I have the following example: "For the function f : A = {a, b, c, d, e} → B = {1, 2,..., 6} defined in Example 10.2 by f = {(a, 3), (b, 5), (c, 2), (d, 3), (e, 6)}, it follows that f −1(3) = {a, d}, f −1({1, 3}) = {a, d}, f −1(4) = ∅ and f −1(B) = A." How does f −1({1, 3}) = {a, d}, looking at f I dont see 1?
I have the following example: "For the function f : A = {a, b, c, d, e} → B = {1, 2,..., 6} defined in Example 10.2 by f = {(a, 3), (b, 5), (c, 2), (d, 3), (e, 6)}, it follows that f −1(3) = {a, d}, f −1({1, 3}) = {a, d}, f −1(4) = ∅ and f −1(B) = A." How does f −1({1, 3}) = {a, d}, looking at f I dont see 1?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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I have the following example:
"For the function f : A = {a, b, c, d, e} → B = {1, 2,..., 6} defined in Example 10.2 by
f = {(a, 3), (b, 5), (c, 2), (d, 3), (e, 6)},
it follows that f −1(3) = {a, d}, f −1({1, 3}) = {a, d}, f −1(4) = ∅ and f −1(B) = A."
How does f −1({1, 3}) = {a, d}, looking at f I dont see 1?
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