+118653 what is the domain and range of $$\sqrt{81-x^2}$$
$$\\81-x^2\ge0\\
81\ge x^2\\
x^2 \le 81\\
-9\le x \le 9 \qquad \mbox{The the domain is [-9,9]} \\$$
The square root of anything is greater or equal to 0 therefore range does not include less than 0.
$$\sqrt{81-x^2} = \sqrt{81-positive\; number} \;\le \sqrt{81}\le9$$
$$Therefore the range is [0,9]$$
+33653 If we exclude complex numbers then the domain is -9 ≤ x ≤ 9 and the range is 0 to 9. If x lies outside this range then we are looking at the square root of a negative number.
+118653 what is the domain and range of $$\sqrt{81-x^2}$$
$$\\81-x^2\ge0\\
81\ge x^2\\
x^2 \le 81\\
-9\le x \le 9 \qquad \mbox{The the domain is [-9,9]} \\$$
The square root of anything is greater or equal to 0 therefore range does not include less than 0.
$$\sqrt{81-x^2} = \sqrt{81-positive\; number} \;\le \sqrt{81}\le9$$
$$Therefore the range is [0,9]$$
