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We are given the function \( f(x) = 4 - 6\sqrt{x} \). We need to find the expression for:

\[
\frac{f(x+h) - f(x)}{h}
\]

and rewrite it in the form:

\[
\frac{A}{\left(\sqrt{Bx + Ch} + \sqrt{x}\right)},
\]

where \(A\), \(B\), and \(C\) are constants. Let's break down the steps:

### Step 1: Expand \( f(x + h) \)
We start with the given function:
\[
f(x) = 4 - 6\sqrt{x}.
\]

Now, calculate \( f(x + h) \):
\[
f(x + h) = 4 - 6\sqrt{x + h}.
\]

### Step 2: Difference Quotient
The difference quotient is:

\[
\frac{f(x+h) - f(x)}{h} = \frac{(4 - 6\sqrt{x+h}) - (4 - 6\sqrt{x})}{h}
= \frac{-6\sqrt{x+h} + 6\sqrt{x}}{h}.
\]

This simplifies to:

\[
\frac{6(\sqrt{x} - \sqrt{x+h})}{h}.
\]

### Step 3: Simplify the Expression
To simplify further, we multiply the numerator and denominator by the conjugate of \( \sqrt{x} - \sqrt{x+h} \), which is \( \sqrt{x} + \sqrt{x+h} \):

\[
\frac{6(\sqrt{x} - \sqrt{x+h})}{h} \cdot \frac{\sqrt{x} + \sqrt{x+h}}{\sqrt{x} + \sqrt{x+h}} = \frac{6 \left[ (\sqrt{x})^2 - (\sqrt{x+h})^2 \right]}{h (\sqrt{x} + \sqrt{x+h})}.
\]

Now simplify the numerator:

\[
(\sqrt{x})^2 - (\sqrt{x+h})^2 = x - (x+h) = -h.
\]

Thus, the expression becomes:

\[
\frac{6(-h)}{h(\sqrt{x} + \sqrt{x+h})} = \frac{-6}{\sqrt{x} + \sqrt{x+h}}.
\]

### Step 4: Rewrite in the Desired Form
We want the expression to match the form:

\[
\frac{A}{\sqrt{Bx + Ch} + \sqrt{x}}.
\]

From our simplified result, we can see that:

- \( A = -6 \),
- \( B = 1 \) (since we have \( \sqrt{x+h} \), so \( Bx + Ch = x + h \)),
- \( C = 1 \) (since \( Ch = h \)).

Thus, the constants are:
- \( A = -6 \),
- \( B = 1 \),
- \( C = 1 \).

### Step 5: Compute the Limit as \( h \to 0 \)
Finally, we are asked to compute:

\[
\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.
\]

From our simplified expression:

\[
\frac{-6}{\sqrt{x+h} + \sqrt{x}},
\]

as \( h \to 0 \), we have \( \sqrt{x+h} \to \sqrt{x} \). Therefore, the limit becomes:

\[
\lim_{h \to 0} \frac{-6}{\sqrt{x+h} + \sqrt{x}} = \frac{-6}{2\sqrt{x}} = \frac{-3}{\sqrt{x}}.
\]

### Final Answer:
- \( A = -6 \)
- \( B = 1 \)
- \( C = 1 \)
- \( \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \frac{-3}{\sqrt{x}} \).

Do you need further clarification or details on any part of this solution?

Here are five related questions to explore further:
1. How does the concept of limits help in defining derivatives?
2. What are the key steps in simplifying expressions involving radicals?
3. How does the conjugate method work in simplifying difference quotients?
4. Why is the limit process essential in determining instantaneous rates of change?
5. What are some common applications of derivatives in physics and engineering?

**Tip:** When working with limits, multiplying by the conjugate is a powerful technique to simplify expressions involving square roots.

Let f(x) = 4 - 6√x. Then the expression (f(x + h) - f(x)) / h can be written in the form A / (√(Bx + Ch) + √x), where A, B, and C are constants. Find the constants A, B, and C. Also, compute the limit as h approaches 0 for (f(x + h) - f(x)) / h.

Explore the process of finding the constants A, B, and C for the function f(x) = 4 - 6√x, rewritten in the form A / (√(Bx + Ch) + √x). Learn step-by-step how to simplify the difference quotient and compute the limit as h approaches 0. Suitable for students studying calculus and limits.

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