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June 12th, 2016, 02:35 AM  #1 
Newbie Joined: Jun 2016 From: Sol System Three Posts: 5 Thanks: 0  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. 
June 12th, 2016, 03:30 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
I wonder how you got the first two without also seeing the last two! First treat the given equation as a quadratic equation for $\displaystyle y^2$: $\displaystyle (y^2)^24(y^2) (x^2+ 4x)= 0$. 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. Last edited by Country Boy; June 12th, 2016 at 03:33 AM. 
June 12th, 2016, 03:30 AM  #3 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,932 Thanks: 1127 Math Focus: Elementary mathematics and beyond 
I list all possible cases: $\displaystyle (y^22)^2=(x+2)^2\implies y^22=x+2\implies y=\pm\sqrt{x+4}$ $\displaystyle (y^22)^2=(x2)^2\implies y^22=x2\implies y=\pm\sqrt{x}$ $\displaystyle (2y^2)^2=(x+2)^2\implies 2y^2=x+2\implies y=\pm\sqrt{x}$ $\displaystyle (2y^2)^2=(x2)^2\implies 2y^2=x2\implies y=\pm\sqrt{x+4}$ 
June 12th, 2016, 04:09 AM  #4  
Newbie Joined: Jun 2016 From: Sol System Three Posts: 5 Thanks: 0  Quote:
( 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 04:31 AM.  
June 12th, 2016, 04:14 AM  #5  
Newbie Joined: Jun 2016 From: Sol System Three Posts: 5 Thanks: 0  Quote:
What I missed would have more easily been seen had I used this method. Last edited by Derric; June 12th, 2016 at 04:48 AM.  
June 12th, 2016, 04:46 AM  #6 
Newbie Joined: Jun 2016 From: Sol System Three Posts: 5 Thanks: 0 
I think it's interesting to note I would have expected f3 and f4 to be: +/ sqrt(x4) 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. 
June 12th, 2016, 10:29 AM  #7  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551  Quote:
 

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