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What is the domain and range of `(y-2)^2/4^2 - (x-1)^2/x^2=1`

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ssdixon | Student, Undergraduate | Salutatorian

Posted May 1, 2012 at 3:10 AM via web

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What is the domain and range of `(y-2)^2/4^2 - (x-1)^2/x^2=1`

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted May 1, 2012 at 4:29 AM (Answer #1)

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The domain and range of the relation `(y - 2)^2/4^2 - (x-1)^2/x^2 = 1` has to be determined.

For a relation of the form y = f(x), the domain is the set of values that x can take for which the values of y are real. The range is all the values that y takes for values of x lying in the domain.

For `(y - 2)^2/4^2 - (x-1)^2/x^2 = 1` , the value of y is real for all values of `x != 0 ` as at x = 0, the term `(x - 1)^2/x^2` is indeterminate.

Also, `(y - 2)^2= 16*(1 + (x-1)^2/x^2)` . The expression on the right can only take on values greater than or equal to 16.

`(y - 2)^2 >= 16`

=> `y - 2 >= 4` or `y - 2 <= -4`

=> `y >= 6` or `y <= -2`

The range of the relation is `(-oo, -2)U(6, oo)`

The required domain is `R - {0}` and the required range is `(-oo, -2)U(6, oo)`

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