if f(x) = x+1/x^2-9 and g(x) = x/x-3, find

a) the domain of f(x)
b) the domain of g(x)
c) (f+g) (x)
d) the domain of (f+g) (x)

assuming that you meant:

f(x) = (x+1)/(x^2-9) and g(x) = x/(x-3)

The domain for both f(x) and g(x) are all values of x except those that make the denominator zero
domain of f(x): all x's except x = ±3
domain of g(x): all x/s except x = 3

(f+g)(x) = (x+1)/(x^2 - 9) + x/(x-3)
the LCD is x^2 - 9 or (x+3)(x-3)
= (x+1 + x(x+3))/(x^2 - 9)
= (x^2 + 4x + 1)/(x^2 - 9(

the domain of that is the same as the domain of f(x)

To find the domain of a function, we need to identify any values that would make the function undefined.

a) To find the domain of f(x), we need to consider the denominator of the function, which is x^2-9. We know that a denominator cannot be zero, so we set x^2-9 ≠ 0 and solve for x.

x^2 - 9 ≠ 0
(x - 3)(x + 3) ≠ 0

From this equation, we see that x ≠ 3 and x ≠ -3. Therefore, the domain of f(x) is all real numbers except x = 3 and x = -3.

b) Similarly, to find the domain of g(x), we need to consider the denominator x - 3. Again, a denominator cannot be zero, so we set x - 3 ≠ 0 and solve for x:

x - 3 ≠ 0
x ≠ 3

Therefore, the domain of g(x) is all real numbers except x = 3.

c) To find (f + g)(x), we need to add f(x) and g(x):

(f + g)(x) = f(x) + g(x)
= (x + 1)/(x^2 - 9) + (x)/(x - 3)

To simplify this expression, we need to find a common denominator:

(f + g)(x) = [(x + 1)(x - 3) + x(x^2 - 9)] / [(x - 3)(x^2 - 9)]

Expanding and simplifying the numerator:

[(x^2 - 2x - 3) + (x^3 - 9x)] / [(x - 3)(x^2 - 9)]
= (x^3 - 2x^2 - 11x - 3) / [(x - 3)(x + 3)(x - 3)]

d) To determine the domain of (f + g)(x), we need to consider the denominator of the expression. In this case, (x - 3)(x + 3)(x - 3) ≠ 0.
Set each term in the denominator equal to zero and solve for x:

x - 3 ≠ 0 (x + 3)(x - 3) ≠ 0
x ≠ 3 x ≠ -3, x ≠ 3

Therefore, the domain of (f + g)(x) consists of all real numbers except x = -3 and x = 3.