To show that $f(x) = 9x^{-\frac{1}{2}} + Ax^{\frac{1}{2}} + B$, let's start with the given function and simplify it.

Given:

$f(x) = \frac{\left(3 - 4\sqrt{x}\right)^2}{\sqrt{x}}, \quad x > 0$

Step 1: Expand the Numerator

The numerator $(3 - 4\sqrt{x})^2$ can be expanded using the formula $(a - b)^2 = a^2 - 2ab + b^2$: $(3 - 4\sqrt{x})^2 = 3^2 - 2 \cdot 3 \cdot 4\sqrt{x} + (4\sqrt{x})^2$ $= 9 - 24\sqrt{x} + 16x$

So, the function $f(x)$ becomes: $f(x) = \frac{9 - 24\sqrt{x} + 16x}{\sqrt{x}}$

Step 2: Separate Terms in the Numerator

Now, divide each term in the numerator by $\sqrt{x}$: $f(x) = \frac{9}{\sqrt{x}} - \frac{24\sqrt{x}}{\sqrt{x}} + \frac{16x}{\sqrt{x}}$

This simplifies each term as follows:

• $\frac{9}{\sqrt{x}} = 9x^{-\frac{1}{2}}$
• $\frac{24\sqrt{x}}{\sqrt{x}} = 24$
• $\frac{16x}{\sqrt{x}} = 16x^{\frac{1}{2}}$

Thus: $f(x) = 9x^{-\frac{1}{2}} + 16x^{\frac{1}{2}} - 24$

Step 3: Compare with the Desired Form

We have: $f(x) = 9x^{-\frac{1}{2}} + 16x^{\frac{1}{2}} - 24$

This matches the form $f(x) = 9x^{-\frac{1}{2}} + Ax^{\frac{1}{2}} + B$, where:

• $A = 16$
• $B = -24$

Final Answer:

$A = 16, \quad B = -24$

Would you like further details on any of the steps?

Here are some related questions:

1. How can we determine the behavior of $f(x)$ as $x \to 0$ or $x \to \infty$?
2. What would the derivative $f'(x)$ of this function look like?
3. How does the presence of $x^{-\frac{1}{2}}$ affect the domain of $f(x)$?
4. Can we integrate $f(x)$ over a specific interval?
5. How would the graph of $f(x) = 9x^{-\frac{1}{2}} + 16x^{\frac{1}{2}} - 24$ appear?

Tip: When simplifying expressions with roots or powers, look for ways to separate terms by division to simplify the overall structure of the expression.