This problem requires finding the second-order Taylor expansion of $\sqrt[3]{1 + 3x}$ around $x = 0$.

Step-by-step solution:

1. Rewrite the expression for expansion: The function is $f(x) = (1 + 3x)^{1/3}$.

2. Use the Taylor series formula: For $f(x)$ centered at $x = 0$, the Taylor expansion up to the second order is: $f(x) \approx f(0) + f'(0)x + \frac{f''(0)}{2}x^2.$

3. Calculate $f(0)$: $f(0) = (1 + 3 \cdot 0)^{1/3} = 1.$

4. Find the first derivative $f'(x)$: Using the chain rule: $f'(x) = \frac{1}{3}(1 + 3x)^{-2/3} \cdot 3 = (1 + 3x)^{-2/3}.$ At $x = 0$: $f'(0) = (1 + 3 \cdot 0)^{-2/3} = 1.$

5. Find the second derivative $f''(x)$: Differentiate $f'(x) = (1 + 3x)^{-2/3}$: $f''(x) = -\frac{2}{3}(1 + 3x)^{-5/3} \cdot 3 = -2(1 + 3x)^{-5/3}.$ At $x = 0$: $f''(0) = -2(1 + 3 \cdot 0)^{-5/3} = -2.$

6. Substitute into the Taylor series: Using $f(0) = 1$, $f'(0) = 1$, and $f''(0) = -2$, the Taylor expansion becomes: $f(x) \approx 1 + 1 \cdot x + \frac{-2}{2}x^2 = 1 + x - x^2.$

Answer:

The correct option is A: $1 + x - x^2$.

Do you want a detailed derivation of any step or more examples?

Related questions:

1. What is the general Taylor expansion formula for any $f(x)$?
2. How would the result change if we expand $\sqrt[3]{1 + 3x}$ to third order?
3. Why do higher-order derivatives beyond second order not appear in this expansion?
4. How does the radius of convergence relate to Taylor expansions like this one?
5. Can you derive the expansion of $(1 + x)^n$ for fractional $n$ in general?

Tip:

For small perturbations around $x = 0$, approximations via Taylor expansion provide powerful simplifications for complex functions.