f(x)= 1/x^2-8x-9
g(x)= 1/x^2+3x+2
What is required to find? :3
Domain
Domain of what?
If the required is their individual domains:
1. For f(x) = 1/(x^2 - 8x - 9)
Note that domain is the set of all possible values of x. In the function, note that the denominator must not be equal to zero because 1/0 is infinity. Therefore, we find the values of x, which the denominator will be zero and restrict the domain to those values:
f(x) = 1/(x^2 - 8x - 9)
f(x) = 1/(x-9)(x+1)
Thus, x is all real numbers except -1 and 9.
2. For g(x) = 1/(x^2 + 3x + 2)
The procedure is pretty much the same. We find the values of x which will make the denominator zero:
g(x) = 1/(x^2 + 3x + 2)
g(x) = 1/(x+1)(x+2)
Thus, x is all real number except -1 and -2.
Hope this helps~ :3
To find the sum of the two functions f(x) and g(x), you need to add the two functions together, term by term. The formula for the sum of two functions is:
h(x) = f(x) + g(x)
Let's calculate h(x) step by step for the given functions f(x) and g(x):
Step 1: Write down the two functions:
f(x) = 1/(x^2 - 8x - 9)
g(x) = 1/(x^2 + 3x + 2)
Step 2: Find the common denominator for the two functions, which is the product of the denominators (x^2 - 8x - 9) and (x^2 + 3x + 2):
Common denominator = (x^2 - 8x - 9) * (x^2 + 3x + 2)
Step 3: Multiply each term of f(x) by the factor that makes its denominator equal to the common denominator from Step 2:
f(x) = (1/(x^2 - 8x - 9)) * (x^2 + 3x + 2)
Step 4: Multiply each term of g(x) by the factor that makes its denominator equal to the common denominator from Step 2:
g(x) = (1/(x^2 + 3x + 2)) * (x^2 - 8x - 9)
Step 5: Add the two fractions together, which will result in the sum function h(x):
h(x) = f(x) + g(x)
= ((1/(x^2 - 8x - 9)) * (x^2 + 3x + 2)) + ((1/(x^2 + 3x + 2)) * (x^2 - 8x - 9))
Note: The resulting function h(x) will be a complex expression due to the denominators in the fractions.
It is important to simplify the expression further, if possible, but without any specific values of x, it may not be possible to simplify the expression further.