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

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.

Thanks

• Square root is by definition the positive square root, so $f(x) \geq 0$ Commented Apr 6, 2020 at 12:46
• $f(x)$ is a square root function.... Commented Apr 6, 2020 at 12:47
• Of course $f(x) \geq 0$ since it is a square root function Commented Apr 6, 2020 at 12:48
• 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 Commented Apr 6, 2020 at 12:51
• Your first step should be $\sqrt{16-x^2} = y\iff16-x^2 = y^2\land y\geq0$. Commented Apr 6, 2020 at 12:55

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 $$y \in [0,4]$$

• I've edited my answer, please check. Commented Apr 6, 2020 at 12:51
• Got it, thanks, I was really really confused... Commented Apr 6, 2020 at 12:53
• You're welcome! Commented Apr 6, 2020 at 12:54

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.