+1348 What is the domain of the function $$\ell(y) = \frac{1}{(y-2)^2+(y-8)^2}~?$$ Express your answer in interval notation.
+129883 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)
+189 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 \((-\infty, \infty)\).
