The image shows a mathematical problem asking to find the difference quotient $\frac{f(x+h) - f(x)}{h}$ for the given function $f(x) = \frac{1}{\sqrt{3x - 2}}$ as $h \to 0$.

Solution:

The difference quotient is a fundamental part of finding the derivative. Let's proceed step-by-step.

1. Function $f(x)$: $f(x) = \frac{1}{\sqrt{3x - 2}}$

2. Function $f(x+h)$: Substitute $x + h$ into the function: $f(x + h) = \frac{1}{\sqrt{3(x + h) - 2}} = \frac{1}{\sqrt{3x + 3h - 2}}$

3. Difference Quotient: Now, apply the formula for the difference quotient: $\frac{f(x + h) - f(x)}{h} = \frac{\frac{1}{\sqrt{3x + 3h - 2}} - \frac{1}{\sqrt{3x - 2}}}{h}$ To simplify this, combine the terms in the numerator: $\frac{f(x + h) - f(x)}{h} = \frac{\sqrt{3x - 2} - \sqrt{3x + 3h - 2}}{h \cdot \sqrt{3x + 3h - 2} \cdot \sqrt{3x - 2}}$ The next step is to simplify the numerator using a technique called "rationalizing the numerator," multiplying both the numerator and the denominator by the conjugate of the numerator: $\text{Conjugate} = \sqrt{3x - 2} + \sqrt{3x + 3h - 2}$ This gives: $\frac{\left( \sqrt{3x - 2} - \sqrt{3x + 3h - 2} \right) \left( \sqrt{3x - 2} + \sqrt{3x + 3h - 2} \right)}{h \cdot \left( \sqrt{3x + 3h - 2} \cdot \sqrt{3x - 2} \right) \left( \sqrt{3x - 2} + \sqrt{3x + 3h - 2} \right)}$ The numerator becomes: $(3x - 2) - (3x + 3h - 2) = -3h$ Now the expression is: $\frac{-3h}{h \cdot \left( \sqrt{3x + 3h - 2} \cdot \sqrt{3x - 2} \right) \left( \sqrt{3x - 2} + \sqrt{3x + 3h - 2} \right)}$ Simplifying: $\frac{-3}{\sqrt{3x + 3h - 2} \cdot \sqrt{3x - 2} \cdot \left( \sqrt{3x - 2} + \sqrt{3x + 3h - 2} \right)}$

4. Taking the Limit as $h \to 0$: As $h \to 0$, $\sqrt{3x + 3h - 2} \to \sqrt{3x - 2}$. So the expression simplifies to: $\frac{-3}{2 \cdot \left( \sqrt{3x - 2} \right)^3}$ Therefore, the derivative of the function $f(x)$ is: $f'(x) = \frac{-3}{2(3x - 2)^{3/2}}$

Would you like more details or have any questions?

Here are some related questions you might find interesting:

1. What is the geometric interpretation of a derivative?
2. How does the difference quotient relate to the concept of a tangent line?
3. Can you apply this same process to a polynomial function?
4. What is the importance of rationalizing the numerator in calculus?
5. How would the derivative change if the function were $f(x) = \frac{1}{\sqrt{ax - b}}$?

Tip: Rationalizing the numerator is an essential algebraic technique to simplify expressions, especially when finding limits and derivatives.