I came up with a wrong solution, can someone tell me where I went wrong?
4 Answers
$\begingroup$Your domain part is correct
Now $-4 \le x \le 4$
Then $0 \le x^2\le 16$
$\Rightarrow -16\le -x^2\le 0$
$\Rightarrow 0\le 16-x^2\le 16$
$\Rightarrow 0\le \sqrt{16-x^2}\le 4$
Hence Range is $[0,4]$
$\endgroup$ $\begingroup$$\sqrt{x}$ is the principal square root function, it takes nonnegative value.
Hence $y \ge 0$.
I don't see how you get $16-y^2 \ge 16$.
We have $$-4 \le x \le 4$$$$0 \le x^2 \le 16$$
$$16-16 \le 16-x^2 \le 16-0$$
$$0 \le \sqrt{16-x^2} \le 4$$
Let $y \in [0,4]$, we let $x=\sqrt{16-y^2}$, we can check that $\sqrt{16-x^2}=\sqrt{16-(16-y^2)}=y.$
$\endgroup$ $\begingroup$$$f(x)=\sqrt{16-x^2}$$note that $$\sqrt{something}\to something \geq0$$so$$16-x^2\geq 0\\-x^2\geq -16\\ \text{multiplying by minus change the sign}\\x^2\leq 16 \\|x|\leq 4\\x\in[-4,4]$$ now try for the range$$y=\sqrt{16-x^2}$$ note again $\sqrt{something}$ to the power of two $$y^2=16-x^2\to x^2+y^2=16$$ it is a circle in center of (0,0) ,radius=4
it seems that $$y \in[-4,4] ,\text{but remember that } \sqrt{something}\geq 0 \\so \\y \in [0,4]$$
$y=\sqrt{16-x^2}>0\Rightarrow x\in[-4,+4]$ is a semicircle of radius $4$ since $y^2=16-x^2\Rightarrow x^2+y^2=4^2$ is a circle of radius $4$, and $y>0$. Therefore, $y\in[0,4]$.
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