f(x) squareroot( (x-5)/(x^2+3x-10) ) the entire problem is under the squareroot. I got [-5,2) U (5, infiniti) not sure if it's right. I know I've asked this a few times but I just feel paranoid that I have the bracket next to -5 in the wrong place, my friend's answer is (-5,2) U [5, infinity)
GADS.I missed thatneed to look more closely when answeringfriend is correct
Your brackets are perfect: One set of parentheses around numerator, and one around denominator, then one around the whole expression to indicate that all of it is under square root. Anyway, I'm assuming you are looking for domain? If so, value under square root must be ≥ 0 (x?5)/(x?+3x?10) ≥ 0 (x?5)/((x+5)(x?2)) ≥ 0 Fraction is ≥ 0 when numerator ≥ 0 and denominator 0 or when numerator ≤ and denominator 0 (x?5) ≥ 0 and (x+5)(x?2) 0 (x ≥ 5) and (x ?5 or x 2) x ≥ 5 (x?5) ≤ 0 and (x+5)(x?2) 0 (x ≤ 5) and (?5 x 2) ?5 x 2 Solution: (?5 x 2) or (x ≥ 5) (?5, 2) U [5, ∞) ————————————————————————— Your friend is correct. The only difference between your answer and your friend's was whether ?5 or 5 is part of solution. So all you have to do is check: If x ?5, then f(x) √((x?5)/(x?+3x?10)) √(?10/(25?15?10)) √(?10/0) f(x) is NOT defined for x ?5, so it is not part of domain If x 5, then f(x) √((x?5)/(x?+3x?10)) √(0/(25+15?10)) √(0/30) 0 f(x) IS defined for x 5, so this value is part of domain
