SOLUTION: The problem is: Evaluate and find the domain. f(x)=1/x^2-9 for x=4,a-5 I am able to evaluate and came up with f(4)=1/7 and f(4)=1/a^2-10a+16. What I don't understand is how

Algebra ->  Functions -> SOLUTION: The problem is: Evaluate and find the domain. f(x)=1/x^2-9 for x=4,a-5 I am able to evaluate and came up with f(4)=1/7 and f(4)=1/a^2-10a+16. What I don't understand is how       Log On


   



Question 156686: The problem is: Evaluate and find the domain. f(x)=1/x^2-9 for x=4,a-5
I am able to evaluate and came up with f(4)=1/7 and f(4)=1/a^2-10a+16. What I don't understand is how to find the domain. I know my book says the answer is: x is not equal to 3, and x is not equal to -3. I just don't know how the domain came to be this, I don't understand how to find the domain. Please help.

Found 2 solutions by mangopeeler07, jim_thompson5910:
Answer by mangopeeler07(462) About Me  (Show Source):
You can put this solution on YOUR website!
f(x)=1/x^2-9

To find the domain, you would have to find [what x can be without making the denominator zero]. So set the denominator, x^2-9, equal to zero:

x^2-9=0

Now factor out the left side.

(x-3)(x+3)=0

Now take each expression separately and set them equal to zero and solve.

x-3=0
x=3

x+3=0
x=-3

So to make the denominator zero, x would have to be 3 or -3. So to get the domain, "what x can be without making the denominator zero", you would say anything but these two values. In other words:

x is not 3 or -3.

Do you follow?
I hope this helped!

-Alani

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Well notice how if x=3, then f%283%29=1%2F%28%283%29%5E2-9%29=1%2F%289-9%29=1%2F0 which is NOT defined. Also, if x=-3, then f%283%29=1%2F%28%28-3%29%5E2-9%29=1%2F%289-9%29=1%2F0 which is also NOT defined. So that's why there are domain restrictions of x%3C%3E-3 and x%3C%3E3