Question

State the domain of the rational function:

[ f(x) = x over x squared +4 ]

f(x) = x
x^2 + 4

A)
(−∞, ∞)
Eliminate

B)
(−∞, 0] ∪ [0, ∞)


C)
(−∞, 0) ∪ (0, ∞)


D)
(−∞, −2) ∪ (−2, 2) ∪ (2, ∞)

Answer 1

The answer is:
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 [D]:   (−∞, −2) ∪ (−2, 2) ∪ (2, ∞)   ;
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Note:  We are given:  f(x) = x / (x² + 4)  ;
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Note that: (x² + 4) ;  factors into: "(x + 2)(x − 2)"
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   So:  f(x) = x / (x² + 4)  ; ↔ f(x) = x / [(x + 2)(x − 2)] ;
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  and since one cannot divide by "0"; the "denominator" cannot equal "0";

so the "denominator", which  is:  "[(x + 2)(x − 2)]" ; cannot equal "0".

Since the "denominator" is the "product of two values", then "neither one nor both of the two values" can equal "0" ; since anything multiplied by "0" equals "0".
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So, since the denominator = [(x + 2)(x − 2)]  ;
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Start with:
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x + 2 = 0 ;  

Subtract "2" from each side ;

x + 2 − 2 = 0 − 2 ; 
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 x  =  -2  ; So, x ≠ -2 ; because if x = -2, the denominator is 0.
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  Then; 
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 x − 2 = 0 ; 

Add "2" to each side;

x − 2 + 2 = 0 + 2 ;

x = 2  ; So, x ≠ 2 ; because when if x = 2 , the denominator is "0".
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So, x ≠ -2 ; because if x = -2, the denominator is 0.

x ≠ 2 ; because when if x = 2 , the denominator is "0".

x < -2 ,  x  can be less than  -2 ; but not equal to -2 .

x  > 2 ,  x can be greater than 2; but not equal to 2.

-2 > x > 2 ; x can be between -2 and 2 ; but not equal to -2 or 2.
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