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?

---

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.