Find the smallest real number $x$ in the domain of the function $$f(x) = \sqrt{(x-3)^2-(x+3)^2}.$$

Answer 1

Simplifying under the square root, we have $$f(x) = \sqrt{x^2 - 6x + 9 -x^2 - 6x -9} = \sqrt{-4x^2-12x}.$$For the expression under the square root to be nonnegative, which it needs to be by definition of real square roots, we need $-4x^2-12x \ge 0$. We can factor this as $-4x(x+3) \ge 0$. The product of two numbers with opposite signs is nonpositive, so $x$ must be negative or $x \le 0$. Of these values of $x$, which is the smallest? The answer is $\boxed{-3}$.