x) = 9-x2/x+3 and g(x) = 3 - x are not equal. Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A Which of the following functions map onto their indicated codomains? Prove each of your answers. Which of the functions in Exercise 1 are one-to-one? Prove each of your answers. Let A = {1, 2, 3,4}. Describe a codomain B and a function f: A rightarrow B such that f is Find sets A, B, C, and functions f : A rightarrow B and g: B rightarrow C such thatExplanation / Answer
1. Only c is a function, as the other two do not have unique outputs for a given input (for instance (b) has both (1,2) and (1,3), and (d) has (0, 0) and (0, 2 pi)) The domain of c is {1,2}. The codomain could be Z or N (or really any set that contains the elements in the range, {1,2})
3. The domain is the set Z - {2}. Noting that for x=/=2 we have y = x+2, the range is Z - {4}. Another codomain could be Q.
4. (b) The domain and range are both R.
(c) The domain is all reals such that (x+pi) > 0, so the domain is (-pi, infinity). Since the denominator can be any positive number, the quotient is also any positive number, so the range is (0, infinity).
(e) For f to be defined, we must have 5-x >= 0 and x-3 >= 0. Thus we require x <= 5 and x >=3. So the domain is [3,5].
Noting that the function is symmetric about x=4, we see that the function attains a maximum value a (4,2) and minima at (3,sqrt(2)) and (5, sqrt(2)). Since the function is continuous on its domain, the range is [sqrt(2), 2]
6. (a) (1,1) and (1, -1) are in R, so R is not a function
(1, 0) and (1, 2pi) are in R, so R is not a function.
12. Note that the functions are equal where they are both defined. However x=-3 is not in the domain of f, whereas it is in the domain of g.
19. I will grudgingly say partially correct, although the claim is definitely wrong. The functions are equal where they are both defined. But x=0 is not in the domain of f, but it is in the domain of g. So the functions are not equal.
1. d) f(x) = x^3 is onto. If y is any real number, then x = (y)^(1/3) is well defined and f(x) = y.
(e) f(x) is not onto since square roots cannot be negative. There is no x in R such that f(x) = -1.
2. (d) f(x) = x^3 is one-to-one. This is easy to see because f is increasing, so whenever x < y we have that f(x) < f(y), so in particular, f(x) =/= f(y).
e) f(x) is not one-to-one. Note that f(-2) = 3 = f(2), so there are two numbers in the domain of f that map to the same point in the range.
4. (a) B = {1}, f(x) = 1 for all x in A.
(b) f(x) = x, B = Z
(c) f(x) = 2x, B = {2,4,6,8}
(d) f(x) = 1, B = Z
8. (a) A = Z, B = {1, 2} , C = {1,2} g(x) = x, f(x) = 1. g maps B onto C, but (g circle f)(x) = g(1) = 1 for all x in A, and thus is not onto.
(b) A = B = C = Z. f(x) = x. g(x) = x^2. Then f is one-to-one, but (g circle f)(x) = g(x) = x^2 is not one-to-one, since (g circle f)(-2) = 4 = (g circle f)(2)
(c) A = B = C = Z, f(x) = x^2, g(x) = x. Then g is one-to-one, but (g circle f) (x) = g(x^2) = x^2 is not one-to-one since (g circle f)(-2) = 4 = (g circle f)(2)
11. I can't tell what relation goes between "|x-y|" and "1", but I'll do the problem assuming it is "less than or equal to."
(a) Is C reflexive? Well, for any real number x, we have |x-x| = 0 <= 1, so (x,x) is in C and thus C is reflexive.
(b) Is C symmetric? Note that |x-y| = |y-x|, so for any two real numbers, if xCy, then |x-y|<= 1. This implies that |y-x| <=1, which means yCx. So C is symmetric.
(c) Is C transitive? No. We have that (3/4, 0) is in C, and (0,-3/4) is in C, but |(3/4) - (-3/4)| = 1.5 > 1, so (3/4, -3/4) is not in C, so C is not transitive.
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