Find the range of $f(x) = \sqrt{16-x^2}$

Question

I solved this question using this method and got $\operatorname{Range}(f) = [-4,4]$

But as per my textbook, the answer is $[0,4]$, please tell me where I'm wrong

$f(x) = \sqrt{16-x^2}$

Let $f(x) = y$

So, $\sqrt{16-x^2} = y$

$16-x^2 = y^2$

$x^2 = 16 – y^2$

So, $x = \sqrt{16-y^2}$

So, $f^{-1}(x) = \sqrt{16 – y^2}$

$\operatorname{Domain}(f^{-1}) = \operatorname{Range}(f)$

$f^{-1}$ is meaningful as long as $16-y^2 \geq 0$

So, $y^2 \leq16$

So, $y ∈ [-4,4]$

So, $\operatorname{Domain}(f^{-1})$ = $[-4,4] = \operatorname{Range}(f)$

I'm pretty sure I made some silly mistake here or my concepts about functions are not clear.

Please let me know about my mistake.

Thanks

Answer 1

First you've let $y = \sqrt{16-x^2} \Rightarrow y \ge 0$

So this puts a limit on $y$ that $y$ should be $\ge 0 $.

So, you have $y\ge 0$ and $y^2 \le 16 \Rightarrow y\ge0 $ and $(-4\le y\le4)$

The common region is $0\le y \le 4$ or $y \in [0,4]$