AP Calculus BC 2016 FRQ — Complete Study Guide

54

Total FRQ Points

6

Questions

9

Pts Each

30%

of Total Score

Question 1 9 pts

Water Tank — Rates & Riemann

🧮 Calculator Required · Estimation · Integration

Question 2 9 pts

Parametric Motion in xy-Plane

🧮 Calculator Required · Speed · Distance

Question 3 9 pts

FTC — Accumulation Function g(x)

📝 No Calculator · Extrema · Concavity

Question 4 9 pts

Differential Equation

📝 No Calculator · Euler's Method · L'Hôpital

Question 5 9 pts

Funnel — Volume & Related Rates

📝 No Calculator · Disk Method · Chain Rule

Question 6 9 pts

Taylor Series

📝 No Calculator · Convergence · Alternating

⚠️ General Exam Strategy

UNITS — Always write units in rates problems. Forgetting = lost point.
JUSTIFY — "Because f' changes sign from − to +" earns the justification point.
SHOW WORK — Correct answer with no work = 0 on setup points.
3 DECIMAL — Round to 3 decimal places unless told otherwise.
IVT / MVT — Know when to invoke these theorems by name.

1

Water Tank — Rates & Riemann Sums

🧮 Calculator Required Riemann Sum IVT 9 pts

Setup: Water pumped IN at $W(t) = 2000e^{-t^2/20}$ L/hr. Water removed at rate $R(t)$ (table given, differentiable & decreasing). At $t=0$, tank has 50,000 L. Time $0 \le t \le 8$ hours.

Table: $R(0)=1340,\ R(1)=1190,\ R(3)=950,\ R(6)=740,\ R(8)=700$ (L/hr)

a

Estimate R′(2). Show work. Indicate units.

▼

Key Concept — Derivative Estimation

$R'(2) \approx \dfrac{R(3) - R(1)}{3 - 1}$ (symmetric difference quotient — use the two table values surrounding $t = 2$)

⚠️ Trap

Don't use $\frac{R(3)-R(0)}{3-0}$. Use the interval symmetric around $t=2$, which is $[1,3]$.

Identify the interval surrounding $t=2$: use $t=1$ and $t=3$ from the table.
Apply the difference quotient: $R'(2) \approx \dfrac{R(3) - R(1)}{3 - 1} = \dfrac{950 - 1190}{3 - 1} = \dfrac{-240}{2} = -120$
Attach units: liters per hour per hour = liters/hour²

Model Answer — Part (a)

$R'(2) \approx \dfrac{R(3)-R(1)}{3-1} = \dfrac{950-1190}{2} = $ $-120$ liters/hour²

b

Left Riemann Sum for total water removed. Overestimate or underestimate?

▼

Key Concept — Left Riemann Sum

$\displaystyle\int_0^8 R(t)\,dt \approx \sum_{i} R(t_{\text{left}}) \cdot \Delta t$

Subintervals from table: $[0,1],\ [1,3],\ [3,6],\ [6,8]$

⚠️ Trap — Unequal Widths!

The subintervals are NOT equal width: $\Delta t = 1, 2, 3, 2$. Multiply each left-endpoint value by its OWN width.

$[0,1]$: width $= 1$, left value $= R(0) = 1340 \Rightarrow 1340 \times 1 = 1340$
$[1,3]$: width $= 2$, left value $= R(1) = 1190 \Rightarrow 1190 \times 2 = 2380$
$[3,6]$: width $= 3$, left value $= R(3) = 950 \Rightarrow 950 \times 3 = 2850$
$[6,8]$: width $= 2$, left value $= R(6) = 740 \Rightarrow 740 \times 2 = 1480$
Sum: $1340 + 2380 + 2850 + 1480 = \mathbf{8050}$ liters
Since $R(t)$ is decreasing, the left endpoints give the largest value on each interval → OVERESTIMATE

Model Answer — Part (b)

$\displaystyle\int_0^8 R(t)\,dt \approx (1)(1340)+(2)(1190)+(3)(950)+(2)(740)$
$= 1340+2380+2850+1480 = $ $8050$ liters

This is an overestimate because $R$ is decreasing, so the left Riemann sum uses values greater than or equal to the function on each subinterval.

c

Estimate total water in tank at end of 8 hours. (Use calculator for W integral)

▼

Key Concept

Total water $= 50000 + \displaystyle\int_0^8 W(t)\,dt - \displaystyle\int_0^8 R(t)\,dt$

Use part (b) for the $R$ integral. Use the calculator for the $W$ integral.

TI-Nspire CX II — Evaluate $\int_0^8 2000e^{-t^2/20}\,dt$

▸ Press menu → 4:Calculus → 3:Integral

▸ Or type directly: ∫(2000·e^(−t²/20), t, 0, 8)

Result ≈ 8264.462 liters

// Keyboard shortcut: press the ∫ key on the touchpad area, or menu→Calculus→Integral

$\int_0^8 W(t)\,dt \approx 8264$ liters (from calculator)
$\int_0^8 R(t)\,dt \approx 8050$ liters (from part b)
Total $= 50000 + 8264 - 8050 = \mathbf{50{,}214}$ liters

Model Answer — Part (c)

$50000 + \displaystyle\int_0^8 2000e^{-t^2/20}\,dt - 8050$
$\approx 50000 + 8264 - 8050 = $ $50{,}214$ liters

d

Is there a time t when pumping rate = removal rate? Explain.

▼

Key Concept — Intermediate Value Theorem (IVT)

Define $h(t) = W(t) - R(t)$. If $h(t)$ changes sign on $[0,8]$, then by IVT (since both $W$ and $R$ are continuous) there exists a $t$ where $h(t) = 0$, i.e., $W(t) = R(t)$.

TI-Nspire CX II — Evaluate W(0) and W(8)

2000·e^(0²/20) → W(0) = 2000

2000·e^(−64/20) → W(8) ≈ 546.95

$W(0) = 2000 < R(0) = 1340$? No — $W(0) = 2000 > R(0) = 1340$, so $h(0) = 2000 - 1340 = 660 > 0$
$W(8) \approx 546.95 < R(8) = 700$, so $h(8) \approx 546.95 - 700 < 0$
$h(0) > 0$ and $h(8) < 0$, and $h$ is continuous → by IVT, $\exists\, t \in (0,8)$ with $h(t) = 0$

Model Answer — Part (d)

Yes. $W(0) = 2000 > 1340 = R(0)$, so $W(0) - R(0) > 0$.
$W(8) \approx 546.95 < 700 = R(8)$, so $W(8) - R(8) < 0$.
Since $W$ and $R$ are both continuous on $[0,8]$, by the Intermediate Value Theorem, there exists a time $t \in (0,8)$ at which $W(t) = R(t)$.

2

Parametric Motion in the xy-Plane

🧮 Calculator Required Parametric Speed 9 pts

Setup: $\dfrac{dx}{dt} = t^2 + \sin(3t^2)$. Graph of $y(t)$ consists of 3 line segments. At $t=0$: position $(5,1)$. From the graph: the $y$ graph goes $(0,1)\to(1,0)\to(2,-1)\to(4,1)$ reading the line segments.

a

Find position of particle at t = 3

▼

Key Concept

$x(3) = x(0) + \displaystyle\int_0^3 \frac{dx}{dt}\,dt = 5 + \int_0^3 [t^2 + \sin(3t^2)]\,dt$

$y(3) = y(0) + \displaystyle\int_0^3 \frac{dy}{dt}\,dt = 1 + \int_0^3 \frac{dy}{dt}\,dt$

For $y$: read $\frac{dy}{dt}$ from the piecewise graph (slope of each segment).

⚠️ Trap — Read Slopes from Graph Carefully

From the graph: segment $[0,1]$: slope $= \frac{0-1}{1} = -1$; segment $[1,2]$: slope $= \frac{-1-0}{1} = -1$; segment $[2,4]$: slope $= \frac{1-(-1)}{2} = 1$. So $y'(t)$ (= $dy/dt$) is the slope of the y-graph.

TI-Nspire CX II — Compute x(3)

∫(t²+sin(3t²), t, 0, 3)

≈ 9.473

// x(3) = 5 + 9.473 ≈ 14.473

$x(3) = 5 + \int_0^3 [t^2+\sin(3t^2)]\,dt \approx 5 + 9.473 = 14.473$
$y$ via geometric areas: from graph, $\int_0^3 y'(t)\,dt$: segment areas = $(-1)(1) + (-1)(1) + (1)(1) = -1$ (using area from 0 to 3)
$y(3) = 1 + (-1) = 0$

Model Answer — Part (a)

$x(3) = 5 + \displaystyle\int_0^3[t^2+\sin(3t^2)]\,dt \approx 5 + 9.473 = $ $14.473$
$y(3) = 1 + \displaystyle\int_0^3 y'(t)\,dt = 1 + (-1-1+1) = $ $0$

Position at $t=3$: $(14.473,\ 0)$

b

Find slope of tangent line at t = 3

▼

Key Concept — Parametric Slope

$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$, evaluated at $t = 3$

$dy/dt$ at $t=3$: from the graph, on segment $[2,4]$, slope $= 1$, so $y'(3) = 1$
$dx/dt\big|_{t=3} = 3^2 + \sin(3\cdot 9) = 9 + \sin(27) \approx 9 + 0.957 = 9.957$
$\text{slope} = \dfrac{1}{9.957} \approx 0.100$

TI-Nspire CX II — Evaluate dx/dt at t=3

9 + sin(27) // make sure mode: RADIANS

≈ 9.957

Model Answer — Part (b)

$\dfrac{dy}{dx}\bigg|_{t=3} = \dfrac{dy/dt}{dx/dt} = \dfrac{1}{9+\sin(27)} \approx $ $0.100$

c

Find the speed of the particle at t = 3

▼

Key Concept — Speed Formula

$\text{speed} = \sqrt{\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2}$

$dx/dt\big|_{t=3} \approx 9.957$, $\quad dy/dt\big|_{t=3} = 1$
Speed $= \sqrt{9.957^2 + 1^2} = \sqrt{99.14 + 1} \approx \sqrt{100.14} \approx 10.007$

Model Answer — Part (c)

Speed $= \sqrt{(9+\sin27)^2 + 1^2} \approx $ $10.007$

d

Total distance traveled from t = 0 to t = 2

▼

Key Concept — Arc Length / Total Distance

$\displaystyle\int_0^2 \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt$

This is total distance, not displacement. Always under the radical.

⚠️ Trap — Distance ≠ Displacement

Do NOT compute $|\Delta x| + |\Delta y|$ separately. Use the arc length integral.

TI-Nspire CX II — Arc Length Integral

// dy/dt from graph: on [0,1] it's −1; on [1,2] it's −1

// But dy/dt is piecewise — split the integral or note it's constant on each interval

∫(√((t²+sin(3t²))²+1), t, 0, 2)

≈ 4.921

// dy/dt = −1 on both [0,1] and [1,2], so (dy/dt)² = 1 throughout [0,2] ✓

Model Answer — Part (d)

$\displaystyle\int_0^2 \sqrt{[t^2+\sin(3t^2)]^2 + [-1]^2}\,dt \approx $ $4.921$

3

FTC — Accumulation Function g(x)

📝 No Calculator FTC Extrema / Concavity 9 pts

Setup: $g(x) = \displaystyle\int_2^x f(t)\,dt$ for $-4 \le x \le 12$. The graph of $f$ is piecewise-linear with key points: $(-4,-4),(-2,0),(4,4),(6,0),(8,-4),(10,0),(12,-4)$.
By FTC: $g'(x) = f(x)$ and $g''(x) = f'(x)$.

a

Relative min, max, or neither at x = 10?

▼

Key Concept — First Derivative Test

$g'(x) = f(x)$. Check sign change of $f$ at $x = 10$.

From the graph: $f(x) < 0$ for $x \in (8,10)$ (below x-axis), $f(x) < 0$ for $x \in (10,12)$ (also below x-axis)
$f(10) = 0$, but $f$ does NOT change sign at $x=10$ (stays negative on both sides)
Therefore $g'$ does not change sign → Neither

Model Answer — Part (a)

$g'(x) = f(x)$. Since $f(x) < 0$ for $8 < x < 10$ and $f(x) < 0$ for $10 < x < 12$, $g'$ does not change sign at $x=10$. Therefore $g$ has neither a relative minimum nor a relative maximum at $x=10$.

b

Does g have a point of inflection at x = 4?

▼

Key Concept — Inflection Point

$g''(x) = f'(x)$. Inflection of $g$ ↔ sign change in $f'$ ↔ $f$ changes from increasing to decreasing (or vice versa) at $x=4$.

From graph: $f$ is increasing on $(-2,4)$ → $f'> 0$
$f$ is decreasing on $(4,8)$ → $f'< 0$
$f'$ changes sign from $+$ to $-$ at $x=4$ → $g''$ changes sign → YES, inflection point

Model Answer — Part (b)

$g''(x) = f'(x)$. Since $f$ is increasing for $x < 4$ and decreasing for $x > 4$ near $x=4$, $f'$ changes sign from positive to negative at $x=4$. Therefore $g''$ changes sign, and g has a point of inflection at $x=4$.

c

Absolute min and max of g on [−4, 12]

▼

Key Concept — Closed Interval Method

Check all critical points (where $g'=f=0$) AND endpoints. Compute $g$ values using geometric areas.

⚠️ Trap — g(2) = 0 (lower limit of integration)

Since $g(x) = \int_2^x f(t)\,dt$, we have $g(2) = 0$. Build all other values from there using triangles/trapezoids.

Critical points where $f = 0$: $x = -2, 2, 6, 10$. Endpoints: $x = -4, 12$.
$g(2) = 0$ (definition)
$g(6) = g(2) + \int_2^6 f\,dt$: triangle with base 2, height 4 → area $= 8$. So $g(6) = 0 + 8 = 8$.
$g(-2) = g(2) + \int_2^{-2} f\,dt = 0 - \int_{-2}^2 f\,dt$. On $[-2,2]$: triangle area $= \frac{1}{2}(4)(4) = 8$... but wait: from $-2$ to $4$ the peak is 4 at $x=4$; from $-2$ to $2$, slope = 1, so the area under $f$ from $-2$ to $2$ = triangle with base 4 height 4... Actually: $f(-2)=0, f(4)=4$, linear; at $x=2$, $f(2)=2$. So $\int_{-2}^2 f\,dt = \frac{1}{2}(4)(2) = 4$ (trapezoid: avg height $\times$ base). $g(-2) = -4$.
$g(-4)$: from $-4$ to $-2$, $f$ goes from $-4$ to $0$: triangle area $= -4$. $g(-4)= g(-2) + \int_{-2}^{-4}f\,dt = -4 + 4 = ...$. Actually $\int_{-4}^{-2}f = -4$ (triangle below axis). $g(-4) = g(-2) - \int_{-2}^{-4}f = -4-(-4) = -8$.
$g(8) = g(6) + \int_6^8 f\,dt = 8 + (-4) = 4$ (triangle base 2, height $-4$: area $= -4$).
$g(10) = g(8) + \int_8^{10} f\,dt = 4 + (-4) = 0$.
$g(12) = g(10) + \int_{10}^{12} f\,dt = 0 + (-4) = -4$.
Summary: $g(-4)=-8,\ g(-2)=-4,\ g(2)=0,\ g(6)=8,\ g(8)=4,\ g(10)=0,\ g(12)=-4$

Model Answer — Part (c)

Absolute maximum value of $g$ is $8$, occurring at $x = 6$.
Absolute minimum value of $g$ is $-8$, occurring at $x = -4$.
(Justify by comparing all critical point and endpoint values.)

d

Find all intervals on [−4, 12] where g(x) ≤ 0

▼

Key Concept

Use the values computed in part (c). $g(x) = 0$ at $x = -4$... wait: $g(-4)=-8 \ne 0$. $g=0$ at $x=2$ and $x=10$.

$g \le 0$ wherever $g$ values are $\le 0$: from the list, that's $[-4, 2]$ and $[10, 12]$.

Model Answer — Part (d)

$g(x) \le 0$ on $-4 \le x \le 2$ and $10 \le x \le 12$.

4

Differential Equation

📝 No Calculator 2nd Derivative L'Hôpital Euler's Method 9 pts

Differential Equation: $\dfrac{dy}{dx} = \dfrac{2x-1}{2y}$

a

Find d²y/dx² in terms of x and y

▼

Key Concept — Implicit Differentiation of dy/dx

Differentiate $\frac{dy}{dx} = \frac{2x-1}{2y}$ with respect to $x$ using the quotient rule. Replace $\frac{dy}{dx}$ with the original expression.

$\dfrac{d^2y}{dx^2} = \dfrac{d}{dx}\left[\dfrac{2x-1}{2y}\right]$
Quotient rule: $= \dfrac{(2)(2y) - (2x-1)(2\,dy/dx)}{(2y)^2}$
Substitute $dy/dx = \dfrac{2x-1}{2y}$:
$= \dfrac{4y - (2x-1)\cdot\frac{2(2x-1)}{2y}}{4y^2} = \dfrac{4y - \frac{(2x-1)^2}{y}}{4y^2}$
Multiply numerator and denominator by $y$: $= \dfrac{4y^2 - (2x-1)^2}{4y^3}$

Model Answer — Part (a)

$\dfrac{d^2y}{dx^2} = \dfrac{4y^2-(2x-1)^2}{4y^3}$

b

Relative min, max, or neither at (2, 8)? Justify.

▼

Key Concept — Second Derivative Test

At $(2,8)$: check $dy/dx = 0$ first (critical point), then check sign of $d^2y/dx^2$.

$\dfrac{dy}{dx}\bigg|_{(2,8)} = \dfrac{2(2)-1}{2(8)} = \dfrac{3}{16} \ne 0$
Since $dy/dx \ne 0$ at $(2,8)$, the point is not a critical point
Therefore $f$ has neither a relative min nor max at $(2,8)$

Model Answer — Part (b)

$\dfrac{dy}{dx}\bigg|_{(2,8)} = \dfrac{3}{16} \ne 0$. Since $f'(2) \ne 0$, the graph of $f$ has neither a relative minimum nor a relative maximum at $(2,8)$.

c

Find lim_{x→1} [g(x)−2] / (3x−3) given g(1) = 2

▼

Key Concept — L'Hôpital's Rule

As $x \to 1$: numerator $g(1)-2 = 2-2 = 0$; denominator $3(1)-3 = 0$. This is $\frac{0}{0}$ form → apply L'Hôpital.

$\lim_{x\to1}\dfrac{g(x)-2}{3x-3} \xrightarrow{\text{L'H}} \lim_{x\to1}\dfrac{g'(x)}{3}$
$g'(x) = \dfrac{dy}{dx} = \dfrac{2x-1}{2y} = \dfrac{2x-1}{2g(x)}$
At $x=1$, $g(1)=2$: $g'(1) = \dfrac{2(1)-1}{2(2)} = \dfrac{1}{4}$
Limit $= \dfrac{1/4}{3} = \dfrac{1}{12}$

Model Answer — Part (c)

$\displaystyle\lim_{x\to 1}\frac{g(x)-2}{3x-3} \overset{L'H}{=} \lim_{x\to 1}\frac{g'(x)}{3} = \frac{g'(1)}{3} = \frac{(2(1)-1)/(2g(1))}{3} = \frac{1/4}{3} = $ $\dfrac{1}{12}$

d

Euler's Method: h(0) = 2, two steps of size 0.5 → approximate h(1)

▼

Key Concept — Euler's Method

$y_{n+1} = y_n + f(x_n, y_n) \cdot \Delta x$ where $f(x,y) = \dfrac{2x-1}{2y}$ and $\Delta x = 0.5$

⚠️ Trap — Two Steps, Each of Size 0.5

"Two steps of equal size" from $x=0$ to $x=1$: each step has $\Delta x = 0.5$. You go $0 \to 0.5 \to 1$.

Step 1: $x_0=0,\ y_0=2$. Slope $= \dfrac{2(0)-1}{2(2)} = \dfrac{-1}{4}$. $\quad y_1 = 2 + (-\tfrac{1}{4})(0.5) = 2 - \tfrac{1}{8} = \dfrac{15}{8}$

Step 2: $x_1=0.5,\ y_1=\tfrac{15}{8}$. Slope $= \dfrac{2(0.5)-1}{2(15/8)} = \dfrac{0}{15/4} = 0$. $\quad y_2 = \dfrac{15}{8} + 0 \cdot (0.5) = \dfrac{15}{8}$

Model Answer — Part (d)

Step 1 ($x=0\to 0.5$): slope $= \dfrac{-1}{4}$; $\quad y_1 = 2 + (-\tfrac{1}{4})(0.5) = \dfrac{15}{8}$
Step 2 ($x=0.5\to 1$): slope $= \dfrac{0}{15/4} = 0$; $\quad y_2 = \dfrac{15}{8} + 0 = \dfrac{15}{8}$

$h(1) \approx$ $\dfrac{15}{8}$

5

Funnel — Volume & Related Rates

📝 No Calculator Disk Method Related Rates 9 pts

Setup: Funnel height 10 inches. At height $h$: $r = \dfrac{1}{20}(3+h^2)$, $\quad 0 \le h \le 10$.

a

Find the average value of the radius

▼

Key Concept — Average Value

$\bar{r} = \dfrac{1}{10-0}\displaystyle\int_0^{10} r(h)\,dh = \dfrac{1}{10}\int_0^{10}\frac{3+h^2}{20}\,dh$

$\bar{r} = \dfrac{1}{10} \cdot \dfrac{1}{20}\displaystyle\int_0^{10}(3+h^2)\,dh$
$= \dfrac{1}{200}\left[3h + \dfrac{h^3}{3}\right]_0^{10} = \dfrac{1}{200}\left(30 + \dfrac{1000}{3}\right)$
$= \dfrac{1}{200}\cdot\dfrac{90+1000}{3} = \dfrac{1090}{600} = \dfrac{109}{60}$

Model Answer — Part (a)

$\bar{r} = \dfrac{1}{10}\displaystyle\int_0^{10}\frac{3+h^2}{20}\,dh = \dfrac{1}{200}\left[3h+\frac{h^3}{3}\right]_0^{10} = \dfrac{1090}{600} = $ $\dfrac{109}{60}$ inches

b

Find the volume of the funnel

▼

Key Concept — Disk Method (rotating about vertical axis)

$V = \displaystyle\int_0^{10} \pi [r(h)]^2\,dh = \pi\int_0^{10}\left[\frac{3+h^2}{20}\right]^2 dh$

$V = \dfrac{\pi}{400}\displaystyle\int_0^{10}(3+h^2)^2\,dh$
Expand: $(3+h^2)^2 = 9 + 6h^2 + h^4$
$\displaystyle\int_0^{10}(9+6h^2+h^4)\,dh = \left[9h+2h^3+\frac{h^5}{5}\right]_0^{10} = 90 + 2000 + 20000 = 22090$
$V = \dfrac{\pi}{400}(22090) = \dfrac{22090\pi}{400} = \dfrac{2209\pi}{40}$ cubic inches

Model Answer — Part (b)

$V = \dfrac{\pi}{400}(22090) = $ $\dfrac{2209\pi}{40}$ cubic inches

c

At h = 3, dr/dt = −1/5 in/sec. Find dh/dt.

▼

Key Concept — Related Rates via Chain Rule

Differentiate $r = \dfrac{1}{20}(3+h^2)$ with respect to $t$:

$\dfrac{dr}{dt} = \dfrac{1}{20}(2h)\dfrac{dh}{dt} = \dfrac{h}{10}\dfrac{dh}{dt}$

At $h=3$: $\dfrac{dr}{dt} = -\dfrac{1}{5}$ in/sec
$-\dfrac{1}{5} = \dfrac{3}{10}\cdot\dfrac{dh}{dt}$
$\dfrac{dh}{dt} = -\dfrac{1}{5}\cdot\dfrac{10}{3} = -\dfrac{2}{3}$ in/sec

Model Answer — Part (c)

$\dfrac{dr}{dt} = \dfrac{h}{10}\dfrac{dh}{dt}$. At $h=3$: $-\dfrac{1}{5} = \dfrac{3}{10}\cdot\dfrac{dh}{dt}$
$\dfrac{dh}{dt} = $ $-\dfrac{2}{3}$ inches per second

6

Taylor Series

📝 No Calculator Power Series Alternating Series 9 pts

Setup: $f(1) = 1$, $f'(1) = -\dfrac{1}{2}$, and for $n \ge 2$: $f^{(n)}(1) = \dfrac{(-1)^n (n-1)!}{2^n}$.

a

Write first four nonzero terms and general term of Taylor series about x = 1

▼

Key Concept — Taylor Series Formula

$f(x) = \displaystyle\sum_{n=0}^{\infty} \dfrac{f^{(n)}(1)}{n!}(x-1)^n$

For $n \ge 2$: $\dfrac{f^{(n)}(1)}{n!} = \dfrac{(-1)^n(n-1)!}{n! \cdot 2^n} = \dfrac{(-1)^n}{n \cdot 2^n}$

$n=0$: $\dfrac{f(1)}{0!}(x-1)^0 = 1$
$n=1$: $\dfrac{f'(1)}{1!}(x-1) = -\dfrac{1}{2}(x-1)$
$n=2$: $\dfrac{(-1)^2 \cdot 1!}{2 \cdot 2^2}(x-1)^2 = \dfrac{1}{8}(x-1)^2 \quad \checkmark$ (uses $\frac{(n-1)!}{n!} = \frac{1}{n}$)
$n=3$: $\dfrac{(-1)^3}{3\cdot2^3}(x-1)^3 = -\dfrac{1}{24}(x-1)^3$
General term ($n \ge 1$): $\dfrac{(-1)^n}{n \cdot 2^n}(x-1)^n$

Model Answer — Part (a)

$f(x) = 1 - \dfrac{1}{2}(x-1) + \dfrac{1}{8}(x-1)^2 - \dfrac{1}{24}(x-1)^3 + \cdots + \dfrac{(-1)^n}{n\cdot 2^n}(x-1)^n + \cdots$

b

Radius of convergence = 2. Find interval of convergence.

▼

Key Concept — Interval of Convergence

Center $= 1$, radius $= 2$: the series converges for $|x-1| < 2$, i.e., $-1 < x < 3$. Must check endpoints separately.

⚠️ Must Check Both Endpoints!

Plug $x = -1$ and $x = 3$ into the general term and test for convergence.

At $x = -1$: $(x-1) = -2$, term $= \dfrac{(-1)^n}{n\cdot2^n}\cdot(-2)^n = \dfrac{(-1)^n \cdot (-1)^n \cdot 2^n}{n\cdot 2^n} = \dfrac{1}{n}$. This is the harmonic series → diverges.
At $x = 3$: $(x-1) = 2$, term $= \dfrac{(-1)^n}{n\cdot2^n}\cdot 2^n = \dfrac{(-1)^n}{n}$. This is the alternating harmonic series → converges (by Alternating Series Test).
Interval: $(-1, 3]$

Model Answer — Part (b)

At $x=-1$: $\displaystyle\sum \frac{1}{n}$ diverges (harmonic series).
At $x=3$: $\displaystyle\sum \frac{(-1)^n}{n}$ converges (Alternating Series Test).
Interval of convergence: $-1 < x \le 3$

c

Approximate f(1.2) using first three nonzero terms

▼

Key Concept

Substitute $x = 1.2$ → $(x-1) = 0.2 = \dfrac{1}{5}$ into the first three terms.

Term 1: $1$
Term 2: $-\dfrac{1}{2}\cdot\dfrac{1}{5} = -\dfrac{1}{10}$
Term 3: $\dfrac{1}{8}\cdot\left(\dfrac{1}{5}\right)^2 = \dfrac{1}{8}\cdot\dfrac{1}{25} = \dfrac{1}{200}$
Sum $= 1 - \dfrac{1}{10} + \dfrac{1}{200} = \dfrac{200-20+1}{200} = \dfrac{181}{200} = 0.905$

Model Answer — Part (c)

$f(1.2) \approx 1 - \dfrac{1}{10} + \dfrac{1}{200} = \dfrac{181}{200} = $ $0.905$

d

Show approximation in (c) is within 0.001 of exact value

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Key Concept — Alternating Series Error Bound

For an alternating series satisfying the Alternating Series Test, the error $|$exact $-$ partial sum$|$ is less than the absolute value of the first omitted term.

The series for $f(1.2)$ at $x=1.2$ is alternating (terms alternate in sign, decrease to 0).
The first omitted term (4th term, $n=3$): $\left|-\dfrac{1}{24}\left(\dfrac{1}{5}\right)^3\right| = \dfrac{1}{24}\cdot\dfrac{1}{125} = \dfrac{1}{3000} < \dfrac{1}{1000} = 0.001$
By the Alternating Series Error Bound, the error $< 0.001$. ∎

Model Answer — Part (d)

The series for $f(1.2)$ is alternating with terms decreasing in absolute value to 0. By the Alternating Series Error Bound, the error is at most the absolute value of the first omitted term:
$\left|\dfrac{(-1)^3}{3\cdot2^3}\left(\dfrac{1}{5}\right)^3\right| = \dfrac{1}{24}\cdot\dfrac{1}{125} = \dfrac{1}{3000} < 0.001$ ✓

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