Question 197249
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 + 2x = 2]


is a quadratic equation.  Defining a domain for it doesn't make any sense.  The equation has a solution set consisting of two values, neither of which is contained in the stated domain.  Now, if what you <i>really</i> meant was:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x) = x^2 + 2x - 2]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x) = x^2 + 2x + 2]


(depending on whether you hit the equals key which is right next to the minus key, or forgot to hold the shift key to type +)


Where the domain is restricted to the set *[tex \LARGE \{1,\,2,\,3\}]


Then what you do is substitute each of the values from your domain set into the function and do the arithmetic required to determine the value of the function for that value of the independent variable.  With three elements in the domain set, in general you will get either 2 or 3 values for the function -- in this case you will get 3 values.  The set consisting of those 3 values is your range.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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