Question

If h(x) is equal to (x^2-4)/(x+2)when x cannot equal to -2, and h(x) is continuous for all real numbers, then what is the value of h(- 2)?

A. 0
B. -2
C. -4
D. 2
E. Impossible

Answer 1

"when x cannot equal to -2"

Answer 2

I'm thinking impossible but not too sure. That is why I'm asking

Answer 3

Well, you are right.

Answer 4

can h(x) actually be continuous for all real number even if it's undefined at x=2?

Answer 5

I suppose so... if the limits are right... nvm

Answer 6

You cannot divide by 0 and that's thAT....

Answer 7

Well I'm working on limits right now

Answer 8

My question was, "does this question make sense?" My first reaction was that if h(x) wasn't defined at x=-2, then it couldn't be continuous at all real x values as described. But I wasn't thinking much.

@estudier is right about the answer to this question though

Answer 9

If the limit is the same and it is not infinity, you can say that it is defined as the limit because you would always be able to change the function to a similar one that does not have the 0 on the bottom, but some people say that if it is writen this way, its because whoever wrote it didn't want it to be defined on -2, but since you are working with limits its probably the first case.

Answer 10

yes, that's the thing I wasn't thinking of at first :) But then I knew I walked into the trap. Thanks for the explanation...

Answer 11

Strictly, if u cancel through it's a different function.

Answer 12

I hate these half baked questions, a function should come equipped with a domain.

Answer 13

Well this is calculus

Answer 14

Yes but what are we to make of the 2 statements:

continuous for all real numbers AND

x cannot equal to -2

?

Is this just obfuscation for the sake of it?

Answer 15

If -2 is not in the domain, the function is not determined there.

Answer 16

well, that was my thought too... I'm with you... I think half these problems are written purposefully to cause confusion.

Answer 17

math is hard enough without poor questions.

Answer 18

if x cannot be -2, then the given function cannot be defined at x=-2. this means that it cannot be continuous for all reals. this means that the situation is impossible

Answer 19

h(-2) has no value because -2 is not part of the domain of the h(x) by definition.

Answer 20

The situation is impossible, I will agree.

It's not "pretty math" to evaluate a function at a point at which it is undefined and then say that the value of the function is "impossible".

That might fly in creative writing class, but it's not a great question.

Answer 21

agreed jake. :)

Answer 22

No you need to look at the first case I said, it might be continuous. We can write the equation of the circle like that: x^2+y^2=r^2 but we know that it is not a function. But we use it as such because is the most convenient. Its a problem of interpretation, and since the subject is calculus, its probably not the case where you just say it does not exist!

Answer 23

Multiply it by (x-2)/(x-2) and you get h(x)=x-2, wich is a perfectly continuous function and is equal -4 in x=-2.

Answer 24

He is saying that it is continuous for all real, that means he is considering -2 in the domain, i had not read that, but know I'm sure, if you say that -2 is in the domain, it is, it doesn't matter how you write the function.

Answer 25

Oh sorry, actually reading again I realized that he is saying that g(x)=that equation and that for x different of -2, f(x)=f(g), but since f(x) is continuous, we can find it using the limit of g.
You were right, -2 is not in the domain of g, but the problem is perfectly possible.

Answer 26

It's just a dumb question.

Answer 27

Its not, its just poorly writen

Answer 28

You say tomato...

Answer 29

remember, that for continuity, the limit must exist and be = to the function value. it is impossible.

Answer 30

If it's continuous, then the limit exists and equals the function value, which implies the function value exists, which it cannot, since the problem says it doesn't, so it's not continuous, except that the problem says it is.

Repeat as desired...

Answer 31

lol. it is a contradiction, bozo no-no in math.

Answer 32

poster has given up and gone to bed........:-)

Answer 33

I wonder sometimes if impossible problems are part of the research they are doing on the OpenStudy community...

But that gives too much credit to the question-authors with some of these curriculum companies.

Answer 34

No, the limit exists and is equal to the value of the function for continuity, the problem does not say that that function is continuous, for the limit to exist, it needs only to be the same when you come from both sides

Answer 35

LOL, proof by repetition...

Answer 36

No, proof by assertion....

Answer 37

Any more hand waving and it will fall off......

Answer 38

You have given the original poster a choice, leave it to him to decide....

Answer 39

No, look:
He is saying that g is equal to that function, and you are right, it is not defined in -2, but f is equal to g ONLY when x is different of -2, in this case, since f is continuous f(-2)= lim g at -2

Answer 40

Question doesn't ask for a limit, it asks for h(-2)

Answer 41

the limit can exist, but it's a point discontinuity at x=-2. So the function is NOT continuous for all real x. Counter example: x = -2, h(x) does not exist because it is only defined for x not including x=-2

Answer 42

\[h(x)\neq \frac{ x^2-4 }{ x+2 }=g(x)\]
Yes h is continuous, but i different from g in -2.
He IS NOT saying that that function is continuous because its not!
He is saying that ANOTHER function is continuous and it is equal to the other in every point BUT x=-2.

Answer 43

And to find h(-2) you take the limit of g, because they are equal in every other point.

Answer 44

h(-2) means evaluate the function at x = -2 but the question says that x cannot be -2.

How many more times?

Answer 45

I've been around this ride so many times, I think I am getting nauseous... maybe time for another activity.

Answer 46

Please, stop repeating the same thing over and over, and try to understand what I am saying. What you are saying is not correct, he doesn't mean evaluate the function at x=-2, he means that the points that belong to h also belong to the points in that formula EXCEPT for x=2. But since h is continuous, even though that function is not continuous, when you take the limit of the formula, you get h(-2)

Answer 47

You want to rewrite the question so as to fit your pov, that's fine.

I am going to answer the question that was asked.

Answer 48

As for understanding it, I understand it now somewhat better than I did 40 years ago....

Answer 49

It is not just my interpretation, its whats writen. Im not reqriting anything, thats the question that was asked and the answer is -4 not impossible. Read carefully the question again and youll se that I'm right, its not a matter of opinion.

Answer 50

We will have to agree to disagree, you have given the poster a choice and it is up to him now.

Answer 51

I'm serious i'm not agreing to disagree, I know I am right, sorry if that sounded arrogant, but, I know it, and I'm asking you to trust me. I'm sorry if I'm being too persistent but I just don't like to leave a subject when I know that the person didn't understood it, i teach, and thats a good thing for teachers, but here its not so great I know it, so sorry, but I can't just walk away.
From now on I'l agree to disagree, just had to say it.

Answer 52

Crap! (I just had to say that too).

Answer 53

I meant it I was not tring to insult anyone. I really had to say.

Answer 54

Look, since about half way up this thread, it has been absolutely clear that we were never going to agree and we keep trying to put this thread to bed..

And every time, you post again and just repeat that you are right, that everyone has not read or has not understood the question, that if we would see your pov, we would realise etc etc..

So let's just leave it, OK?

Answer 55

Ok, I already said I was going to agree to disagree.