Why is the range of $y=\sqrt{9-x^2}$ $[0,3]$ instead of $[-3,3]$? Solving it algebraically, it is my understanding that the answer would be the second one. But that is apparently not the right answer. Could someone explain to me why this is? Thanks.
O shoot I'm sorry. thought the domain of the function was [0,3].
What real value can you pick so that $\sqrt{9-x^2} = -1$? After all, on the real numbers, $\sqrt{\cdot}$ maps non-negative reals into non-negative reals...