If f(x) = x^2 - 81 and
g(x) = (x - 9)^-1 ( x + 9)
find g(x) x f(x)
a. (x+9)^2
b. (x-9)^2
c. 2(x-9)
d. (x+9)(x-9)
i know how to set this up, i just don't understand how to do it with two parenthesis on the g(x) side and the -1 exponent is throwing me off, provide a brief explanation including what the -1 does .

Since I'm on my phone, I can't write the solution in LaTeX, but I can try to explain the approach as best as I can.

We are given two separate functions, namely f(x) and g(x). Since f(x) is a difference of two squares (x^2 is a squared term and 81 is a squared term), we can rewrite it as (x + 9)(x - 9).

Now, this will help us when we multiply through by g(x). Let's discuss the solution to g(x). In exponent laws, any number to the power of a negative number will be reciprocated. That is, the function will turn into fractional notation.

Example: Rewrite x^-1 in positive index form.

Since the power is negative, we will need to reciprocate x^1 to get 1/(x^1).

Example: Rewrite 3x^-2 in positive index form.

Since the x term is only affected by the power, the 3 is left unchanged and stays on the numerator. Now, since the power is negative, we need to reciprocate x^2 to yield: 3/(x^2).

Using the same process, we can rewrite g(x) as being (x + 9)/(x - 9).

Now, we want f(x) * g(x) so that becomes:

(x + 9)(x - 9) *(x + 9)/(x - 9)

(x - 9) cancels in the numerator and denominator to yield: (x + 9)^2

Ais the answer for your question