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#1 2008-02-06 23:53:59
- Identity
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- Registered: 2007-04-18
- Posts: 934
Range
What is the range of
Thankyou!
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#2 2008-02-07 04:14:19
- JaneFairfax
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- Registered: 2007-02-23
- Posts: 6,868
Re: Range
Hence f(x) can take any value except
.Last edited by JaneFairfax (2008-02-07 04:22:44)
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#3 2008-02-07 04:18:37
- mathsyperson
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- Registered: 2005-06-22
- Posts: 4,900
Re: Range
I get
.Doesn't change the final answer though.
Why did the vector cross the road?
It wanted to be normal.
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#4 2008-02-07 04:29:59
- Ricky
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- Registered: 2005-12-04
- Posts: 3,791
Re: Range
Typically, the way you would approach such a question in pre-calculus is to look at the parent graph:
You should immediately know what this graph looks like. Now your graph has a vertical asymptote at -3/2 and a horizontal asymptote at 1/2. To get the first, solve for the denominator = 0. To get the second, divide the term with the largest coefficient in the numerator by the largest in the denominator.
With that vertical asymptote, the graph assumes all values above an below 1/2. So the only value not in it is 1/2.
Personally, I like Jane's method better. But this is how I was taught in precalc.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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#5 2008-02-07 18:16:45
- Identity
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- Registered: 2007-04-18
- Posts: 934
Re: Range
@ Jane: I understand how you did it, but I don't understand why that method actually works? Does it have something to do with taking limits? Limits are good, but the problem is sometimes the limits can be found elsewhere in the graph as actual values.
@ Ricky: I also don't understand the method, but it seems imilar to Jane's.
Thanks
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#6 2008-02-08 01:08:42
- TheDude
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- Registered: 2007-10-23
- Posts: 361
Re: Range
Which part of Jane's method is confusing you? The rearranging of the function or determining the range from the new expression? If it's the latter then the explanation is simple. The expression
can take any value except 0. This can be rather easily shown by noting that the graph is continuous everywhere except at x = -3/2 and finding the limits as x approaches -infinity, -3/2 from the left, -3/2 from the right, and +infinity. Once you've convinced yourself of this fact you can see that the added 1/2 makes the only unreachable value 1/2.
Wrap it in bacon
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#7 2008-02-11 18:36:31
- Identity
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- Registered: 2007-04-18
- Posts: 934
Re: Range
Thanks, I get it now, now it seems so obvious
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