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June 12th, 2016, 03:35 AM   #1
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Separating a Relation into Multiple Functions

Hi I am reviewing some precalculus material and have run into some thing I am having trouble understanding.

The problem is to characterize the relation

y^4 - 4y^2 = x^2 + 4x

I found x intercept at (0,-4) and y intercepts at (0,0) (0,-2) and (0,2),

And determined the relation is symmetric about the x axis.

I solved for y and got

f1(x) = sqrt( x+4 )
f2(x) = - sqrt( x+4 )

The answer key in the book also lists,

f3(x) = sqrt( -x )
f4(x) = -sqrt( -x )

I'm having trouble seeing how to derive f3(x) and f4(x), and the book provides only the function listing without further explanation.
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June 12th, 2016, 04:30 AM   #2
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I wonder how you got the first two without also seeing the last two!

First treat the given equation as a quadratic equation for $\displaystyle y^2$: $\displaystyle (y^2)^2-4(y^2)- (x^2+ 4x)= 0$.

Use the quadratic formula to solve that equation:
$\displaystyle y^2= \frac{4\pm \sqrt{16+ 4x^2+ 16x}}{2}= \frac{4\pm 2\sqrt{x^2+ 4x+ 4}}{2}$
Of course $\displaystyle x^2+ 4x+ 4= (x+ 2)^2$ so we have
$\displaystyle y^2= \frac{4\pm 2(x+ 2)}{2}= 2\pm (x+ 2)$

Taking the positive sign, $\displaystyle y^2= x+ 4$ so $\displaystyle y= \pm\sqrt{x+ 4}$ and taking the negative sign $\displaystyle y^2= -x$ so $\displaystyle y= \pm \sqrt{-x}$.

If we are dealing only with real numbers, the first two solutions are valid for all x greater than or equal to -4 and the second two only for x less than or equal to 0.
Thanks from Derric

Last edited by Country Boy; June 12th, 2016 at 04:33 AM.
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June 12th, 2016, 04:30 AM   #3
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I list all possible cases:

$\displaystyle (y^2-2)^2=(x+2)^2\implies y^2-2=x+2\implies y=\pm\sqrt{x+4}$

$\displaystyle (y^2-2)^2=(-x-2)^2\implies y^2-2=-x-2\implies y=\pm\sqrt{-x}$

$\displaystyle (2-y^2)^2=(x+2)^2\implies 2-y^2=x+2\implies y=\pm\sqrt{-x}$

$\displaystyle (2-y^2)^2=(-x-2)^2\implies 2-y^2=-x-2\implies y=\pm\sqrt{x+4}$
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June 12th, 2016, 05:09 AM   #4
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Quote:
Originally Posted by greg1313 View Post
I list all possible cases:

$\displaystyle (y^2-2)^2=(x+2)^2\implies y^2-2=x+2\implies y=\pm\sqrt{x+4}$

$\displaystyle (y^2-2)^2=(-x-2)^2\implies y^2-2=-x-2\implies y=\pm\sqrt{-x}$

$\displaystyle (2-y^2)^2=(x+2)^2\implies 2-y^2=x+2\implies y=\pm\sqrt{-x}$

$\displaystyle (2-y^2)^2=(-x-2)^2\implies 2-y^2=-x-2\implies y=\pm\sqrt{x+4}$
This is how I worked the problem, and I see what I missed now. That is:

( y^2 - 2 )^2 = ( +/- ( x+2) )^2

and the subsequent results. All I did was the positive case.

Last edited by Derric; June 12th, 2016 at 05:31 AM.
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June 12th, 2016, 05:14 AM   #5
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Quote:
Originally Posted by Country Boy View Post
I wonder how you got the first two without also seeing the last two!

First treat the given equation as a quadratic equation for $\displaystyle y^2$: $\displaystyle (y^2)^2-4(y^2)- (x^2+ 4x)= 0$.

Use the quadratic formula to solve that equation:
$\displaystyle y^2= \frac{4\pm \sqrt{16+ 4x^2+ 16x}}{2}= \frac{4\pm 2\sqrt{x^2+ 4x+ 4}}{2}$
Of course $\displaystyle x^2+ 4x+ 4= (x+ 2)^2$ so we have
$\displaystyle y^2= \frac{4\pm 2(x+ 2)}{2}= 2\pm (x+ 2)$

Taking the positive sign, $\displaystyle y^2= x+ 4$ so $\displaystyle y= \pm\sqrt{x+ 4}$ and taking the negative sign $\displaystyle y^2= -x$ so $\displaystyle y= \pm \sqrt{-x}$.

If we are dealing only with real numbers, the first two solutions are valid for all x greater than or equal to -4 and the second two only for x less than or equal to 0.
Well, I completed the squares rather than taking this approach. I do like having more than one way to approach a problem.

What I missed would have more easily been seen had I used this method.

Last edited by Derric; June 12th, 2016 at 05:48 AM.
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June 12th, 2016, 05:46 AM   #6
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I think it's interesting to note I would have expected f3 and f4 to be:

+/- sqrt(-x-4) rather then just +/- sqrt(-x).

Thus in the interval from 0 to -4 there are four separate functions. I suppose for the purpose of learning math that is irrelevant, although I'm trying to think of an actual case were this equation might be modeling something, then a decision might have to be made which function or pair of functions would be relevant in that interval.
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June 12th, 2016, 11:29 AM   #7
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Quote:
Originally Posted by Derric View Post
I think it's interesting to note I would have expected f3 and f4 to be:

+/- sqrt(-x-4) rather then just +/- sqrt(-x).

Thus in the interval from 0 to -4 there are four separate functions. I suppose for the purpose of learning math that is irrelevant, although I'm trying to think of an actual case were this equation might be modeling something, then a decision might have to be made which function or pair of functions would be relevant in that interval.
Frequently when you are modeling something, there are constraints on the value a valid answer may have. For example, non-negativity constraints are very common. The math may give multiple answers, but the extra constraints imposed by the nature of what is being modeled reduce that number.
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