Question

find domain
a/(a^2-9) + 3/(9-a^2)

Answer 1

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.

Answer 2

simplify it

Answer 3

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)).