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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 $[3,7]$.

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    $\begingroup$ $0$ is clearly the minimum nonnegative value of $16-(x-3)^2$. What about the maximum value? $\endgroup$
    – user137731
    Oct 12, 2016 at 0:49
  • $\begingroup$ Note the discrininant, $16 -(x-3)^2\le 16$ and never higher. So $o (x) \le 7s never more. $\endgroup$
    – fleablood
    Oct 12, 2016 at 0:55

2 Answers 2

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$$ -1 \leq x \leq 7 \Leftrightarrow \\ \\ -4 \leq x-3 \leq 4 \Rightarrow \\ \\ 0 \leq (x-3)^2 \leq 16 \Leftrightarrow \\ \\ -16 \leq -(x-3)^2 \leq 0 \Leftrightarrow \\ \\ 0 \leq 16-(x-3)^2 \leq 16 \Leftrightarrow \\ \\ 0 \leq \sqrt{16-(x-3)^2} \leq 4 \Leftrightarrow \\ \\ 3 \leq 3+\sqrt{16-(x-3)^2} \leq 7 $$

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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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