Let S = {−1, 0, 2, 4, 7}, find f(S) if
1. f(x) = 1
2. f(x) = 2x + 1
3. f(x) = bx/5c
4. f(x) = d(x 2 + 1)/3e
1. f(x) = 1
Explanation: Output is 1 irrespective of of the value of x.
2. f(x) = 2x+1
Explanation: Put each value in set and evaluate.
x = -1 --> 2 * ( - 1 ) + 1 --> - 2 + 1 --> -1
x = 0 --> 2 * ( 0 ) + 1 --> 0 + 1 and so on for other values in S.
3. f(x) = bx/5c
Explanation : Here b and c are contant values as will be provided.
x = -1 --> b * (-1) / 5c --> -b / 5c
x = 0 --> b * (0) /5c --> 0 and so on for other values.
4. f(x) = d(x^2+1)/3e
Explanation : Here d and e are contant values.
x = -1 --> d ( (-1)^2 +1 ) / 3e --> d( 1 + 1) / 3e --> 2d/3e and so on for other values.
The table for all the values in S is as shown below along with mapping as per f(x).
x; x E S | f(x) = 1 | f(x) = 2x + 1 | f(x) = bx/5c | f(x) = d(x^2 + 1)/3e |
-1 | 1 | -1 | -b/5c | 2d/3e |
0 | 1 | 1 | 0 | d/3e |
2 | 1 | 5 | 2b/5c | 5d/3e |
4 | 1 | 9 | 4b/5c | 17d/3e |
7 | 1 | 15 | 7b/5c | 50d/3e |
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