Find domain of the function $$f(x) = sqrt(x) + 2/x + sqrt(-x).$$ Express your answer in interval notation.
The expression $\sqrt{x}$ is defined only for $x \geq 0$, the expression $\frac{2}{x}$ is defined for $x \neq 0$, and the expression $\sqrt{-x}$ is defined only for $-x \geq 0$, or $x \leq 0$. Therefore, the domain of $f(x)$ is the intersection of these interval: $x \in \boxed{(-\infty, 0] \cup (0, \infty)}$.
To find the domain of the function, we need to consider any restrictions on the variables that would make the function undefined.
Let's break down the given function into three parts.
1. $$\sqrt{x}:$$
The square root function is defined only for non-negative values of $$x$$ since you can't take the square root of a negative number.
2. $$\frac{2}{x}:$$
Here, the only restriction is that $$x$$ cannot be equal to zero, as division by zero is undefined.
3. $$\sqrt{-x}:$$
The square root of a negative number is not a real number. Therefore, $$-x$$ must be non-negative for the square root function to be defined.
Combining these restrictions, we have:
1. $$\sqrt{x}:$$ $$x \geq 0$$
2. $$\frac{2}{x}:$$ $$x \neq 0$$
3. $$\sqrt{-x}:$$ $$-x \geq 0 \Rightarrow x \leq 0$$
To find the overall domain, we need to find the intersection of these three individual domains.
The domain is given by $$x$$ such that $$x \in (-\infty, 0] \cup (0, \infty)$$, expressed in interval notation.
Therefore, the domain of the function $$f(x) = \sqrt{x} + \frac{2}{x} + \sqrt{-x}$$ is $$(-\infty, 0] \cup (0, \infty)$$.
To find the domain of a function, we need to determine the values of x for which the function is defined.
Let's analyze each term in the function separately.
First, we have the term $$\sqrt{x}.$$ The square root function is defined only for non-negative numbers, so we need to ensure that $$x$$ is greater than or equal to zero.
Next, we have the term $$\frac{2}{x}.$$ The denominator cannot be equal to zero because division by zero is undefined. Therefore, we need to exclude the value $$x = 0$$ from the domain.
Lastly, we have the term $$\sqrt{-x}.$$ The square root of a negative number is not a real number, so we need to ensure that $$-x$$ is non-negative. In other words, $$-x \geq 0$$, which means $$x \leq 0$$.
Now, let's combine all the conditions we found:
- $$x \geq 0$$ (for $$\sqrt{x}$$)
- $$x \neq 0$$ (for $$\frac{2}{x}$$)
- $$x \leq 0$$ (for $$\sqrt{-x}$$)
To express the domain in interval notation, we can write it as the intersection of the intervals that satisfy each of the conditions.
- For $$x \geq 0$$, we can use the interval $$[0, +\infty)$$.
- For $$x \neq 0$$, we exclude the value $$x = 0$$, so the interval is $$(-\infty, 0) \cup (0, +\infty)$$.
- For $$x \leq 0$$, we can use the interval $$(-\infty, 0]$$.
Taking the intersection of these intervals, we find that the domain of the function $$f(x)$$ is $$(-\infty, 0] \cup [0, +\infty)$$.