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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

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    $\begingroup$ Square root is by definition the positive square root, so $f(x) \geq 0$ $\endgroup$
    – Gareth Ma
    Commented Apr 6, 2020 at 12:46
  • $\begingroup$ $f(x)$ is a square root function.... $\endgroup$ Commented Apr 6, 2020 at 12:47
  • $\begingroup$ Of course $f(x) \geq 0$ since it is a square root function $\endgroup$ Commented Apr 6, 2020 at 12:48
  • $\begingroup$ Yeah, so $\sqrt{16-x^2} \geq 0$, which means that $16 \geq x^2$ which is true for all elements of $[-4,4]$, right? So, why not $Range (f) = [-4,4]$, I'm sorry if I'm being irritating, I just don't get it $\endgroup$ Commented Apr 6, 2020 at 12:51
  • $\begingroup$ Your first step should be $\sqrt{16-x^2} = y\iff16-x^2 = y^2\land y\geq0$. $\endgroup$ Commented Apr 6, 2020 at 12:55

2 Answers 2

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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 $y \in [0,4]$

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  • $\begingroup$ I've edited my answer, please check. $\endgroup$
    – 19aksh
    Commented Apr 6, 2020 at 12:51
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    $\begingroup$ Got it, thanks, I was really really confused... $\endgroup$ Commented Apr 6, 2020 at 12:53
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    $\begingroup$ You're welcome! $\endgroup$
    – 19aksh
    Commented Apr 6, 2020 at 12:54
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It is clear that $$16-x^2\le16,$$ which is tight.

Then a square root is non-negative, and you immediately get

$$0\le\sqrt{16-x^2}\le\sqrt{16}=4.$$


All values in that range can be reached, because the equation

$$y=\sqrt{16-x^2}$$ has solutions for all $y\in[0,4]$ (one such solution is $x=\sqrt{16-y^2}$).


Notice that this approach does not require to worry about the domain of the function.

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