# Thread: Find f(S) where S=[1,2]

1. ## Find f(S) where S=[1,2]

I'm completely lost on this. My first answer was f(S)=[0,3]. How are they getting [1,3]? If f is the set of images of the points in the interval [1,2], then don't I simply plug in 1 and 2 into f(x) to find the new set?

2. ## Re: Find f(S) where S=[1,2]

Originally Posted by divinelogos
I'm completely lost on this. My first answer was f(S)=[0,3]. How are they getting [1,3]? If f is the set of images of the points in the interval [1,2], then don't I simply plug in 1 and 2 into f(x) to find the new set?
If $f(x)=4-x^2$ then $f([1,2])=[0,3]$.
I do not agree with the posted image.

3. ## Re: Find f(S) where S=[1,2]

well, no you don't simply "plug in the values of the endpoints" unless your function is non-decreasing/non-increasing on the interval. as it so happens, 4 - x2 is decreasing on [1,2], so in this instance you can do that. f(1) = 3, and f(2) = 0, so the image f([1,2]) = [0,3] (we have to reverse the order of the endpoints because f is decreasing).

on the other hand, we couldn't do this for [-2,2], since f(-2) = f(2) = 0, but clearly the image of this interval contains more than the point {0}.

i'm not surprised that the book is wrong, since its definition of f(S) is likewise incorrect, it should read:

$f(S) = \{f(s) | s\in S = [1,2]\}$, although perhaps this is a bit nit-picky on my part.