Answer : Finding the preimage (s) of a value a by a function f is equivalent to solving equation f(x)=a f ( x ) = a . Finding the preimage (s) of a value a by a function f , which has a known curve, is equivalent to find the abscissae of the intersection(s) of the curve with the ordinate line y=a
In the following list of properties, f: A → B indicates than f is function (or map) from A to B:
If f : A → B then
f [A] = ∅
f -1[B] = ∅.
If f : A → B and X, Y ⊆ B then f-1[X ∩ Y] = f-1[X] ∩ f-1[Y].
If f: A → B and X, Y ⊆ B then f-1[X ∪ Y] = f-1[X] ∪ f-1[Y]
If f: A → B and X, Y, ⊆ A then
f [A ∪ B] = f|A| ∪ f |B|
f [A ∩ B] ⊆ f|A| ∩ f |B|In the following list of properties, f: A → B indicates than f is function (or map) from A to B:
If f : A → B then
f [A] = ∅
f -1[B] = ∅.
If f : A → B and X, Y ⊆ B then f-1[X ∩ Y] = f-1[X] ∩ f-1[Y].
If f: A → B and X, Y ⊆ B then f-1[X ∪ Y] = f-1[X] ∪ f-1[Y]
If f: A → B and X, Y, ⊆ A then
f [A ∪ B] = f|A| ∪ f |B|
f [A ∩ B] ⊆ f|A| ∩ f |B|
The image of 5 is √5.
The preimage of 5 is 25; for ℕ(the set of natural numbers) it is the set of all perfect squares in ℕ [2].
Explanation: