How Do You Determine the Image and Pre-Image of Intervals in Functions?

[missavvy]

Homework Statement

Define f(x) = |x| - 1, what is f((-2,3])?
What is f^-1((-2,3])?

Homework Equations

The Attempt at a Solution

[-2,1] is the answer for the first part.
I don't understand why though... so you have
f((-2,3]) = |(-2,3]| - 1

I don't quite get how to take the absolute value of an interval. Could someone maybe explain it to me?

:)
THANKS!

 

[lanedance]

[-2,1] is the answer for the first part.

this isn't correct, think of the function operating on an interval to give all points defined as
f{(a,b)} = {f(x) | x is in (a,b)}

now what is the minimum f(x) = |x| - 1 can be for any x?
what is f(-2), f(3), and f(0)

 

[HallsofIvy]

You don't "take the absolute value of an interval".

For set A, f(A) means the set of all values of f(x) for x in A: $f(A)= \{f(x)| x\in A\}$.

For x between -2 and 0, |x|= -x so f(x)= |x|- 1= -x- 1, a linear function. f(-2)= 2- 1= 1 and f(0)= 0- 1= -1 so for x between -2 and 0, f(x) takes on all values between -1 and 1.
For x between 0 and 3, |x|= x so f(x)= |x|- 1= x- 1, a linear function. f(0)= -1 and f(3)= 2 so for x between 0 and 1, f(x) takes on all values between -1 and 2.

That is, for all x between -2 and 3, f(x) takes on all values between -1 and 2. Notice that -2 is not included in (-2, 3] but f(-2)= 1= f(2) so f((-2, 3])= [-1, 2].

$f^{-1}(B)$ is set of all x values such that $f(x)\in B$.

Notice that, even though the function f(x)= |x|- 1 is NOT one-to-one and so does NOT have an inverse (that is, $f^{-1}(x)$ is not defined) $f^{-1}(-2,3])$, applied to a set, is defined. If we look at x< 0, we see that f(x)= |x|- 1= -x- 1= 3 when x= -4. f(x) is never equal to -2 but does go between -1 and 3 (and so is in (-2,3]) for x between -4 and 0. For x> 0, f(x)= |x|- 1= x- 1= 3 when x= 4. Again it does not go down to -2 but for every x between 0 and 4, lies in the set (-2, 3]. That is, $f^{-1}((-2, 3])= [-4, 4].$