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, ∞)

The answer is:

 [D]:   (−∞, −2) ∪ (−2, 2) ∪ (2, ∞)   ;


Note:  We are given:  f(x) = x / (x² + 4)  ;


Note that: (x² + 4) ;  factors into: "(x + 2)(x − 2)"

   So:  f(x) = x / (x² + 4)  ; ↔ f(x) = x / [(x + 2)(x − 2)] ;

  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".

So, since the denominator = [(x + 2)(x − 2)]  ;

Start with:

x + 2 = 0 ;  

Subtract "2" from each side ;

x + 2 − 2 = 0 − 2 ; 

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

  Then; 

 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".

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.

Sin(a)+cos(a)+cot(a)-1=2(csc(x)+1)