2025 AP AP Calculus BC Practice Paper with Answers

Part A:
The average biomass over the interval \(0 \le t \le 5\) is given by the average value formula:
\[\text{Average Value} = \frac{1}{5 - 0} \int_0^5 B(t)\,dt = \frac{1}{5} \int_0^5 18 \ln(1 + 0.5t^2)\,dt\]
Using a graphing calculator to evaluate the integral:
\[\frac{1}{5} (120.18498) = 24.036996\]
The average biomass of the bloom is \(24.037\) metric tons.

Part B:
The average rate of change of \(B\) over \(0 \le t \le 6\) is:
\[\frac{B(6) - B(0)}{6 - 0} = \frac{18 \ln(1 + 0.5(6)^2) - 18 \ln(1)}{6} = \frac{18 \ln(19) - 0}{6} = 3\ln(19) \approx 8.833317\]
Set the instantaneous rate of change equal to the average rate of change:
\[B'(t) = \frac{18t}{1 + 0.5t^2} = 8.833317\]
Using a graphing calculator to solve this equation for \(0 \le t \le 6\) yields:
\[t = 0.571 \quad \text{and} \quad t = 3.505\]

Part C:
Differentiating \(B'(t)\) with respect to \(t\):
\[B''(t) = \frac{18(1 + 0.5t^2) - 18t(t)}{(1 + 0.5t^2)^2} = \frac{18 - 9t^2}{(1 + 0.5t^2)^2}\]
At \(t = 2\):
\[B''(2) = \frac{18 - 9(2)^2}{(1 + 0.5(2)^2)^2} = \frac{18 - 36}{(1 + 2)^2} = \frac{-18}{9} = -2\]
Because \(B''(2) = -2 < 0\), the rate of change of the biomass, \(B'(t)\), is decreasing at time \(t = 2\).

Part D:
By the Fundamental Theorem of Calculus:
\[W'(t) = B'(t) - (0.4t + 1) = \frac{18t}{1 + 0.5t^2} - (0.4t + 1)\]
Setting \(W'(t) = 0\) on \(6 \le t \le 14\):
\[\frac{18t}{1 + 0.5t^2} = 0.4t + 1 \implies t \approx 8.18128\]
To determine the absolute maximum on \([6, 14]\), evaluate \(W(t)\) at the critical point and at the endpoints:

- \(W(6) = B(6) - \int_6^6 (0.4x + 1)\,dx = 18\ln(19) \approx 53.000\)
- \(W(8.181) = B(8.181) - \int_6^{8.181} (0.4x + 1)\,dx \approx 63.719 - 8.367 = 55.352\)
- \(W(14) = B(14) - \int_6^{14} (0.4x + 1)\,dx \approx 82.712 - 40.000 = 42.712\)

Therefore, \(W\) attains its absolute maximum value at time \(t = 8.181\) (or \(8.182\)) days.

Part A (2 points):
- P1: Average value formula setup: \(\frac{1}{5-0} \int_0^5 B(t)\,dt\) (or \(\frac{1}{5} \int_0^5 18 \ln(1 + 0.5t^2)\,dt\))
- P2: Correct answer (\(24.037\) or \(24.036\))

Scoring notes:
- Correct integral with evidence of division by 5 earns P1.
- A reported numerical answer accurate to three decimal places earns P2.

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Part B (2 points):
- P3: Setup using average rate of change: \(\frac{B(6) - B(0)}{6 - 0}\) (or \(3\ln(19)\) or \(8.833\))
- P4: Correct values of \(t\) with supporting equation: \(t = 0.571\) and \(t = 3.505\)

Scoring notes:
- Both values of \(t\) are required to earn P4, supported by setting \(B'(t) = \frac{B(6)-B(0)}{6}\).

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Part C (2 points):
- P5: Value of \(B''(2) = -2\)
- P6: Conclusion with reason (decreasing because \(B''(2) < 0\))

Scoring notes:
- P6 requires explicitly referencing \(B''(2) < 0\) or that the derivative of \(B'(t)\) is negative at \(t = 2\).

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Part D (3 points):
- P7: Considers \(W'(t) = 0\) (or \(B'(t) - (0.4t + 1) = 0\))
- P8: Justification (Candidates Test evaluating \(W(6)\), \(W(8.181)\), and \(W(14)\), or valid global sign analysis)
- P9: Answer with supporting work: \(t = 8.181\) (or \(8.182\))

Scoring notes:
- P7 is not earned by merely presenting the numerical value \(t = 8.181\).
- P8 requires evaluating the function at the critical point and both endpoints, correct to at least one decimal place, or a complete sign analysis of \(W'(t)\) on the entire interval \([6, 14]\).

### Part A
The distance from the origin to a point on curve \(C\) is given by \(r(\theta) = 2 + \cos(2\theta)\).
The rate of change of \(r\) with respect to \(\theta\) is:
\[ \frac{dr}{d\theta} = -2\sin(2\theta) \]
Evaluating at \(\theta = 1.2\):
\[ \left.\frac{dr}{d\theta}\right|_{\theta = 1.2} = -2\sin(2.4) \approx -1.351003 \]
Thus, the rate of change is \(-1.351\).

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### Part B
To find the intersection points of curve \(C\) and the circle \(r = 2\) for \(0 \le \theta \le \pi\):
\[ 2 + \cos(2\theta) = 2 \implies \cos(2\theta) = 0 \implies 2\theta = \frac{\pi}{2}, \frac{3\pi}{2} \implies \theta = \frac{\pi}{4}, \frac{3\pi}{4} \]
Curve \(C\) lies outside the circle \(r = 2\) when \(\cos(2\theta) > 0\), which occurs on \(\left[0, \frac{\pi}{4}\right]\) and \(\left[\frac{3\pi}{4}, \pi\right]\).

The area is:
\[ \text{Area} = \frac{1}{2} \int_0^{\pi/4} \left( (2+\cos(2\theta))^2 - 2^2 \right) d\theta + \frac{1}{2} \int_{3\pi/4}^\pi \left( (2+\cos(2\theta))^2 - 2^2 \right) d\theta \]
Using symmetry:
\[ \text{Area} = 2 \cdot \frac{1}{2} \int_0^{\pi/4} \left( (2+\cos(2\theta))^2 - 4 \right) d\theta = \int_0^{\pi/4} \left(4\cos(2\theta) + \cos^2(2\theta)\right) d\theta = 2 + \frac{\pi}{8} \approx 2.392699 \]
The area is \(2.393\) (or \(2.392\)).

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### Part C
The \(y\)-coordinate of a point on curve \(C\) is:
\[ y(\theta) = r(\theta)\sin\theta = (2 + \cos(2\theta))\sin\theta \]
At \(\theta = \frac{\pi}{3}\):
\[ y\left(\frac{\pi}{3}\right) = \left(2 + \cos\left(\frac{2\pi}{3}\right)\right)\sin\left(\frac{\pi}{3}\right) = \left(2 - \frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right) = \frac{3\sqrt{3}}{4} \approx 1.299 \]
The derivative of \(y\) with respect to \(\theta\) is:
\[ \frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta + r(\theta)\cos\theta = -2\sin(2\theta)\sin\theta + (2+\cos(2\theta))\cos\theta \]
Evaluating at \(\theta = \frac{\pi}{3}\):
\[ \left.\frac{dy}{d\theta}\right|_{\theta = \pi/3} = -2\sin\left(\frac{2\pi}{3}\right)\sin\left(\frac{\pi}{3}\right) + \left(2+\cos\left(\frac{2\pi}{3}\right)\right)\cos\left(\frac{\pi}{3}\right) = -2\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{3}{2}\right)\left(\frac{1}{2}\right) = -\frac{3}{2} + \frac{3}{4} = -0.75 \]
Since \(y\left(\frac{\pi}{3}\right) > 0\) and \(\left.\frac{dy}{d\theta}\right|_{\theta = \pi/3} < 0\), the \(y\)-coordinate is positive and decreasing toward \(0\). Therefore, the point is moving toward the \(x\)-axis.

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### Part D
The \(x\)-coordinate is:
\[ x(\theta) = r(\theta)\cos\theta = (2+\cos(2\theta))\cos\theta \]
By the chain rule:
\[ \frac{dx}{dt} = \frac{dx}{d\theta} \cdot \frac{d\theta}{dt} \]
where
\[ \frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta - r(\theta)\sin\theta = -2\sin(2\theta)\cos\theta - (2+\cos(2\theta))\sin\theta \]
Evaluating at \(\theta = \frac{\pi}{6}\):
\[ \left.\frac{dx}{d\theta}\right|_{\theta = \pi/6} = -2\sin\left(\frac{\pi}{3}\right)\cos\left(\frac{\pi}{6}\right) - \left(2+\cos\left(\frac{\pi}{3}\right)\right)\sin\left(\frac{\pi}{6}\right) = -2\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) - (2.5)(0.5) = -1.5 - 1.25 = -2.75 \]
Therefore:
\[ \left.\frac{dx}{dt}\right|_{\theta = \pi/6} = (-2.75)(4) = -11 \]

Part A (1 point):
- Point 1 (P1): Earned for answer with setup: \(\left.\frac{dr}{d\theta}\right|_{\theta=1.2} = -1.351\) (or \(-2\sin(2.4)\)).

Part B (3 points):
- Point 2 (P2): Limits of integration (\(\theta = 0, \frac{\pi}{4}\) and \(\theta = \frac{3\pi}{4}, \pi\), or equivalent use of symmetry).
- Point 3 (P3): Correct integrand \((2+\cos(2\theta))^2 - 2^2\).
- Point 4 (P4): Correct area answer \(2.393\) (or \(2.392\), or \(2 + \frac{\pi}{8}\)).

Part C (3 points):
- Point 5 (P5): Correct \(y\)-coordinate: \(y\left(\frac{\pi}{3}\right) = \frac{3\sqrt{3}}{4} \approx 1.299\).
- Point 6 (P6): Evaluates \(\left.\frac{dy}{d\theta}\right|_{\theta=\pi/3} = -0.75\).
- Point 7 (P7): Conclusion ("toward the \(x\)-axis") with reason referencing the signs of \(y\) and \(\frac{dy}{d\theta}\).

Part D (2 points):
- Point 8 (P8): Chain rule setup: \(\frac{dx}{dt} = \frac{dx}{d\theta} \cdot \frac{d\theta}{dt}\).
- Point 9 (P9): Correct numerical answer \(-11\).

Part A:

The average rate of change of \(H\) over the interval \([3, 5]\) provides an approximation for \(H'(4)\):
$$H'(4) \approx \frac{H(5) - H(3)}{5 - 3} = \frac{10 - 14}{5 - 3} = \frac{-4}{2} = -2^\circ\text{C/min}^2$$

Part B:

Because \(H\) is twice-differentiable on \([0, 12]\), \(H\) is differentiable on \((0, 12)\) and continuous on \([0, 12]\).

The average rate of change of \(H\) over \([0, 12]\) is:
$$\frac{H(12) - H(0)}{12 - 0} = \frac{3 - 18}{12} = \frac{-15}{12} = -\frac{5}{4}$$

Therefore, by the Mean Value Theorem, there exists at least one value \(c\) in the interval \(0 < c < 12\) such that \(H'(c) = -\frac{5}{4}\).

Part C:

The trapezoidal sum with the three given subintervals is:
$$\int_0^{12} H(t)\,dt \approx \frac{H(0) + H(3)}{2}(3 - 0) + \frac{H(3) + H(5)}{2}(5 - 3) + \frac{H(5) + H(12)}{2}(12 - 5)$$
$$= \frac{18 + 14}{2}(3) + \frac{14 + 10}{2}(2) + \frac{10 + 3}{2}(7)$$
$$= 16(3) + 12(2) + 6.5(7) = 48 + 24 + 45.5 = 117.5$$

In the context of the problem, \(\int_0^{12} H(t)\,dt\) represents the total decrease in the temperature of the liquid in the vat, in degrees Celsius, from time \(t = 0\) to \(t = 12\) minutes.

Part D:

Let \(T(t)\) denote the temperature of the second vat at time \(t\). By the Fundamental Theorem of Calculus:
$$T(9) = T(0) + \int_0^9 G(t)\,dt = 25 + \int_0^9 \left(4 + 6\cos\left(\frac{\pi}{6}t\right)\right)dt$$

Evaluating the definite integral:
$$\int_0^9 \left(4 + 6\cos\left(\frac{\pi}{6}t\right)\right)dt = \left[ 4t + 6 \cdot \frac{6}{\pi} \sin\left(\frac{\pi}{6}t\right) \right]_0^9 = \left[ 4t + \frac{36}{\pi} \sin\left(\frac{\pi}{6}t\right) \right]_0^9$$
$$= \left(4(9) + \frac{36}{\pi} \sin\left(\frac{3\pi}{2}\right)\right) - \left(0 + \frac{36}{\pi}\sin(0)\right)$$
$$= 36 + \frac{36}{\pi}(-1) - 0 = 36 - \frac{36}{\pi}$$

Thus, the temperature at \(t = 9\) minutes is:
$$T(9) = 25 + \left(36 - \frac{36}{\pi}\right) = 61 - \frac{36}{\pi}^\circ\text{C}$$

Part A (2 points):
- 1 point (P1): Difference quotient setup using data from the table (e.g., \(\frac{10-14}{5-3}\) or \(-2\)).
- 1 point (P2): Correct units of measure (\(^\circ\text{C/min}^2\) or degrees Celsius per minute per minute).

Part B (2 points):
- 1 point (P3): States that \(H\) is continuous because \(H\) is differentiable, and computes the average rate of change \(\frac{H(12)-H(0)}{12-0} = -\frac{5}{4}\).
- 1 point (P4): Concludes "yes" by citing the Mean Value Theorem with correct hypotheses verified.

Part C (2 points):
- 1 point (P5): Trapezoidal sum setup showing three terms with the correct subinterval widths.
- 1 point (P6): Correct value (\(117.5\)) and correct contextual explanation with units (total decrease in temperature of 117.5 degrees Celsius from \(t=0\) to \(t=12\) minutes).

Part D (3 points):
- 1 point (P7): Integral expression with initial condition \(25 + \int_0^9 G(t)\,dt\).
- 1 point (P8): Correct antiderivative \(4t + \frac{36}{\pi}\sin\left(\frac{\pi}{6}t\right)\).
- 1 point (P9): Correct numerical answer: \(61 - \frac{36}{\pi}\) (or unsimplified equivalent).

### Part A
To find \( g(5) \):
\[ g(5) = \int_{-2}^5 f(t)\, dt = \int_{-2}^2 f(t)\, dt + \int_2^5 f(t)\, dt \]
The region from \( t = -2 \) to \( t = 2 \) is a semicircle of radius \( 2 \):
\[ \int_{-2}^2 f(t)\, dt = \frac{1}{2} \pi (2)^2 = 2\pi \]
The region from \( t = 2 \) to \( t = 5 \) is a triangle below the \( t \)-axis with base \( 3 \) and height \( 3 \):
\[ \int_2^5 f(t)\, dt = -\frac{1}{2}(3)(3) = -\frac{9}{2} \]
Therefore, \( g(5) = 2\pi - \frac{9}{2} \).

To find \( g(-4) \):
\[ g(-4) = \int_{-2}^{-4} f(t)\, dt = -\int_{-4}^{-2} f(t)\, dt \]
The region from \( t = -4 \) to \( t = -2 \) is a triangle below the \( t \)-axis with base \( 2 \) and height \( 2 \):
\[ \int_{-4}^{-2} f(t)\, dt = -\frac{1}{2}(2)(2) = -2 \]
Thus, \( g(-4) = -(-2) = 2 \).

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### Part B
By the Fundamental Theorem of Calculus, \( g'(x) = f(x) \).
On the interval \([5, 8]\), the line segment has slope \( m = \frac{3 - (-3)}{8 - 5} = \frac{6}{3} = 2 \).
The equation of the line segment is \( f(x) = 2(x - 5) - 3 = 2x - 13 \).
Thus:
\[ g'(7) = f(7) = 2(7) - 13 = 1 \]
Since \( g''(x) = f'(x) \) and \( f(x) \) is linear with slope \( 2 \) on \((5, 8)\):
\[ g''(7) = f'(7) = 2 \]

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### Part C
The graph of \( g \) has a point of inflection where \( g''(x) = f'(x) \) changes sign, which corresponds to where \( f(x) \) changes from increasing to decreasing or from decreasing to increasing.
- On \((-4, 0)\), \( f \) is increasing, and on \((0, 5)\), \( f \) is decreasing. Thus, \( f \) changes from increasing to decreasing at \( x = 0 \).
- On \((0, 5)\), \( f \) is decreasing, and on \((5, 8)\), \( f \) is increasing. Thus, \( f \) changes from decreasing to increasing at \( x = 5 \).
Therefore, the graph of \( g \) has points of inflection at \( x = 0 \) and \( x = 5 \).

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### Part D
By the Extreme Value Theorem, since \( g \) is continuous on \([-4, 8]\), the absolute maximum must occur at a critical point where \( g'(x) = f(x) = 0 \) or at an endpoint.
Setting \( g'(x) = f(x) = 0 \) gives the critical points:
- \( x = -2 \)
- \( x = 2 \)
- For \( x \in [5, 8] \), \( 2x - 13 = 0 \implies x = 6.5 \)

Evaluating \( g(x) \) at the candidates:
- \( g(-4) = 2 \)
- \( g(-2) = 0 \)
- \( g(2) = 2\pi \approx 6.283 \)
- \( g(6.5) = g(5) + \int_5^{6.5} f(t)\, dt = 2\pi - 4.5 - \frac{1}{2}(1.5)(3) = 2\pi - 6.75 \approx -0.467 \)
- \( g(8) = g(6.5) + \int_{6.5}^8 f(t)\, dt = 2\pi - 6.75 + \frac{1}{2}(1.5)(3) = 2\pi - 4.5 = 2\pi - \frac{9}{2} \approx 1.783 \)

Comparing these values, the absolute maximum value of \( g \) is \( 2\pi \) (at \( x = 2 \)).

Part A (2 points):
- 1 point (P1): \(g(5) = 2\pi - \frac{9}{2}\)
- 1 point (P2): \(g(-4) = 2\)

Part B (2 points):
- 1 point (P3): \(g'(7) = 1\)
- 1 point (P4): \(g''(7) = 2\)

Part C (2 points):
- 1 point (P5): Identifies \(x = 0\) and \(x = 5\) (and no other values)
- 1 point (P6): Reason tied to \(f\) changing from increasing to decreasing or decreasing to increasing

Part D (3 points):
- 1 point (P7): Considers \(g'(x) = f(x) = 0\) (identifies critical points)
- 1 point (P8): Justification via Candidates Test evaluating \(g(x)\) at critical points and endpoints
- 1 point (P9): Correct absolute maximum value \(2\pi\)

### Part A
To find \( \frac{d^2y}{dx^2} \), differentiate \( \frac{dy}{dx} = (1 - 2x)y^2 \) with respect to \( x \) using the product rule and chain rule:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(1 - 2x) \cdot y^2 + (1 - 2x) \cdot \frac{d}{dx}(y^2) = -2y^2 + (1 - 2x)\left(2y \frac{dy}{dx}\right) \]
At the point \( (0, 2) \):
\[ \left.\frac{dy}{dx}\right|_{(0,2)} = (1 - 2(0))(2)^2 = (1)(4) = 4 \]
Substitute \( x = 0 \), \( y = 2 \), and \( \frac{dy}{dx} = 4 \) into the second derivative:
\[ f''(0) = \left.\frac{d^2y}{dx^2}\right|_{(0,2)} = -2(2)^2 + (1 - 0)(2)(2)(4) = -8 + 16 = 8 \]

### Part B
The second-degree Taylor polynomial for \( f \) about \( x = 0 \) is:
\[ P_2(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 \]
Given \( f(0) = 2 \), \( f'(0) = 4 \), and \( f''(0) = 8 \):
\[ P_2(x) = 2 + 4x + \frac{8}{2}x^2 = 2 + 4x + 4x^2 \]

### Part C
By the Lagrange error bound, the error in approximating \( f(0.1) \) with \( P_2(0.1) \) satisfies:
\[ |f(0.1) - P_2(0.1)| \le \frac{\max_{0 \le x \le 0.1} |f'''(x)|}{3!} |0.1 - 0|^3 \]
Since \( |f'''(x)| \le 48 \) for all \( x \in [0, 0.1] \):
\[ |f(0.1) - P_2(0.1)| \le \frac{48}{6} (0.1)^3 = 8 \cdot 0.001 = 0.008 \]
Thus, the approximation differs from \( f(0.1) \) by at most \( 0.008 \).

### Part D
With two steps of equal size starting at \( x = 0 \) and ending at \( x = 0.4 \), the step size is:
\[ \Delta x = \frac{0.4 - 0}{2} = 0.2 \]
Step 1: \( (x_0, y_0) = (0, 2) \)
\[ \left.\frac{dy}{dx}\right|_{(0,2)} = (1 - 0)(2)^2 = 4 \]
\[ f(0.2) \approx f(0) + \Delta x \left.\frac{dy}{dx}\right|_{(0,2)} = 2 + 0.2(4) = 2 + 0.8 = 2.8 \]
Step 2: \( (x_1, y_1) = (0.2, 2.8) \)
\[ \left.\frac{dy}{dx}\right|_{(0.2, 2.8)} = (1 - 2(0.2))(2.8)^2 = (0.6)(2.8)^2 = 0.6(7.84) = 4.704 \]
\[ f(0.4) \approx f(0.2) + \Delta x \left.\frac{dy}{dx}\right|_{(0.2, 2.8)} = 2.8 + 0.2(4.704) = 2.8 + 0.9408 = 3.7408 \]

### Part A (3 points)
- Point 1 (P1): Product rule in differentiation
- Point 2 (P2): Chain rule in implicit differentiation
- Point 3 (P3): \( f''(0) = 8 \) with correct supporting work

Scoring notes:
- \( \frac{d^2y}{dx^2} = -2y^2 + (1 - 2x)2y \frac{dy}{dx} \) earns P1 and P2.
- An unsimplified correct expression such as \( -2(2)^2 + (1)(2)(2)(4) \) earns P3.

### Part B (2 points)
- Point 4 (P4): Two terms correct
- Point 5 (P5): Remaining term correct

Scoring notes:
- \( P_2(x) = 2 + 4x + 4x^2 \) or \( 2 + 4x + \frac{8}{2}x^2 \) earns both P4 and P5.
- P4 and P5 can be earned with answers consistent with imported values from part A.
- Terms of degree > 2 or "+ ..." do not earn P5.

### Part C (2 points)
- Point 6 (P6): Form of error bound
- Point 7 (P7): Correct analysis and explicit connection to 0.008

Scoring notes:
- P6 is earned for presenting \( \frac{48}{3!}(0.1)^3 \) or \( \frac{\max |f'''(x)|}{3!}(0.1)^3 \).
- To earn P7, a response must have earned P6 and conclude with \( \text{Error} \le 0.008 \) or equivalent.

### Part D (2 points)
- Point 8 (P8): First step of Euler's method \( f(0.2) \approx 2 + (0.2)(4) = 2.8 \)
- Point 9 (P9): Final approximation \( 3.7408 \) (or unsimplified equivalent \( 2.8 + 0.2(0.6)(2.8)^2 \)) with supporting work

### Part A
Using the Ratio Test:
\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(x-2)^{n+1}}{(n+1)4^{n+1}} \cdot \frac{n \cdot 4^n}{(x-2)^n} \right| = \lim_{n \to \infty} \left( \frac{|x-2|}{4} \cdot \frac{n}{n+1} \right) = \frac{|x-2|}{4} \]

The series converges absolutely when:
\[ \frac{|x-2|}{4} < 1 \implies |x-2| < 4 \implies -4 < x - 2 < 4 \implies -2 < x < 6 \]

Now, test the endpoints:
- When \( x = -2 \):
\[ \sum_{n=1}^{\infty} \frac{(-2-2)^n}{n \cdot 4^n} = \sum_{n=1}^{\infty} \frac{(-4)^n}{n \cdot 4^n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \]
This is an alternating series where \( u_n = \frac{1}{n} > 0 \), \( u_{n+1} < u_n \), and \( \lim_{n \to \infty} \frac{1}{n} = 0 \). Therefore, the series converges by the Alternating Series Test.

- When \( x = 6 \):
\[ \sum_{n=1}^{\infty} \frac{(6-2)^n}{n \cdot 4^n} = \sum_{n=1}^{\infty} \frac{4^n}{n \cdot 4^n} = \sum_{n=1}^{\infty} \frac{1}{n} \]
This is the harmonic series (or a \( p \)-series with \( p = 1 \)), which diverges.

Therefore, the interval of convergence is \( -2 \le x < 6 \) (or \( [-2, 6) \)).

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### Part B
Differentiating the series for \( f(x) \) term-by-term with respect to \( x \):
\[ f'(x) = \frac{d}{dx}\left( \frac{x-2}{4} + \frac{(x-2)^2}{2 \cdot 4^2} + \frac{(x-2)^3}{3 \cdot 4^3} + \cdots + \frac{(x-2)^n}{n \cdot 4^n} + \cdots \right) \]
\[ f'(x) = \frac{1}{4} + \frac{2(x-2)}{2 \cdot 4^2} + \frac{3(x-2)^2}{3 \cdot 4^3} + \cdots + \frac{n(x-2)^{n-1}}{n \cdot 4^n} + \cdots \]
\[ f'(x) = \frac{1}{4} + \frac{x-2}{16} + \frac{(x-2)^2}{64} + \cdots + \frac{(x-2)^{n-1}}{4^n} + \cdots \]

The first three nonzero terms are \( \frac{1}{4} + \frac{x-2}{16} + \frac{(x-2)^2}{64} \).
The general term is \( \frac{(x-2)^{n-1}}{4^n} \) for \( n \ge 1 \) (or \( \frac{(x-2)^k}{4^{k+1}} \) for \( k \ge 0 \)).

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### Part C
The Taylor series for \( f' \) is a geometric series with first term \( a = \frac{1}{4} \) and common ratio \( r = \frac{x-2}{4} \).
For all \( x \) where \( |r| < 1 \), the sum of the series is:
\[ f'(x) = \frac{a}{1 - r} = \frac{\frac{1}{4}}{1 - \frac{x-2}{4}} = \frac{\frac{1}{4}}{\frac{4 - (x-2)}{4}} = \frac{1}{6 - x} \]

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### Part D
At \( x = -2 \), the common ratio of the geometric series is:
\[ r = \frac{-2-2}{4} = -1 \]
A geometric series converges if and only if \( |r| < 1 \). Because \( |r| = |-1| = 1 \ge 1 \), the series diverges.
(Alternatively: At \( x = -2 \), the terms of the series are \( \frac{1}{4} - \frac{1}{4} + \frac{1}{4} - \frac{1}{4} + \cdots \). Since \( \lim_{n \to \infty} a_n \ne 0 \), the series diverges by the \( n \)th-term test for divergence.)

Therefore, the Taylor series for \( f' \) does not converge at \( x = -2 \).

### Part A: 5 points
- Point 1 (P1): Sets up ratio \( \left| \frac{(x-2)^{n+1}}{(n+1)4^{n+1}} \cdot \frac{n \cdot 4^n}{(x-2)^n} \right| \).
- Point 2 (P2): Computes limit of ratio \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \frac{|x-2|}{4} \).
- Point 3 (P3): Identifies interior of interval of convergence \( (-2, 6) \) or \( -2 < x < 6 \).
- Point 4 (P4): Considers both endpoints \( x = -2 \) and \( x = 6 \).
- Point 5 (P5): Endpoint analysis and final interval of convergence \( [-2, 6) \) or \( -2 \le x < 6 \).

### Part B: 2 points
- Point 6 (P6): First three nonzero terms: \( \frac{1}{4} + \frac{x-2}{16} + \frac{(x-2)^2}{64} \) (or equivalent unsimplified form).
- Point 7 (P7): General term: \( \frac{(x-2)^{n-1}}{4^n} \) for \( n \ge 1 \) or \( \frac{(x-2)^k}{4^{k+1}} \) for \( k \ge 0 \).

### Part C: 1 point
- Point 8 (P8): Verification: Shows that the sum of the geometric series \( \frac{\frac{1}{4}}{1 - \frac{x-2}{4}} \) simplifies to \( \frac{1}{6-x} \).

### Part D: 1 point
- Point 9 (P9): Answer with reason: Concludes "no" because \( |r| = 1 \ge 1 \) for a geometric series, or by using the \( n \)th-term test for divergence.