To find the slope of the tangent line to the function \( f(x) = x\sqrt[3]{2x+1} \) at \( x = 1.5 \), we need to evaluate the derivative \( f'(x) \) at \( x = 1.5 \).

### Recap of the Derivative
The derivative of the function \( f(x) = x(2x+1)^{\frac{1}{3}} \) is given by:

\[
f'(x) = (2x+1)^{\frac{1}{3}} \left( 1 + \frac{2x}{3(2x+1)} \right)
\]

### Evaluate the Derivative at \( x = 1.5 \)

1. **Calculate \( (2(1.5) + 1)^{\frac{1}{3}} \):**
   \[
   2(1.5) + 1 = 3 + 1 = 4
   \]
   \[
   4^{\frac{1}{3}} \approx 1.5874
   \]

2. **Calculate \( \frac{2(1.5)}{3(2(1.5) + 1)} \):**
   \[
   \frac{2(1.5)}{3(4)} = \frac{3}{12} = 0.25
   \]

3. **Add the results:**
   \[
   1 + 0.25 = 1.25
   \]

4. **Multiply to get \( f'(1.5) \):**
   \[
   f'(1.5) = 1.5874 \times 1.25 \approx 1.9843
   \]

### Final Answer

The slope of the tangent line at \( x = 1.5 \) is approximately **1.9843**.

Would you like further details or have any other questions?

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**Related Questions:**

1. How does the slope of the tangent line change as \( x \) increases?
2. What is the geometric interpretation of the derivative at a specific point?
3. Can we approximate the value of \( f(x) \) using the tangent line equation near \( x = 1.5 \)?
4. What are the potential errors in numerical approximations of roots?
5. How would the result change if the function were \( f(x) = x\sqrt{2x+1} \) instead?

**Tip:** When evaluating derivatives at non-integer points, pay attention to simplifying and approximating roots accurately for the most precise results.

Learn how to find the slope of the tangent line to the function f(x) = x√(2x + 1) at x = 1.5. Detailed step-by-step calculation of the derivative and evaluation at the point x = 1.5 using the power rule and chain rule. Suitable for advanced high school students studying calculus.

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