SOLUTION: The domain of the function: (1/ (sqrt (2x + 15)) and ( (x+2) / (x^2 - 225) ). I don't understand how to write it in interval notation.

Question 661586: The domain of the function: (1/ (sqrt (2x + 15)) and ( (x+2) / (x^2 - 225) ). I don't understand how to write it in interval notation.
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
The domain of the function: (1/ (sqrt (2x + 15)) and ( (x+2) / (x^2 - 225) ). I don't understand how to write it in interval notation.
**
(1/ (sqrt (2x + 15))
(1/√(2x+15)
2 restrictions apply here:
radican≥0
and
denominator≠0
..
for radican:
2x+15>0
2x>-15
x>-15/2
..
for denominator:
2x+15=0
2x=-15
x≠-15/2
domain: (-15/2,∞)
..
( (x+2) / (x^2 - 225) )
(x+2)/(x^2-225)
(x+2)/(x^2-15^2)
(x+2)/(x+15)(x-15)
one restriction applies here: denominator≠0
(x+15)(x-15)=0
x≠15
and
x≠-15
domain: (-∞,-15) U (-15,15) U (15,∞)

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