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1. Why is f not a function from R to R ifa) f (x) = 1/x? b) f (x) =x ? c) f (x) =2 1x  ?2. Determine whether f is a function from Z to R ifa) f (n) = ±n b) 2 1f n n  c) 214f n n 3. Find these valuesa)1.1   b)0.1   c)4   d)3.2  e)5.2   e)2   e)1 22 3      4. Determine whether each of these functions from {a, b, c, d} to itself is one-to-one (onto)a) f (a) = b, f (b) = a, f (c) = c, f (d) = db) f (a) = b, f (b) = b, f (c) = d, f (d) = cc) f (a) = d, f (b) = b, f (c) = c, f (d) = d5. Determine whether each of these functions from Z to Z is one-to-one (onto)a) f (n) = n − 1 b) f (n) = n2 + 1 c) f (n) = n3 d) 2nf n     6. Determine whether f : Z × Z → Z is onto ifa) f (m, n) = 2m − n b) f (m, n) = m2 − n2 c) f (m, n) = m + n + 1d) f (m, n) = |m| − |n| e) f (m, n) = m2 − 4 f) f (m, n) = m + n7. Determine whether each of these functions is a bijection from R to R.a) f (x) = −3x + 4 b) f (x) = −3x2 + 7 c) f (x) = (x + 1)/(x + 2) d) f (x) = x5 + 18. Let S = {−1, 0, 2, 4, 7}. Find f (S) ifa) f (x) = 1 b) f (x) = 2x + 1 c) 5xf x     

Question

  1. Why is f not a function from R to R ifa) f (x) = 1/x? b) f (x) =x ? c) f (x) =2 1x  ?2. Determine whether f is a function from Z to R ifa) f (n) = ±n b) 2 1f n n  c) 214f n n 3. Find these valuesa)1.1   b)0.1   c)4   d)3.2  e)5.2   e)2   e)1 22 3      4. Determine whether each of these functions from {a, b, c, d} to itself is one-to-one (onto)a) f (a) = b, f (b) = a, f (c) = c, f (d) = db) f (a) = b, f (b) = b, f (c) = d, f (d) = cc) f (a) = d, f (b) = b, f (c) = c, f (d) = d5. Determine whether each of these functions from Z to Z is one-to-one (onto)a) f (n) = n − 1 b) f (n) = n2 + 1 c) f (n) = n3 d) 2nf n     6. Determine whether f : Z × Z → Z is onto ifa) f (m, n) = 2m − n b) f (m, n) = m2 − n2 c) f (m, n) = m + n + 1d) f (m, n) = |m| − |n| e) f (m, n) = m2 − 4 f) f (m, n) = m + n7. Determine whether each of these functions is a bijection from R to R.a) f (x) = −3x + 4 b) f (x) = −3x2 + 7 c) f (x) = (x + 1)/(x + 2) d) f (x) = x5 + 18. Let S = {−1, 0, 2, 4, 7}. Find f (S) ifa) f (x) = 1 b) f (x) = 2x + 1 c) 5xf x     
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Solution

The question seems to be a mix of text and mathematical symbols, which makes it difficult to understand. However, I'll try to answer the parts that I can decipher:

  1. a) f(x) = 1/x is not a function from R to R because it's undefined for x=0. b) f(x) = x is a function from R to R because for every x in R, there is a unique y in R. c) I'm unable to decipher the function.

  2. a) f(n) = ±n is not a function from Z to R because it doesn't assign a unique output to each input. b) f(n) = n^2 + 1 is a function from Z to R because for every n in Z, there is a unique y in R. c) I'm unable to decipher the function.

  3. I'm unable to decipher the values you want me to find.

  4. a) The function is one-to-one (injective) but not onto (surjective). b) The function is not one-to-one (not injective) and not onto (not surjective). c) The function is one-to-one (injective) and onto (surjective).

  5. a) f(n) = n - 1 is one-to-one and onto. b) f(n) = n^2 + 1 is not one-to-one but it's onto. c) f(n) = n^3 is one-to-one and onto. d) I'm unable to decipher the function.

  6. I'm unable to decipher the functions.

  7. a) f(x) = -3x + 4 is a bijection. b) f(x) = -3x^2 + 7 is not a bijection. c) f(x) = (x + 1)/(x + 2) is not a bijection. d) f(x) = x^5 + 1 is a bijection.

  8. a) f(S) = {1} b) f(S) = {-1, 1, 5, 9, 15} c) I'm unable to decipher the function.

Please provide a clearer question for a more accurate response.

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Similar Questions

7. Determine whether each of these functions is a bijection from R to R.a) f (x) = −3x + 4 b) f (x) = −3x2 + 7 c) f (x) = (x + 1)/(x + 2) d) f (x) = x5 + 1

9. Let f be the function from R to R defined by f (x) = x2. Finda) f −1({1}) b) f −1({x | 0 < x < 1}) c) f −1({x | x > 4})

5. Determine whether each of these functions from Z to Z is one-to-one (onto)a) f (n) = n − 1 b) f (n) = n2 + 1 c) f (n) = n3 d) 2nf n     6. Determine whether f : Z × Z → Z is onto ifa) f (m, n) = 2m − n b) f (m, n) = m2 − n2 c) f (m, n) = m + n + 1d) f (m, n) = |m| − |n| e) f (m, n) = m2 − 4 f) f (m, n) = m + n7. Determine whether each of these functions is a bijection from R to R.a) f (x) = −3x + 4 b) f (x) = −3x2 + 7 c) f (x) = (x + 1)/(x + 2) d) f (x) = x5 + 18. Let S = {−1, 0, 2, 4, 7}. Find f (S) ifa) f (x) = 1 b) f (x) = 2x + 1 c) 5xf x     

Which of the following equations is a function?a.)b.)c.)d.)

For each of the following, state whether it is possible to have a function meeting the criteria given, explain why orwhy not, and if it is possible, give two examples.(a) A function f : N≥0 → N≥0 which is not surjective.(b) A function f : N≥0 → Z which is injective.(c) A function f : Q → Q which is bijective.

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