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What is the domain of the function $$\ell(y) = \frac{1}{(y-2)^2+(y-8)^2}~?$$ Express your answer in interval notation.

 Aug 9, 2023
 #1
avatar+129197 
+1

The domain will be all y's that DON'T make the  denominator =  0

 

So

 

(y -2)^2  + (y - 8)^2  =  0

 

(y - 2)^2  = - (y  - 8)^2 

 

y^2 - 4y + 4  = -y^2 + 16y - 64

 

2y^2 -20y + 68  =  

 

y^2 -10y + 34 =  0

 

y^2 -10y   =   -34                      complete the square on y

 

y^2 -10y + 25  = -34 + 25

 

)y - 5)^2  = -9        take  both  roots

 

y - 5  =  3i , -3i

 

y = 3i+5    and  y = -3i + 5

 

So....no  real  numbers make the  denominator 0  so  the  domain is (-inf, inf)

 

 

cool cool cool

 Aug 9, 2023
 #2
avatar+177 
-1

This method is valid, but I see a shortcut that takes less time than trying to solve a quadratic.

 

As Cphill mentioned, the only possible restriction of the domain is any y-value such that \((y - 2)^2 + (y - 8)^2 = 0\). Upon close examination, the denominator is written as the sum of two squares. The sum of 2 squares yields 0 only if both quantities are 0. In other words, \(y-2\) and \(y - 8\) have to equal 0 for the same value of y. Is this the case? Well, let's examine.

 

Examine y - 2. y - 2 = 0 when y = 2, which is the only candidate that could be a restriction on the domain. However, y - 8 does not equal 0 when y = 2. Therefore, the only candidate does not make the denominator 0.

 

Therefore, the domain is \((-\infty, \infty)\).

The3Mathketeers  Aug 10, 2023

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