



 
Hi Sharon, I am going to use two similar functions
Looking at this expression I realize that x^{2} must be at least 3 for otherwise x^{2}  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 x^{2} ≥ 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
I hope this helps,  


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