To find the coefficient of the term containing $(x+1)^2$ in the Taylor polynomial of $f(x) = 6x^4 + 2x^3$ centered at $x = -1$, we need to compute the second derivative of $f(x)$ and evaluate it at $x = -1$.

Step 1: Compute the first and second derivatives of $f(x)$

The function is: $f(x) = 6x^4 + 2x^3$

First derivative: $f'(x) = \frac{d}{dx}\left(6x^4 + 2x^3\right) = 24x^3 + 6x^2$

Second derivative: $f''(x) = \frac{d}{dx}\left(24x^3 + 6x^2\right) = 72x^2 + 12x$

Step 2: Evaluate the second derivative at $x = -1$

Substitute $x = -1$ into $f''(x)$: $f''(-1) = 72(-1)^2 + 12(-1) = 72(1) - 12 = 72 - 12 = 60$

Step 3: Use the Taylor series formula

The coefficient of the $(x + 1)^2$ term in the Taylor polynomial is given by: $\frac{f''(-1)}{2!}$

Since $f''(-1) = 60$ and $2! = 2$, the coefficient is: $\frac{60}{2} = 30$

Final Answer

The coefficient for the term containing $(x+1)^2$ is:

(Choice A) 30

Would you like more details or have any questions? Here are 5 related questions that could further your understanding:

1. How do you find the general formula for the nth term in a Taylor series?
2. What is the significance of centering a Taylor series at a point $x = a$?
3. How do higher-order derivatives affect the accuracy of a Taylor polynomial?
4. Can you calculate the Taylor series of $f(x)$ centered at a different point, such as $x = 0$?
5. What happens if we center the Taylor series at a point where the function or its derivatives are undefined?

Tip: Always ensure that you evaluate derivatives carefully at the point of expansion when calculating Taylor series coefficients.