SOLUTION: 1. Find the domain of the expression x/(x^2 - 9).
x^2 - 9 = 0
x^2 = 9
sqrtx^2} = sqrt{9}
x = -3, and x = 3.
Domain = {x | x cannot be -3 and 3}.
2. Find the dom
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-> SOLUTION: 1. Find the domain of the expression x/(x^2 - 9).
x^2 - 9 = 0
x^2 = 9
sqrtx^2} = sqrt{9}
x = -3, and x = 3.
Domain = {x | x cannot be -3 and 3}.
2. Find the dom
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Question 1208404: 1. Find the domain of the expression x/(x^2 - 9).
x^2 - 9 = 0
x^2 = 9
sqrtx^2} = sqrt{9}
x = -3, and x = 3.
Domain = {x | x cannot be -3 and 3}.
2. Find the domain of the expression
(-9x^2 - x + 1)/(x^3 + x).
x^3 + x =
x(x^2 + 1) = 0
Setting x^2 + 1 to zero and solving for x leads to complex roots. I will disregard complex roots for now. The only value that x cannot be is 0.
Domain = {x | x cannot be 0}
What do y I say?
x^2-9 = 0 had been correctly solved to get x = -3 and x = 3.
And you are correct that x cannot equal either of these values; otherwise, we get a division by zero error.
Another way to solve is:
x^2-9 = 0
x^2-3^2 = 0
(x-3)(x+3) = 0 .... difference of squares rule
x-3 = 0 or x+3 = 0
x = 3 or x = -3
The domain in set-builder notation is
Or you can slightly condense things to get
The portion translates to "x is in the set of real numbers" i.e. "x is a real number".
The interval notation would look like this
This is basically us starting with the entire number line and then poking two holes at x = -3 and x = 3 on the number line.
Each "U" refers to a set union joining or gluing together the disjoint intervals.
Curved parenthesis exclude each endpoint.
(