The problem is asking to explain why the square root of $-x$, represented as $\sqrt{-x}$, is not necessarily undefined and how that affects the domain and range of the function $f(x) = \sqrt{-x}$.

Explanation:

1. The square root of a negative number, such as $\sqrt{-144}$, is undefined in the real number system because there is no real number that, when squared, gives a negative result. In the real number system, only positive numbers have real square roots.

2. However, for $\sqrt{-x}$, the expression depends on x. When $x$ is positive, $-x$ becomes negative. For example, if $x = 144$, then $\sqrt{-144}$ is undefined (as already stated).

3. Now, if we examine $f(x) = \sqrt{-x}$, to ensure that the expression is defined for real numbers, we must ensure that the expression inside the square root, $-x$, is non-negative (since square roots of negative numbers are undefined in the real number system). This means: $-x \geq 0 \implies x \leq 0.$ Therefore, the domain of $f(x)$ consists of values of $x$ that are less than or equal to zero (i.e., $x \leq 0$).

4. In this case, the square root of $-x$ is defined for all non-positive values of $x$ (i.e., $x \leq 0$) because when $x$ is non-positive, $-x$ becomes non-negative, and we can take the square root of a non-negative number.

Domain and Range of $f(x) = \sqrt{-x}$:

• Domain: As we concluded, the function is only defined when $x \leq 0$. Therefore, the domain is $(-\infty, 0]$.

• Range: Since $\sqrt{-x}$ results in a non-negative number (because square roots of non-negative numbers are non-negative), the range of $f(x)$ is $[0, \infty)$.

Conclusion:

The function $f(x) = \sqrt{-x}$ is defined for $x \leq 0$, and its range is all non-negative real numbers.

Do you want any further clarification or details?

Related Questions:

1. What is the difference between the square root of a negative number and a complex number solution?
2. How would the domain change if the function were $f(x) = \sqrt{x^2 - 4}$?
3. What is the geometric interpretation of the function $f(x) = \sqrt{-x}$?
4. How does the concept of imaginary numbers relate to the square root of negative numbers?
5. How would the range change if the function were $f(x) = -\sqrt{-x}$?

Tip:

Remember, the square root of a negative number is undefined in the real number system but can be expressed in the complex number system as an imaginary number (e.g., $\sqrt{-1} = i$).