To solve the given function $f(x) = \frac{(x^2 - 3x)^{\frac{1}{2}}}{x^2}$, let's start by simplifying the expression.

Step-by-Step Solution:

1. Simplify the expression inside the square root:

$f(x) = \frac{\sqrt{x^2 - 3x}}{x^2}$

2. Factor the numerator:

$x^2 - 3x = x(x - 3)$

So,

$f(x) = \frac{\sqrt{x(x - 3)}}{x^2}$

3. Simplify the square root of the product:

$\sqrt{x(x - 3)} = \sqrt{x} \cdot \sqrt{x - 3}$

Thus,

$f(x) = \frac{\sqrt{x} \cdot \sqrt{x - 3}}{x^2}$

4. Separate the fractions:

$f(x) = \frac{\sqrt{x}}{x^2} \cdot \sqrt{x - 3}$

5. Simplify the fraction involving $x$:

$\frac{\sqrt{x}}{x^2} = \frac{x^{1/2}}{x^2} = x^{1/2 - 2} = x^{-3/2}$

So,

$f(x) = x^{-3/2} \cdot \sqrt{x - 3}$

6. Combine the terms:

$f(x) = \frac{\sqrt{x - 3}}{x^{3/2}}$

Thus, the simplified form of the given function is:

$f(x) = \frac{\sqrt{x - 3}}{x^{3/2}}$

Domain of the Function:

To determine the domain of $f(x)$, consider the conditions under which the expression is defined:

1. The expression inside the square root, $x - 3$, must be non-negative.
2. The denominator $x^{3/2}$ must be non-zero.

This implies: $x - 3 \geq 0$ $x \neq 0$

So, $x \geq 3$ and $x \neq 0$.

Hence, the domain of $f(x)$ is $x \in (3, \infty)$.

Do you have any questions or need further details?

Here are 5 related questions you might want to explore:

1. How do you find the derivative of $f(x) = \frac{\sqrt{x - 3}}{x^{3/2}}$?
2. What is the integral of $f(x) = \frac{\sqrt{x - 3}}{x^{3/2}}$?
3. How do you determine the critical points of $f(x) = \frac{\sqrt{x - 3}}{x^{3/2}}$?
4. What are the asymptotes of $f(x) = \frac{\sqrt{x - 3}}{x^{3/2}}$?
5. How do you find the limit of $f(x) = \frac{\sqrt{x - 3}}{x^{3/2}}$ as $x$ approaches infinity?

Tip: When simplifying expressions involving square roots and exponents, it often helps to rewrite the radicals in terms of fractional exponents.