# 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 $[3,7]$.

• $0$ is clearly the minimum nonnegative value of $16-(x-3)^2$. What about the maximum value? – user137731 Oct 12 '16 at 0:49