1
$\begingroup$

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 $[3,7]$.

$\endgroup$
  • 1
    $\begingroup$ $0$ is clearly the minimum nonnegative value of $16-(x-3)^2$. What about the maximum value? $\endgroup$ – user137731 Oct 12 '16 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 '16 at 0:55
3
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.