Hi Sharon,

I am going to use two similar functions

f(x) = √(x - 5)
g(x) = x² + 2

fog(x) = f(g(x)) = f( x² + 2) = √(x² + 2 - 5) = √(x² - 3)

Looking at this expression I realize that x² must be at least 3 for otherwise x² - 3 is negative and you can't take the square root of a negative number. Hence the domain of fog(x) is all x for which x² ≥ 3. That is all x for which either x ≥ √3 or x ≤ -√3.

gof(x) = g(√(x - 5)) = (√(x - 5))² + 2

It's tempting to write (√(x - 5))² + 2 as x - 5 + 2 = x - 3 but you need to be careful. For example if x = 0 then x - 3 evaluates to -3 but (√(x - 5))² + 2 fails to yield a result since you can't take the square root of -5. In fact to evaluate (√(x - 5))² + 2 you must have x ≥ 5. That is the domain of gof(x) is all x ≥ 5. In that case you can simplify to x - 3 so I would say

gof(x) = x - 3 with the domain of fog(x) being all x ≥ 5.

I hope this helps,
Penny