Little help with understanding intervals

Probability

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Jan 26, 2012
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In my text book I have an example of a graph, g(x) = 1/x (1< x < 2)

I understand the real numbers above in brackets are the domain of x which satisfies the domain if I got that right!

If I draw the above graph I have x values as the domain above, which are also understood as closed intervals, {a,b}

The image set I understand to be values on the y axis, and in the above example would be 1/2 and 1, {1/2, 1}

What I don't seem to be able to get to grips with is the understanding of how to use the real numbers of the domain to find intervals, and if different how the image set values are found?

Could somebody give me an explantion using the above example please.
 
My first and second impressions were that I didn't understand the question.

"...use the real numbers of the domain to find intervals."

Seems to me like you found [1,2] and [1/2,1] okay
 
My first and second impressions were that I didn't understand the question.

"...use the real numbers of the domain to find intervals."

Seems to me like you found [1,2] and [1/2,1] okay

Thank you for replying:-D, I am probably not explaining my misunderstanding of this subject clearly?

I have a graph of f(x) = 3(x + 1)^2 - 12

I want to find the image set of the function f and I want to write this in interval notation.

I think the image set is the domian, and the values are {-12, infinity) but am unsure?

Any help appreciated:confused:
 
You seem to have it, but the braces are a bit unusual. It's usually [] if you have the endpoints, and () if you don't.

Of course, you never have the endpoint if it is unbounded, such as your example.

In your example, "-12" is "-12". There isn't much we can do about that. But the other term, let's just ask what's the least value we can get out of it?
 
You seem to have it, but the braces are a bit unusual. It's usually [] if you have the endpoints, and () if you don't.

Of course, you never have the endpoint if it is unbounded, such as your example.

In your example, "-12" is "-12". There isn't much we can do about that. But the other term, let's just ask what's the least value we can get out of it?


Are you saying that {- 12, infinity} should be {- 12, 0}?

Sorry about the incorrect use of the brackets, I do understand them but have not got the facility to produce them on my keyboard, and latex I don't properly understand how to use it.

:confused:
 
Are you saying that {- 12, infinity} should be {- 12, 0}?

No. Where did I say that? But that zero is important. This estabilshes the Minimum value in the Range, -12.

Now, think about the maximum value.
 
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