SOLUTION: How do you solve this function of operations: Given:f(x) = x^2 - 4 (x squared minus 4) and g(x) = radical 2x + 4 Solve: Domain of f(g(x)) I understand that I have to plug in

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: How do you solve this function of operations: Given:f(x) = x^2 - 4 (x squared minus 4) and g(x) = radical 2x + 4 Solve: Domain of f(g(x)) I understand that I have to plug in       Log On


   

Question 6671: How do you solve this function of operations:
Given:f(x) = x^2 - 4 (x squared minus 4) and g(x) = radical 2x + 4
Solve: Domain of f(g(x))
I understand that I have to plug in g(x) into f(x) but I don't understand the part about the Domain. Please help me. Thanks.
Answer by prince_abubu(198) About Me  (Show Source):
You can put this solution on YOUR website!
I'm thinking that you meant +g%28x%29+=+sqrt%282x+%2B+4%29+. If you plug that into f(x), you'll get:



+f%28g%28x%29%29+=+2x+ <---- The domain would have to be the set of values that you could plug into f(g(x)) legally. In other words, what values of x can we put into the function so that none of those values will make a denominator zero, or whatever inside the square root "house" negative. In this case, you can choose any x value you want, since you can multiply 2 times any number.

Say that you meant +g%28x%29+=+sqrt%282x%29+%2B+4+. If you plug that into f(x), you'll get:

+f%28g%28x%29%29+=+%28g%28x%29%29%5E2+-+4+=+%28sqrt%282x%29+%2B+4%29%5E2+-+4+

+f%28g%28x%29%29+=+%28sqrt%282x%29+%2B+4%29%28sqrt%282x%29+%2B+4%29+-+4+

+f%28g%28x%29%29+=+%28sqrt%282x%29%29%5E2+%2B+8sqrt%282x%29+%2B+16+-+4+ <--- Used FOIL

+f%28g%28x%29%29+=+2x+%2B+8sqrt%282x%29+%2B+12+ <---- Simplified.

Now, what's the domain of f(g(x)) in that case? It would be all real numbers x greater than or equal to zero. You can't plug in negative numbers into that function because doing so will force the 2x inside +8sqrt%282x%29+ to be a negative number, which you can't do inside square roots.