<s>[INST] The range and domain of $o(x)=3+\sqrt{16-(x-3)^2}$ What is the domain and range of $$o(x)=3+\sqrt{16-(x-3)^2}\tag1$$

For the domain, I know that the expression under the radical has to be larger than or equal to $0$. So therefore, I get this:$$16-(x-3)^2\geq 0\\(7-x)(x+1)\geq 0\\\therefore x\in [-1,7]$$
But for the range, I thought of the function as a square root function; because it is in the form $y=\sqrt x+b$. Since the square root of something cannot be less than $0$, I thought the range was $[3,+\infty)$. Which is wrong (according to the book).
The range is supposedly $[3,7]$.
 [/INST] You are right that $\sqrt{x}+b $ can not be less than $b $ so $3 +\sqrt {16- (x-3)^2} $ can't be less than $3$.
But note: $\sqrt {a - x} $ can not ever be more than $\sqrt {a} $.  So $b \le b +\sqrt {x -a} \le b + \sqrt a$.
So $16 - (x-3)^2 \le 16$ so $0 \le \sqrt {16 - (x-3)^2}\le 4$ so $3 \le 3+\sqrt {16 - (x-3)^2}\le 7$ so 
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