Question 6671
I'm thinking that you meant {{{ g(x) = sqrt(2x + 4) }}}. If you plug that into f(x), you'll get:


{{{ f(g(x)) = (g(x))^2 - 4 = (sqrt(2x + 4))^2 - 4 = 2x + 4 - 4 = 2x }}}


{{{ f(g(x)) = 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(x) = sqrt(2x) + 4 }}}. If you plug that into f(x), you'll get:


{{{ f(g(x)) = (g(x))^2 - 4 = (sqrt(2x) + 4)^2 - 4 }}}


{{{ f(g(x)) = (sqrt(2x) + 4)(sqrt(2x) + 4) - 4 }}}


{{{ f(g(x)) = (sqrt(2x))^2 + 8sqrt(2x) + 16 - 4 }}} <--- Used FOIL


{{{ f(g(x)) = 2x + 8sqrt(2x) + 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(2x) }}} to be a negative number, which you can't do inside square roots.