What Are the Images and Pre-Images of Intervals Under a Squaring Function?

[twoski]

Homework Statement

If A and B are sets and f : A → B, then for any subset S of A we define:

f(S) = {b ∈ B : b = f(a) for some a ∈ S}

Similarly, for any subset T of B we define the pre-image of T as:

f$^{-1}$(T) = {a ∈ A : f(a) ∈ T}

Note that f$^{-1}$(T) is well defined even if f does not have an inverse.

Now let f : R → R be defined as f(x) = $x^{2}$

Let S1 denote the closed interval [−2, 1], and let S2 be the open interval (−1, 2). Also let T1 = S1 and T2 = S2.

Determine:

f(S1 ∪ S2)

f(S1) ∪ f(S2)

f(S1 ∩ S2)

f(S1) ∩ f(S2)

f$^{-1}$(T1 ∪ T2)

f$^{-1}$(T1) ∪ f$^{-1}$(T2)

f$^{-1}$(T1 ∩ T2)

f$^{-1}$(T1) ∩ f$^{-1}$(T2)

The Attempt at a Solution

I think I'm being asked to verify that each subset satisfies "some b = f(a) for some a ∈ S" where S is the subset i have to compute.

For example, the first question is f(S1 U S2) = f( [-2,2) ).

My question is, what do i have to do next? I can't plug in every value x where -2 =< x < 2 since I'm using real numbers.

 

[haruspex]

For example, the first question is f(S1 U S2) = f( [-2,2) ).

Yes.

I can't plug in every value x where -2 =< x < 2 since I'm using real numbers.

Of course you can't, but if you sketch a graph of the function it will suggest an answer which you can then set about proving.

 

[twoski]

Well since my first answer has a range spanning from -2 up to but not including 2, it is one-to-one. The question wants me to answer whether or not every value passed to the function will map to one value, i think.

It's easy for me to see that f( [-2,2) ) is valid, but how would i go about showing it? The question doesn't explicitly ask me to solve anything, it just says "determine" so maybe the answer is just f( [-2,2) )... Bah, now I'm all paranoid that I'm missing something...

 

[haruspex]

Well since my first answer has a range spanning from -2 up to but not including 2, it is one-to-one.

How so? 1-1 would mean that no two values in [-2,2) get mapped by f to the same value. False for f(x)=x², surely.

The question wants me to answer whether or not every value passed to the function will map to one value, i think.

No, it's asking for the set of values obtained by applying the function to all the numbers in the range [-2,2).

 

[twoski]

Could i denote the set of values as [0,4]? Since we're working with real numbers it wouldn't be feasible to list all of the possible values.

 

[micromass]

Could i denote the set of values as [0,4]? Since we're working with real numbers it wouldn't be feasible to list all of the possible values.

Yes. That is the correct answer. But can you justify it?? Can you actually prove that $f([-2,2])=[0,4]$??

 

[twoski]

Would it be sufficient to say that for every value x in (-2,2], f(x) will be no less than 0 and no greater than 4? I mean, there doesn't seem to be any other way to put it. I can't invert the function and plug in 0 and 4, so all i can really do is plug in the important values in my domain (ie. -1.99999, 0, 2)

I finished the rest of them...

f(S1) U f(S2) = f( [-2,1] ) U f( (-1,2) ) = [1,4] U (1,4) = [1,4]

f(S1 ∩ S2) = f( (-1,1] ) = [0,1]

f(S1) ∩ f(S2) = f( [-2,1] ) ∩ f( (-1,2) ) = [1,4] ∩ (1,4) = (1,4)

I'm starting to wonder whether i have to actually prove my answer for each question, it seems like it would be a bit redundant... Maybe i am just supposed to show the range and that's it?

The second part with the inverse is confusing to me.

f'(T1 U T2) = f'( [-2,2) )

So the result of this would be every number x where $x^{2} \in [-2,2)$ or in other words, the square roots of every number in [-2,2), i think. Which would give me a messy result.

 

[haruspex]

Would it be sufficient to say that for every value x in (-2,2], f(x) will be no less than 0 and no greater than 4?

That shows f(S1 U S2) $\subseteq$ [0,4]. Also need to show the converse.

f(S1) U f(S2) = f( [-2,1] ) U f( (-1,2) ) = [1,4] U (1,4) = [1,4]

No. Please do sketch the graph in each case. What does f look like over [-2,1]?

f(S1 ∩ S2) = f( (-1,1] ) = [0,1]

Yes.

f(S1) ∩ f(S2) = f( [-2,1] ) ∩ f( (-1,2) ) = [1,4] ∩ (1,4) = (1,4)

No.

The second part with the inverse is confusing to me.
f'(T1 U T2) = f'( [-2,2) )
So the result of this would be every number x where $x^{2} \in [-2,2)$

Every real number x, that is

or in other words, the square roots of every number in [-2,2)

No, that's not the same. There are two differences. First, you don't have to have a square root for every number. There is no real number for which the square is -1, so the answer will not include a square root of -1. Secondly, 'the square root' of a positive number is generally defined to mean the positive square root, but here we want every real number that has a square in the desired range.
This question illustrates the difference between a pre-image function, as can be applied to sets, and an inverse function as applied to the values.

 

[twoski]

No, that's not the same. There are two differences. First, you don't have to have a square root for every number. There is no real number for which the square is -1, so the answer will not include a square root of -1. Secondly, 'the square root' of a positive number is generally defined to mean the positive square root, but here we want every real number that has a square in the desired range.
This question illustrates the difference between a pre-image function, as can be applied to sets, and an inverse function as applied to the values.

So the answer to f$^{-1}$( [-2,2) ) is the set of real numbers obtained through $x^{2}$ in [-2,2)

Would i do this with set builder notation? The answer can't be [0,1.4142]..

 

[haruspex]

So the answer to f$^{-1}$( [-2,2) ) is the set of real numbers obtained through $x^{2}$ in [-2,2)

Would i do this with set builder notation? The answer can't be [0,1.4142]..

No, but that's nearly right. Check the bounds.

 

[twoski]

So if i were to write it as [0, √2], would that technically be correct? This solution doesn't account for negative numbers, which throws me off.

 

[haruspex]

So if i were to write it as [0, √2], would that technically be correct? This solution doesn't account for negative numbers, which throws me off.

As I said, that's nearly right. Are the squares of all numbers in that range contained within [-2,2)?