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Guest
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 Posted: Tue, 7 Jun 2005 19:59:24 UTC Post subject: domain of a curve |
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What is the domain of
y = +- x/sqrt(x^2 - 4) ? |
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Guest
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| Posted: Tue, 7 Jun 2005 20:01:30 UTC Post subject: This is meant as a challenge to the tutors and others. |
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| [No additional message.] |
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Shadow
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Joined: 30 Mar 2005
Posts: 4953
Location: Urbana, IL
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| Posted: Tue, 7 Jun 2005 20:32:59 UTC Post subject: |
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_________________
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This is Bunny. Copy Bunny into your signature to help him on his way to world domination |
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stapel
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| Posted: Tue, 7 Jun 2005 21:24:45 UTC Post subject: |
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Last edited by stapel on Thu, 20 Oct 2005 07:44:02 UTC; edited 1 time in total |
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Guest
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| Posted: Tue, 7 Jun 2005 21:48:47 UTC Post subject: No, for instance x not = -1 and x not = 1. --> |
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| The equation has no real values of y for these x values. |
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Guest
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| Posted: Tue, 7 Jun 2005 21:53:57 UTC Post subject: stapel, you can have negatives inside square roots if they.. |
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are in the denominator and zero is in the numerator. The result is the real
value of zero for y. |
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skeeter
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Location: North TX
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| Posted: Tue, 7 Jun 2005 21:59:02 UTC Post subject: |
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| to re-iterate Eliz's response ... the value x^2 - 4 must be > 0. |
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Shadow
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| Posted: Tue, 7 Jun 2005 22:08:57 UTC Post subject: |
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oh....REAL solutions....
right
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Last edited by Shadow on Wed, 8 Jun 2005 12:08:00 UTC; edited 3 times in total |
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Guest
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| Posted: Tue, 7 Jun 2005 22:24:09 UTC Post subject: skeeter, LISTEN up, stapel never stated that x^2 - 4 > 0. |
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She asked questions is all. So there IS nothing for YOU to re-iterate.
Anyway, you're wrong. The point (0,0) satisfies the equation form
without the square root in it, so it must also satisfy the equation where y
is solved in terms of x. That is because both of these equations are
equivalent. |
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dmoran
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Joined: 01 Jan 2004
Posts: 2705
Location: Norman, Oklahoma
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| Posted: Tue, 7 Jun 2005 22:35:25 UTC Post subject: Re: skeeter, LISTEN up, stapel never stated that x^2 - 4 > |
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| treeline wrote: | She asked questions is all. So there IS nothing for YOU to re-iterate.
Anyway, you're wrong. The point (0,0) satisfies the equation form
without the square root in it, so it must also satisfy the equation where y
is solved in terms of x. That is because both of these equations are
equivalent. |
She asked those questions for you to think about. Anyway, the function is undefined for -2<=x<=2. So x=0 does not work, as you imply.
Dave |
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skeeter
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Location: North TX
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| Posted: Tue, 7 Jun 2005 23:11:30 UTC Post subject: |
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| stapel wrote: | The "domain" is the allowed x-values. What values are allowed for this function? (Can you divide by zero?NO Can you have negatives inside square roots?again, NO)
Eliz. |
the answer to Eliz's questions is the inequality x^2 - 4 > 0 ... which gives you the domain you originally asked for ... smart ass. |
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Matt
Member of the 'S.O.S. Math' Hall of Fame
Joined: 01 Oct 2003
Posts: 8506
Location: Sacramento, CA
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| Posted: Wed, 8 Jun 2005 00:14:52 UTC Post subject: Re: skeeter, LISTEN up, stapel never stated that x^2 - 4 > |
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Hi treeline,
| treeline wrote: | The point (0,0) satisfies the equation form
without the square root in it... |
No, the point (0,0) is not on the curve, because y is undefined when x = 0 (you would have to take the square root of -4 when x = 0). To find the domain of y, we must determine the values of x for which y is defined (i.e., nothing funny happens, such as dividing by 0).
First, the expression under the square root symbol must be positive (it can't be negative, for otherwise we would be taking the square root of a negative number, and it can't be 0, for otherwise we would have a division by 0 error). So to (dare I say) re-iterate what was said above, we require:
which, when translated into interval notation, makes the domain:
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Guest
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| Posted: Wed, 8 Jun 2005 02:01:59 UTC Post subject: Matthew, recall that the equivalent form of the equation is: |
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x^2y^2 - 4y^2 - x^2 = 0. When the point (0,0) is substituted in this
equation, you get a true statement which means that (0,0) is a point of the
graph.
To educate you more about this, go to "Schaum's Outline of Theory and
Problems of Differential and Integral Calculus Second Edition."
On pages 126 and 127 is discussed the curve y^2(x^2 - 4) = x^4.
The origin, which is the only intercept, is a type of "isolated point" because
for x near 0, y is imaginary.
This curve has the equivalent form y = x^2/sqrt(x^2 - 4). |
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stapel
Member of the 'S.O.S. Math' Hall of Fame
Joined: 10 Nov 2003
Posts: 1110
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| Posted: Wed, 8 Jun 2005 02:30:07 UTC Post subject: |
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Last edited by stapel on Thu, 20 Oct 2005 07:43:56 UTC; edited 1 time in total |
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Matt
Member of the 'S.O.S. Math' Hall of Fame
Joined: 01 Oct 2003
Posts: 8506
Location: Sacramento, CA
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| Posted: Wed, 8 Jun 2005 02:45:44 UTC Post subject: Re: Matthew, recall that the equivalent form of the equation |
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| treeline wrote: | x^2y^2 - 4y^2 - x^2 = 0. When the point (0,0) is substituted in this
equation, you get a true statement which means that (0,0) is a point of the
graph. |
If you were asking about the function
then I repeat, (0,0) is not a point on the graph for the reason I stated above. However, if your question referred to the implicitly defined function:
then you are right, (0,0) is a point on the curve. So whether or not (0,0) is a point on the curve depends on which equation you meant to write. |
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