Day one calculus question - finding domains

Factual Questions
1

Although I have for many years used many calculus techniques in electrical engineering and finding areas and volumes of irregular shapes (manual integration), I have never had an actual calculus class.

So, I thought I would teach myself. Example problem two already disagrees with with what I would expect. Is the lesson wrong, or do I need to update my reasoning on this?



                  x - 2
      f(x) = ----------
                x^2 - 4

I recognize the bottom can be factored immediately and come to:

                  (x - 2)
      f(x) = ----------
                (x-2)(x+2)

and then:

                     1
      f(x) = ----------  
                 (x+2)

so… I would consider the domain all reals excluding x = -2.

The lesson says all real exluding x = 2 and -2. The don’t remove the (x-2) terms from the top and bottom. Is there a reason for this? I would think if they weren’t gonna do this there would be an explanation. Leaving that term in has a 0 numerator when the denominator goes to zero. Is that a problem?

2

Maybe because a 2 in to the original f(x) makes the denominator 0 ?

3

At the same time the numerator is zero. Both forms of the function should be indentical by everything I have been taught. 0/0 is not generally undefined.

4

0/0 is always undefined. If this were a limit, it would be a different matter. But, your book is right, and it goes back to the fact that division by 0 is undefined.

5

Note that the reductions you applied to get from the original function are only valid when the denominator is not zero. That’s what you’re getting caught by.

6

Note that the reductions you applied to get from the original function are only valid when the denominator is not zero. That’s what you’re getting caught by.

At the risk of double posting, let me clarify that. When x equals 2, 1/(x + 2) is equal to 1/4. But (x - 2)/(x[sup]2[/sup] - 4) is equal to 0/0 (which is undefined). And there’s no good reason why 1/4 should equal 0/0.

7

Sorry. My previous post is probably a little unclear. Let me try to explain it another way. The last step of Algebra you perform in the OP is not allowable. Specifically, you can’t do it when x = 2. If x = 2, then you are dividing the numerator and the denominator by 0, which as you know is not allowed. So, the proper way to do the last step is:

f(x) = 1 / (x + 2), when x is not equal to 2.

So you have to exclude x = 2 before you even get to this step.

Now, as I said, it would be a completely different problem if the function were defined thus:

f(x) = lim (y -> x) of (y - 2) / (y[sup]2[/sup] - 4)

In that case, your Algebra would be more correct.

8

Thanks for the answer. I see this is correct, but you would think the lesson might mention that the operation that is normally not an issue when solving a regular algebra equation IS an issue when it is a function.

In summary, in normal algebra, doing this simplication is fine as you know you are solving for a specific real value?

9

No. Dividing by zero, or a quantity equal to zero, is always forbidden. If you divide by (x - a), then you have to specify that x is not equal to a.

10

scotth,

Find the error in the following “proof” that 1 = 2, and I think you’ll understand why you can’t just cancel like that:

Let a = b
a = b           (Given)
ab = b[sup]2[/sup]        (multiply both sides by b)
ab - a[sup]2[/sup] = b[sup]2[/sup] - a[sup]2[/sup]    (subtract b[sup]2[/sup] from both sides)
a(b - a) = (b + a)(b - a)     (factor)
a = b + a          (cancel common factor – (b - a) )
a = a + a        (substitution)
a = 2a       (combine)
1 = 2         (cancel common factor – a)