Find the domain of the function $$f(x) = \frac{\sqrt{x}}{\sqrt{x^2}}.$$ Express your answer as an interval or as a union of intervals.
For this function to be defined, both the numerator and denominator must be defined. The numerator is defined for all nonnegative values of $x$, and the denominator is defined for all nonzero values of $x$. Therefore, the domain of the function is the set of all nonnegative real numbers excluding zero, or $\boxed{(0,\infty)}$.
To find the domain of a function, we need to identify any values of x that would make the function undefined.
In this case, the function is a fraction, specifically the square root of x divided by the square root of x squared. Remember that the square root of x is defined only for x values greater than or equal to zero, since we cannot take the square root of a negative number in the real number system.
In addition, the denominator, the square root of x squared, will always be defined because squaring any real number will result in a non-negative value. Hence, we don't need to worry about the denominator being undefined.
Therefore, the only restriction comes from the numerator, which is the square root of x. In conclusion, the domain of the function is all x-values greater than or equal to zero.
Expressing this as an interval, we can write the domain as [0, ∞).
To find the domain of a function, we need to determine the values of $x$ for which the function is defined. In this case, we have the function $$f(x) = \frac{\sqrt{x}}{\sqrt{x^2}}.$$ The function will be defined if the denominator, $\sqrt{x^2}$, is not equal to zero because division by zero is undefined.
Since $\sqrt{x^2}$ is always nonnegative for all real numbers $x$, the only way it could be zero is if $x = 0$. Therefore, we need to exclude $x = 0$ from the domain.
The domain of the function $f(x)$ is all real numbers $x$ excluding $0$. We can express this as the union of two intervals:
$$(-\infty, 0) \cup (0, \infty)$$