find domain
a/(a^2-9) + 3/(9-a^2)
To find the domain of the function a/(a^2-9) + 3/(9-a^2), we need to identify any values of "a" that would cause the denominator(s) of the function to equal zero.
For the first term, a/(a^2-9), the denominator (a^2-9) equals zero when a = -3 or a = 3.
For the second term, 3/(9-a^2), the denominator (9-a^2) equals zero when a = -3 or a = 3.
Therefore, the values a = -3 and a = 3 would make the function undefined. Hence, the domain of the function a/(a^2-9) + 3/(9-a^2) is all real numbers except for a = -3 and a = 3.
simplify it
To simplify the expression a/(a^2-9) + 3/(9-a^2), we can combine the fractions by finding a common denominator.
The denominator of the first term is (a^2-9), which can be factored as (a-3)(a+3).
The denominator of the second term is (9-a^2), which can be factored as (3+a)(3-a).
So, the least common denominator (LCD) is (a-3)(a+3)(3+a)(3-a).
Now, we can rewrite the expression with a common denominator:
(a/(a^2-9)) + (3/(9-a^2)) = (a/(a-3)(a+3)) + (3/(3+a)(3-a))
To combine the fractions, we can multiply the numerator and denominator of the first term by (3+a), and the numerator and denominator of the second term by (a-3):
(a(3+a) + 3(a-3)) / ((a-3)(a+3)(3+a)(3-a))
Expanding and simplifying the numerator, we get:
(3a + a^2 + 3a - 9) / ((a-3)(a+3)(3+a)(3-a))
Combining like terms in the numerator, we have:
(a^2 + 6a - 9) / ((a-3)(a+3)(3+a)(3-a))
Now, we can simplify further by canceling out common factors:
(a^2 + 6a - 9) / ((a-3)(a+3)(3-a)) = (a^2 + 6a - 9) / (-(a-3)(a+3)(a-3))
Finally, simplifying the negative sign and factoring out the common factor (a-3), we get:
(a^2 + 6a - 9) / (-1(a-3)^2(a+3))
So, the simplified form of the expression a/(a^2-9) + 3/(9-a^2) is (a^2 + 6a - 9) / (-1(a-3)^2(a+3)).
To find the domain of the given expression, we need to consider the values of "a" that make the expression undefined. In other words, we will look for any real numbers that would result in a zero denominator.
In this case, the expression has two denominators: (a^2 - 9) and (9 - a^2).
Firstly, let's examine the denominator (a^2 - 9):
The expression is undefined if (a^2 - 9) = 0, because division by zero is undefined. Therefore, we solve for "a" to find the values that make the denominator zero:
a^2 - 9 = 0
(a - 3)(a + 3) = 0
From this equation, we can see that the denominator (a^2 - 9) becomes zero when a = 3 or a = -3. Therefore, we should exclude these values from the domain.
Secondly, let's consider the denominator (9 - a^2):
Similarly, we set (9 - a^2) equal to zero:
9 - a^2 = 0
a^2 = 9
Taking the square root of both sides, we find:
a = ±3
So here, the value of "a" must not be equal to ±3.
Now we combine the results for both denominators:
The domain of the given expression is all real numbers except for a = 3, -3, and ±3.
In interval notation, we can represent the domain as (-∞, -3) ∪ (-3, 3) ∪ (3, ∞).