Find the smallest real number $x$ in the domain of the function $$f(x) = \sqrt{(x-3)^2-(x+3)^2}.$$
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 $\boxed{-3}$.
To find the smallest real number $x$ in the domain of the function $f(x)$, we need to determine where the expression inside the square root is nonnegative.
The expression inside the square root is $(x-3)^2-(x+3)^2$. To simplify this expression, we can expand it:
$(x-3)^2-(x+3)^2 = x^2-6x+9-(x^2+6x+9) = x^2-6x+9-x^2-6x-9 = -12x.$
Since we want this expression to be nonnegative, we have:
$-12x \ge 0.$
To solve this inequality, we divide both sides by $-12$, remembering to flip the direction of the inequality because we are dividing by a negative number:
$x \le 0.$
Hence, the smallest real number $x$ in the domain of the function $f(x)$ is $x = 0$.
To find the smallest real number $x$ in the domain of the function $f(x)$, we need to determine the values of $x$ for which the function is defined.
In this case, we have the expression $\sqrt{(x-3)^2-(x+3)^2}$ under the square root. For a square root to be defined, the expression inside it must be non-negative.
Therefore, we need to solve the inequality:
$$(x-3)^2 - (x+3)^2 \geq 0.$$
Let's simplify this inequality step by step:
Expand both squares:
$$x^2 - 6x + 9 - (x^2 + 6x + 9) \geq 0.$$
Combine like terms:
$$-12x \geq 0.$$
Divide by $-12$ (Note: Dividing by a negative number will reverse the inequality):
$$x \leq 0.$$
So, the smallest real number $x$ in the domain of the function $f(x)$ is $x = 0$.