For 2 sets X={1,2,3,4},Y={−15,0,30},
f is a surjective function from X to Y. If f(1)=−15 and f(2)=f(4)=0,
what is f(3) ?
argggh,, must admit this is a new one for me
http://www.mathsisfun.com/sets/injective-surjective-bijective.html
every Y must be in f(X).
Since -15 and 0 are taken, we still need something to map to 30.
So, f(3) = 30.
To determine the value of f(3), we need to understand the concept of a surjective function and use the given information.
A function is surjective (or onto) if every element in the codomain (Y in this case) has at least one pre-image in the domain (X). In simpler terms, it means that for every element in Y, there is at least one element in X that maps to it.
Given that f is a surjective function, we can conclude that every element in Y should have a pre-image in X.
From the given information:
- f(1) = -15: This tells us that 1 maps to -15.
- f(2) = f(4) = 0: This tells us that both 2 and 4 map to 0.
To find f(3), we can examine the elements in Y and check if any of them are missing their pre-images from X.
Looking at Y, the values are -15, 0, and 30.
We already know that -15 has a pre-image of 1, and 0 has pre-images of 2 and 4.
Hence, the only element remaining in Y without a pre-image is 30. This means that f(3) must be 30.
Therefore, f(3) = 30.