Few questions about function domains and properties

[acrash]

(There is a better formatted word version of all of this in the attachments)

Hi
I was trying to solve some problems in the website AoSP when I came across this problem
https://artofproblemsolving.com/wiki/index.php?title=1990_AHSME_Problems/Problem_8

which made me question few things in my understanding of functions and their domains and solutions.
The question asks about the number of real solutions to the equation |X-2| + |X-3| = 1
and my process was this: -

1) We are trying to find the real solutions to the equation whose range is all the real numbers (no undefined points), So we are going to have to check all of R (real numbers)

2) since both absolute values have the same variable in them, then the same ranges caused by the absolute value should apply on that variable as follows:

A) When [X≥2 and X≥3] = [X≥2 ∩ X≥3] = [ X≥3] (which is possible), we have:

X-2 + X-3 = 1 => 2X-5 = 1 => X = 3, which is valid because X=3 ∈ X≥3

B) When [X≥2 and X≤3] = [X≥2 ∩ X≤3] = [2≤X≤3] (which is possible), we have:

X-2 + 3-X = 1 => 1 = 1 => X = any value, which means X=any value ∈ 2≤X≤3

C) When [X≤2 and X≤3] = [X≤2 ∩ X≤3] = [X≤2] (which is possible), we have:

2-X + 3-X = 1 => -2X+5 = 1 => X = 2, which is valid because X=2 ∈ X≤2

D) When [X≤2 and X≥3] = [X≤2 ∩ X≥3] = [3≤X≤2] = Φ (because it is IMPOSSIBLE and therefore will cause a contradiction), we have:

2-X + X-3 = 1 => -1 = 1 => This case can’t be.

So we have the sets of solutions:
A(X≥3) = {3}, B(2≤X≤3) = {all values of X such that 2≤X≤3}, C(X≤2) ={2}, D(3≤X≤2) = Φ

So the solution set to the original problem is
A ∪ B ∪ C (all of R)= {all values of X such that 2≤X≤3}.
Which means that there are infinite solutions.

But that got me thinking. What if the original problem was like this |X-2| + |X-3| = T
Where T is ANY element in the set {1,-1}.
My thinking it would be like this (please correct me if I’m wrong):

A) When [X≥2 and X≥3] = [X≥2 ∩ X≥3] = [ X≥3] (which is possible), we have

X-2 + X-3 = T => 2X-5 = T =>

1) if T=1 => 2X-5 = 1 => X = 3, valid because X=3 ∈ X≥3

2) if T=-1 => 2X-5 = -1 => X = 2 not valid because X=2 ∉ X≥3

B) When [X≥2 and X≤3] = [X≥2 ∩ X≤3] = [2≤X≤3] (which is possible), we have

X-2 + 3-X = 1 => 1 = T => X = any value, since 1 ∈ T, which means X=any value ∈ 2≤X≤3

C) When [X≤2 and X≤3] = [X≤2 ∩ X≤3] = [X≤2] (which is possible), we have

2-X + 3-X = T => -2X+5 = T =>

1) if T=1 => -2X+5 = 1 => X = 2, valid because X=2 ∈ X≤2

2) if T=-1 => -2X+5 = -1 => X = 3 not valid because X=3 ∉ X≤2

D) When [X≤2 and X≥3] = [X≤2 ∩ X≥3] = [3≤X≤2] = Φ (because it is IMPOSSIBLE but …… let’s see)

2-X + X-3 = T => -1 = T => X = any value, since -1 ∈ T, which means X=any value ∈ 3≤X≤2

So we have the sets of solutions:
A(X≥3) = {3}, B(2≤X≤3) = {all values of X such that 2≤X≤3}, C(X≤2) = {2}, D(3≤X≤2) = {all values of X such that 3≤X≤2}

So the solution set to the original problem is

A ∪ B ∪ C ∪ D (all of R)= {all values of X such that 2≤X≤3}. Because there are no real values of X such that 3≤X≤2

I know that an easier way of looking at this problem is by noticing that two positive real numbers can never add up to a negative and that way the problem goes back to the original one, but I think that this way is still valid (correct me if I’m wrong please)
Which got me thinking about the equation at the step D and how it had infinite solutions but none of them valid because we were taking into account the domain we restricted its useful solutions in (the range of useful answers). How 2-X + X-3 = -1 has infinite solutions, but the fact that the domain 3≤X≤2 doesn’t exist made all the solutions invalid to be incorporated into the set of solutions for the equation |X-2| + |X-3| = T where T is ANY element in the set {1,-1}, even though that 2-X + X-3 = -1 is a form extrapolated from |X-2| + |X-3| = T and utilized in order to solve it.
That in turn made me think about how important it is to determine the domain of a function and the range of the useful answers. So here is my question:
If we have an equation like (2x^2)/x=0 we say that that is equivalent to 2X = 0 which would mean that X = 0 is the solution.

But wouldn’t we say that the original equation (2x^2)/x=0 is not defined at X=0 because we can’t divide by zero and that it actually has no solutions, but the equation 2X = 0 is actually a different equation that has a solution at X=0.

Or can we say that (x^2-1)/(x-1)=x+1 by factoring the numerator and canceling one of the factors with denominator, even though the domain of the LHS equation can’t include X=1 but the domain of the RHS can.

Another example is the equation √x= -2 I can square both sides and get X = 4 but that wouldn’t be the solution to the original equation since even though it is true that -2 and 2 can be squared and get 4, the square root function doesn’t produce any negative values since it just maps positive real numbers to other positive real numbers. But at the same time, I don’t understand what would make the equation X= √16 any different, since most of the time we say that in this case X= ± 4

My way of thinking is this (and please tell me what’s wrong with it). I get a problem or an equation and try to extract some properties about it and about the domain of answers that I expect to limit my self by. I do all the transformations that I need to do on the problem or equation in order to solve it, but then check whether the original problem accepts that answers that I have produced, for example: are the answers within the domain of the function or problem or equation or not. The simplest (but not necessarily mathematically imposed) example would something like finding the side length of a square with a giving area, which can never be negative.

My last question is for an equation like x^2+x=x/3 and the equation x+1=1/3 , the first equation has two solutions {0, -2/3} while the second has only the one solution {-2/3} even though we can transform the first equation into the second one by dividing the two sides by x . So, it seems like dividing by x is somehow destructive. How do I know what kind of transformations preserve the properties of the original equation and what others won’t?

 

[Mark44]

In the future, please limit your posts to one problem.

(There is a better formatted word version of all of this in the attachments)

Hi
I was trying to solve some problems in the website AoSP when I came across this problem
https://artofproblemsolving.com/wiki/index.php?title=1990_AHSME_Problems/Problem_8

which made me question few things in my understanding of functions and their domains and solutions.
The question asks about the number of real solutions to the equation |X-2| + |X-3| = 1
and my process was this: -

1) We are trying to find the real solutions to the equation whose range is all the real numbers (no undefined points), So we are going to have to check all of R (real numbers)

2) since both absolute values have the same variable in them, then the same ranges caused by the absolute value should apply on that variable as follows:

A) When [X≥2 and X≥3] = [X≥2 ∩ X≥3] = [ X≥3] (which is possible), we have:

X-2 + X-3 = 1 => 2X-5 = 1 => X = 3, which is valid because X=3 ∈ X≥3

B) When [X≥2 and X≤3] = [X≥2 ∩ X≤3] = [2≤X≤3] (which is possible), we have:

X-2 + 3-X = 1 => 1 = 1 => X = any value, which means X=any value ∈ 2≤X≤3

C) When [X≤2 and X≤3] = [X≤2 ∩ X≤3] = [X≤2] (which is possible), we have:

2-X + 3-X = 1 => -2X+5 = 1 => X = 2, which is valid because X=2 ∈ X≤2

D) When [X≤2 and X≥3] = [X≤2 ∩ X≥3] = [3≤X≤2] = Φ (because it is IMPOSSIBLE and therefore will cause a contradiction), we have:

2-X + X-3 = 1 => -1 = 1 => This case can’t be.

So we have the sets of solutions:
A(X≥3) = {3}, B(2≤X≤3) = {all values of X such that 2≤X≤3}, C(X≤2) ={2}, D(3≤X≤2) = Φ

So the solution set to the original problem is
A ∪ B ∪ C (all of R)= {all values of X such that 2≤X≤3}.
Which means that there are infinite solutions.

Yes.
There are three intervals to consider: x < 2, 2 < x < 3, and x > 3.
If x < 2, then x - 2 < 0 and x - 3 < 0, so the equation becomes -(x - 2) - (x - 3) = 1. From this equation you get 1 solution.
Next, if 2 < x < 3, then x - 2 > 0, and x - 3 < 0, so the equation becomes x - 2 - (x - 3) = 1, or -2 + 3 = 1, which is identically true for all x in the interval (2, 3).
Finally, if x > 3, then x - 2 > 0, and x - 3 > 0, so the equation becomes x - 2 + x - 3 = 1. This produces 1 solution.

In my work above, I used the idea that if x < 0, then |x| = -x.

We should also consider the cases with x = 2 and x = 3.

<snip>

That in turn made me think about how important it is to determine the domain of a function and the range of the useful answers. So here is my question:
If we have an equation like (2x^2)/x=0 we say that that is equivalent to 2X = 0 which would mean that X = 0 is the solution.

No, this equation has no solutions. If x = 0, the denominator is 0, and division by 0 is not defined.
For all values of x other than 0, the graphs of $y = \frac {2x^2} x$ and $y = 2x$ are identical, but the first one will have a "hole" at the point (0, 0).

But wouldn’t we say that the original equation (2x^2)/x=0 is not defined at X=0 because we can’t divide by zero and that it actually has no solutions, but the equation 2X = 0 is actually a different equation that has a solution at X=0.

Or can we say that (x^2-1)/(x-1)=x+1 by factoring the numerator and canceling one of the factors with denominator, even though the domain of the LHS equation can’t include X=1 but the domain of the RHS can.

Similar to the above.

Another example is the equation √x= -2 I can square both sides and get X = 4 but that wouldn’t be the solution to the original equation since even though it is true that -2 and 2 can be squared and get 4, the square root function doesn’t produce any negative values since it just maps positive real numbers to other positive real numbers. But at the same time, I don’t understand what would make the equation X= √16 any different, since most of the time we say that in this case X= ± 4

No, we don't. $x = \sqrt{16} = 4$, not $\pm 4$.

My way of thinking is this (and please tell me what’s wrong with it). I get a problem or an equation and try to extract some properties about it and about the domain of answers that I expect to limit my self by. I do all the transformations that I need to do on the problem or equation in order to solve it, but then check whether the original problem accepts that answers that I have produced, for example: are the answers within the domain of the function or problem or equation or not. The simplest (but not necessarily mathematically imposed) example would something like finding the side length of a square with a giving area, which can never be negative.

My last question is for an equation like x^2+x=x/3 and the equation x+1=1/3 , the first equation has two solutions {0, -2/3} while the second has only the one solution {-2/3} even though we can transform the first equation into the second one by dividing the two sides by x . So, it seems like dividing by x is somehow destructive. How do I know what kind of transformations preserve the properties of the original equation and what others won’t?

Yes, by dividing both sides by zero, you are losing the possibility that x = 0. You pretty much never want to divide by a variable if that variable could possibly be zero in the original equation.