1
Given: C(t) = 7.6 · arctan(0.2t) (acres affected, t = weeks)
Given derivative: C′(t) = 38 / (25 + t²)
Make sure your calculator is in RADIAN MODE.
Given derivative: C′(t) = 38 / (25 + t²)
Make sure your calculator is in RADIAN MODE.
Part A
Find the average number of acres affected from t = 0 to t = 4.
▼
Core Concept
Average Value of a Function — not average rate of change!
Average Value = (1/(b−a)) · ∫[a to b] f(t) dt
⚠ Common Trap
"Average number of acres" = average VALUE of C(t), not slope. Use the average value formula, NOT [C(4)−C(0)] / 4.
🖩 TI-Nspire CX II Steps
1
Press Menu → 4: Calculus → 3: Integral
2
Enter: ∫(7.6·arctan(0.2·t), t, 0, 4)
3
Multiply result by 1/4 (since b−a = 4−0 = 4)
4
Or do it all at once: (1/4)·∫(7.6·arctan(0.2·t), t, 0, 4)
Model Answer Setup + Final Answer
(1/4) · ∫₀⁴ 7.6·arctan(0.2t) dt
≈ (1/4) · (5.765...) ≈ 1.441 acres
Write the integral setup clearly before calculator use. Answer ≈ 1.441 acres.
Part B
Find t when instantaneous rate of change = average rate of change over [0, 4].
▼
Core Concept
Mean Value Theorem (MVT): There exists t in (0,4) such that C′(t) = [C(4)−C(0)] / (4−0).
Set C′(t) = [C(4) − C(0)] / 4, then solve for t
1
Compute average rate of change: [C(4) − C(0)] / 4
C(4) = 7.6·arctan(0.8) ≈ 7.6·(0.6747) ≈ 5.128
C(0) = 7.6·arctan(0) = 0
→ Avg ROC ≈ 5.128/4 ≈ 1.282
C(4) = 7.6·arctan(0.8) ≈ 7.6·(0.6747) ≈ 5.128
C(0) = 7.6·arctan(0) = 0
→ Avg ROC ≈ 5.128/4 ≈ 1.282
2
Set C′(t) equal to that:
38/(25 + t²) = 1.282...
38/(25 + t²) = 1.282...
3
Solve using calculator
🖩 TI-Nspire CX II Steps
1
Store avg ROC: compute (7.6·arctan(0.8)−0)/4 → store as k
2
Menu → 3: Algebra → 1: Solve
Type: solve(38/(25+t²) = k, t)
Type: solve(38/(25+t²) = k, t)
3
Take the positive solution in (0, 4)
Model Answer Setup + Final Answer
38/(25 + t²) = [C(4) − C(0)] / 4
t ≈ 2.195 weeks
Part C
Write a limit expression for the end behavior of C′(t). Evaluate it.
▼
Core Concept
End behavior of C′(t) as t → ∞ means: what happens to the RATE as time goes to infinity?
lim[t → ∞] C′(t) = lim[t → ∞] 38/(25 + t²)
⚠ Common Trap
The question asks for end behavior of C′(t), not C(t). Write the derivative in the limit, not the original function.
1
As t → ∞, the denominator 25 + t² → ∞
2
Numerator stays at 38 (constant)
3
Therefore the fraction → 0
Model Answer
lim[t → ∞] 38/(25 + t²) = 0
The rate of spread approaches 0 — the species stops spreading.
Part D
A(t) = C(4) − ∫₄ᵗ 0.1·ln(x) dx. At what t in [4.3, 6] does A attain its maximum?
▼
Core Concept
FTC Part 1: A′(t) = −0.1·ln(t) (derivative of integral with variable upper limit).
A′(t) = 0 − 0.1·ln(t) = −0.1·ln(t)
Find where A′(t) = 0, then confirm it's a maximum using sign analysis.
⭐ Key Insight
A′(t) = −0.1·ln(t). For t > 1: ln(t) > 0 → A′(t) < 0. So A is always decreasing on [4.3, 6]. Maximum must be at the left endpoint t = 4.3.
⚠ Common Trap
Don't forget to justify your answer! Just stating "t = 4.3" without reasoning gets no justification credit. You must show A′(t) < 0 on the interval.
🖩 TI-Nspire CX II — Verify A′(t) sign
Compute −0.1·ln(4.3) and −0.1·ln(6) — both negative → confirms A is decreasing.
Model Answer with Justification
A attains its maximum at t = 4.3
Justification: A′(t) = −0.1·ln(t). For all t in [4.3, 6], ln(t) > 0, so A′(t) < 0. A is strictly decreasing on the entire interval; therefore the maximum occurs at the left endpoint t = 4.3.
2
Curve C: r² = sin(2θ), for 0 ≤ θ ≤ π → r = √(sin(2θ))
Also given: dx/dθ = −sin(θ)·(4sin²(θ)−3) / (2·sin(2θ)^(2/3)) for dθ/dt = 1/5
Note: This is a lemniscate-like petal curve. RADIAN MODE required.
Also given: dx/dθ = −sin(θ)·(4sin²(θ)−3) / (2·sin(2θ)^(2/3)) for dθ/dt = 1/5
Note: This is a lemniscate-like petal curve. RADIAN MODE required.
Part A
Find dr/dθ at the point on curve C where θ = 1.3.
▼
Core Concept
Differentiate r² = sin(2θ) implicitly, then solve for dr/dθ:
2r · (dr/dθ) = 2cos(2θ) → dr/dθ = cos(2θ) / r
Need r at θ = 1.3: r = √(sin(2·1.3))
🖩 TI-Nspire CX II Steps
1
Compute r: √(sin(2·1.3)) → store as r
2
Compute: cos(2·1.3) / r
3
Or use Menu → 4: Calculus → 1: Derivative
d/dθ [√(sin(2θ))] at θ = 1.3
d/dθ [√(sin(2θ))] at θ = 1.3
Model Answer Setup + Final Answer
dr/dθ = cos(2θ)/r = cos(2·1.3) / √(sin(2·1.3))
≈ cos(2.6) / √(sin(2.6)) ≈ (−0.8569) / (0.9218) ≈ −0.929
Part B
Find the area inside curve C but outside r = 1/2.
▼
Core Concept — Polar Area
Area = (1/2)∫[α to β] (r_outer² − r_inner²) dθ
Need to find intersection of r² = sin(2θ) and r = 1/2, i.e., where sin(2θ) = 1/4.
⭐ Strategy
The curve r² = sin(2θ) exists only when sin(2θ) ≥ 0 → 0 ≤ θ ≤ π/2 (first petal, in 0 ≤ θ ≤ π).Find θ₁, θ₂ where sin(2θ) = 1/4 in [0, π/2].
🖩 TI-Nspire CX II Steps
1
Find intersections: solve(sin(2θ) = 1/4, θ) on [0, π/2]
→ θ₁ ≈ 0.1268, θ₂ ≈ 1.4440
→ θ₁ ≈ 0.1268, θ₂ ≈ 1.4440
2
Area integral: (1/2)·∫(sin(2θ)−1/4, θ, θ₁, θ₂)
Plus symmetric second petal if applicable
Plus symmetric second petal if applicable
Model Answer Setup
(1/2)∫[θ₁ to θ₂] [sin(2θ) − (1/2)²] dθ × 2 (for both petals)
= ∫[θ₁ to θ₂] [sin(2θ) − 1/4] dθ ≈ 0.454
Write the full integral expression with limits before computing. Show the intersection equations.
Part C
For 0 ≤ θ ≤ π/2, find θ corresponding to the point farthest from the y-axis.
▼
Core Concept
Farthest from y-axis = maximize |x| = |r·cos(θ)|.x = r·cos(θ) = √(sin(2θ))·cos(θ)
Maximize x by setting dx/dθ = 0. The given formula for dx/dθ is:dx/dθ = [−sin(θ)(4sin²(θ)−3)] / [2·sin^(2/3)(2θ)]
1
Set numerator = 0 (denominator ≠ 0 in interior): −sin(θ)(4sin²(θ)−3) = 0
2
sin(θ) = 0 → θ = 0 (endpoint), or 4sin²(θ) = 3 → sin(θ) = √3/2 → θ = π/3
3
Check sign change of dx/dθ at θ = π/3 to confirm maximum of x.
⭐ Sign Analysis for Justification
For θ slightly less than π/3: 4sin²(θ) < 3 → factor (4sin²θ−3) < 0 → dx/dθ > 0 (increasing x)For θ slightly greater than π/3: 4sin²(θ) > 3 → factor > 0 → dx/dθ < 0 (decreasing x)
→ x has a local maximum at θ = π/3
Model Answer with Justification
θ = π/3
dx/dθ changes from positive to negative at θ = π/3, so x (distance from y-axis) is maximized there.
Part D
dθ/dt = 1/5. Find dr/dt when θ = 1.3. Show setup.
▼
Core Concept — Chain Rule (Related Rates)
dr/dt = (dr/dθ) · (dθ/dt)
dr/dθ was found in Part A ≈ −0.929; dθ/dt = 1/5.
⚠ Common Trap
Make sure to use the exact same setup from Part A for dr/dθ. Don't recompute from scratch without referencing Part A's method.
Model Answer Setup + Final Answer
dr/dt = (dr/dθ)|[θ=1.3] · (1/5)
= (cos(2·1.3)/√(sin(2·1.3))) · (1/5) ≈ (−0.929) · (0.2) ≈ −0.186 units/time
Negative means the particle is moving closer to the origin at θ = 1.3.
3
R(t) = reading rate (words/min), differentiable on [0, 10].
Teacher's model: W(t) = −(3/10)t² + 8t + 100
| t (min) | 0 | 2 | 8 | 10 |
|---|---|---|---|---|
| R(t) (wpm) | 90 | 100 | 150 | 162 |
Part A
Approximate R′(1) using average rate of change over [0, 2]. Include units.
▼
Core Concept
Approximate the derivative using a difference quotient (secant slope):
R′(1) ≈ [R(2) − R(0)] / (2 − 0)
⚠ Don't Forget Units!
R is in words/min, t is in min. So R′ has units: (words/min) / min = words/min²
Model Answer
R′(1) ≈ (100 − 90) / (2 − 0) = 10/2 = 5 words/min²
Part B
Must there be a value c in (0, 10) such that R(c) = 155? Justify.
▼
Core Concept — Intermediate Value Theorem (IVT)
IVT: If f is continuous on [a, b], and k is between f(a) and f(b), then ∃ c ∈ (a,b) with f(c) = k.
⭐ IVT Checklist — Must mention ALL THREE
① R is differentiable on [0,10], therefore continuous on [0,10].② R(8) = 150 < 155 < 162 = R(10) — so 155 is between R(8) and R(10).
③ By IVT, ∃ c ∈ (8, 10) ⊂ (0, 10) such that R(c) = 155. ✓
⚠ Common Mistake
Saying "R is continuous" without explaining WHY. Must say "R is differentiable → continuous." Also, use a specific subinterval like (8, 10) rather than just (0, 10).
Model Answer
Yes. Since R is differentiable on [0,10], R is continuous on [0,10].
R(8) = 150 < 155 < 162 = R(10). By the Intermediate Value Theorem, there must exist a value c in (8, 10) ⊂ (0, 10) such that R(c) = 155.
Part C
Use a trapezoidal sum with three subintervals to approximate ∫₀¹⁰ R(t) dt.
▼
Core Concept — Trapezoidal Sum
For each subinterval [a, b]: area = (b−a)/2 · [R(a) + R(b)]
T = Σ (Δt/2)·[R(tₖ) + R(tₖ₊₁)]
Subintervals from data: [0,2], [2,8], [8,10]
⚠ Common Mistake — Unequal Widths!
The widths are NOT equal: Δt₁ = 2, Δt₂ = 6, Δt₃ = 2. Each trapezoid has its OWN width. Don't use a single Δt for all.
1
Trap 1 [0,2]: (2/2)·[90 + 100] = 1·190 = 190
2
Trap 2 [2,8]: (6/2)·[100 + 150] = 3·250 = 750
3
Trap 3 [8,10]: (2/2)·[150 + 162] = 1·312 = 312
4
Total: 190 + 750 + 312 = 1252
Model Answer
≈ (2/2)(90+100) + (6/2)(100+150) + (2/2)(150+162) = 190 + 750 + 312 = 1252 words
Part D
How many total words does the teacher read? W(t) = −(3/10)t² + 8t + 100.
▼
Core Concept
Total words = ∫₀¹⁰ W(t) dt — integrate the rate over [0, 10].
1
∫ W(t) dt = ∫ (−3t²/10 + 8t + 100) dt = −t³/10 + 4t² + 100t + C
2
Evaluate from 0 to 10:
= [−(10)³/10 + 4(10)² + 100(10)] − [0]
= [−(10)³/10 + 4(10)² + 100(10)] − [0]
3
= [−100 + 400 + 1000] = 1300
Model Answer
∫₀¹⁰ (−3t²/10 + 8t + 100) dt = [−t³/10 + 4t² + 100t]₀¹⁰ = 1300 words
4
f is defined on [−6, 12], consists of two semicircles and one line segment.
g(x) = ∫₋₆ˣ f(t) dt
From the graph: f appears to have zeros at x = −6, 0, 6, 12; positive humps and negative portions.
Key graph readings needed: f(−6) = 0, semicircle of radius 3 on [−6,0] (area = 9π/2, above axis), line on [0,6] going negative, semicircle on [6,12] below axis.
g(x) = ∫₋₆ˣ f(t) dt
From the graph: f appears to have zeros at x = −6, 0, 6, 12; positive humps and negative portions.
Key graph readings needed: f(−6) = 0, semicircle of radius 3 on [−6,0] (area = 9π/2, above axis), line on [0,6] going negative, semicircle on [6,12] below axis.
Part A
Find g′(8). Give a reason.
▼
Core Concept — FTC Part 1
g′(x) = f(x) (by FTC)
So g′(8) = f(8). Read f(8) from the graph.
⭐ Reading the graph at x = 8
x = 8 is in the semicircle on [6, 12] that lies below the x-axis. The center is at (9, 0), radius 3. At x = 8: distance from center = |8−9| = 1, so f(8) = −√(3²−1²) = −√8 = −2√2.
Model Answer
g′(8) = f(8) = −2√2
Reason: By the Fundamental Theorem of Calculus, g′(x) = f(x) for all x in the domain.
Part B
Find all x in (−6, 12) where g has an inflection point. Give a reason.
▼
Core Concept
g′(x) = f(x), so g″(x) = f′(x).Inflection point of g ↔ g″ changes sign ↔ f′ changes sign ↔ f changes from increasing to decreasing (or vice versa).
Inflection of g where f changes monotonicity
⭐ Reading the graph for f′ sign changes
Look where f has a local max or min (= where f′ = 0 and changes sign):• At x = −3: f changes from increasing to decreasing → local max → f′ changes + to − → g″ changes + to − → inflection of g
• At x = 9: f changes from decreasing to increasing → local min → f′ changes − to + → g″ changes − to + → inflection of g
Model Answer
g has inflection points at x = −3 and x = 9
Reason: g″(x) = f′(x). At x = −3, f changes from increasing to decreasing, so g″ changes sign from positive to negative. At x = 9, f changes from decreasing to increasing, so g″ changes sign from negative to positive. g has inflection points at both locations.
Part C
Find g(12) and g(0). Label your answers.
▼
Core Concept — Geometric Areas
g(x) = ∫₋₆ˣ f(t) dt = net signed area from −6 to x under f.
1
g(0): Area of upper semicircle on [−6, 0], radius 3.
Area = (1/2)π(3)² = 9π/2. Since f > 0 here, g(0) = 9π/2.
Area = (1/2)π(3)² = 9π/2. Since f > 0 here, g(0) = 9π/2.
2
g(12): g(0) + area on [0,6] + area on [6,12].
[0,6]: line segment from (0,0) to (6,−6) (triangle below axis) → area = −(1/2)(6)(6) = −18
[6,12]: lower semicircle radius 3 → area = −(1/2)π(3)² = −9π/2
g(12) = 9π/2 + (−18) + (−9π/2) = −18
[0,6]: line segment from (0,0) to (6,−6) (triangle below axis) → area = −(1/2)(6)(6) = −18
[6,12]: lower semicircle radius 3 → area = −(1/2)π(3)² = −9π/2
g(12) = 9π/2 + (−18) + (−9π/2) = −18
⚠ Common Mistake
Areas below x-axis are NEGATIVE in accumulation functions. Don't use absolute area values.
Model Answer
g(0) = 9π/2 | g(12) = −18
Part D
Find the x where g attains its absolute minimum on [−6, 12]. Justify.
▼
Core Concept — Candidates Test
g′(x) = f(x). Critical points where f(x) = 0: x = 0 and x = 6 (where f crosses x-axis).Also check endpoints x = −6 and x = 12.
1
g(−6) = 0 (lower limit = upper limit)
2
g(0) = 9π/2 ≈ 14.14
3
g(6) = g(0) + (area of triangle [0,6]) = 9π/2 − 18 ≈ −3.86
4
g(12) = −18
⭐ Compare All Candidates
g(−6) = 0, g(0) ≈ 14.14, g(6) ≈ −3.86, g(12) = −18Minimum is g(12) = −18
Model Answer with Justification
g attains its absolute minimum at x = 12
Justification: The candidates for the absolute minimum are the critical points (where g′ = f = 0: x = 0, x = 6) and endpoints (x = −6, x = 12). Computing: g(−6) = 0, g(0) = 9π/2, g(6) = 9π/2 − 18, g(12) = −18. The smallest value is g(12) = −18.
5
DE: dy/dx = 3x(y−2)², with initial condition f(1) = −1.
Find the particular solution y = f(x).
Find the particular solution y = f(x).
Part A
Find f″(1), the second derivative at the point (1, −1).
▼
Core Concept
Differentiate dy/dx = 3x(y−2)² implicitly with respect to x using product rule + chain rule.
d²y/dx² = 3(y−2)² + 3x · 2(y−2) · (dy/dx)
1
At (1, −1): dy/dx = 3(1)(−1−2)² = 3(9) = 27
2
d²y/dx² = 3(y−2)² + 6x(y−2)(dy/dx)
= 3(−3)² + 6(1)(−3)(27)
= 27 + (−486) = −459
= 3(−3)² + 6(1)(−3)(27)
= 27 + (−486) = −459
⚠ Common Mistake
Forgetting the product rule: d/dx[3x·(y−2)²] = 3(y−2)² + 3x · d/dx[(y−2)²]. Both terms are required.
Model Answer
f″(1) = 3(−3)² + 6(1)(−3)(27) = 27 − 486 = −459
Part B
Write the second-degree Taylor polynomial for f about x = 1.
▼
Core Concept — Taylor Polynomial
P₂(x) = f(1) + f′(1)(x−1) + f″(1)/2! · (x−1)²
f(1) = −1, f′(1) = 27, f″(1) = −459
Model Answer
P₂(x) = −1 + 27(x−1) − (459/2)(x−1)²
= −1 + 27(x−1) − 229.5(x−1)²
Part C
Use Lagrange error bound to show the approximation of f(1.1) differs by at most 0.01.
▼
Core Concept — Lagrange Error Bound
|Error| ≤ M/3! · |x − a|³ = M/6 · |x−1|³
where M = max|f‴(x)| on [1, 1.1]. Given: |f‴(x)| ≤ 60. x − a = 1.1 − 1 = 0.1.
⭐ This is a SHOW problem — show EVERY step
Don't just write the answer; show the bound calculation and compare to 0.01.
1
Lagrange bound: |f(1.1) − P₂(1.1)| ≤ (60/6) · |1.1 − 1|³
2
= 10 · (0.1)³ = 10 · 0.001 = 0.01
3
Therefore the error ≤ 0.01. ✓
Model Answer
|f(1.1) − P₂(1.1)| ≤ (60/3!) · |1.1−1|³ = (60/6)(0.1)³ = 10 · 0.001 = 0.01
Since the Lagrange error bound gives ≤ 0.01, the approximation differs from f(1.1) by at most 0.01. ∎
Part D
Use Euler's method, 2 equal steps from x=1, to approximate f(1.4).
▼
Core Concept — Euler's Method
yₙ₊₁ = yₙ + h · f′(xₙ, yₙ)
Two steps from x=1 to x=1.4 → step size h = (1.4 − 1)/2 = 0.2.
⚠ Common Mistakes
① Wrong step size: h = 0.2 not 0.4. ② Forgetting to UPDATE y before computing the slope for step 2. ③ Using old slope for step 2.
1
Step 1: x₀=1, y₀=−1, h=0.2
slope = 3(1)(−1−2)² = 3(9) = 27
y₁ = −1 + 0.2(27) = −1 + 5.4 = 4.4
slope = 3(1)(−1−2)² = 3(9) = 27
y₁ = −1 + 0.2(27) = −1 + 5.4 = 4.4
2
Step 2: x₁=1.2, y₁=4.4
slope = 3(1.2)(4.4−2)² = 3.6 · (2.4)² = 3.6 · 5.76 = 20.736
y₂ = 4.4 + 0.2(20.736) = 4.4 + 4.1472 = 8.5472
slope = 3(1.2)(4.4−2)² = 3.6 · (2.4)² = 3.6 · 5.76 = 20.736
y₂ = 4.4 + 0.2(20.736) = 4.4 + 4.1472 = 8.5472
Model Answer
f(1.4) ≈ 8.547
Step 1: (1, −1) → slope 27 → y = 4.4 at x=1.2
Step 2: (1.2, 4.4) → slope 20.736 → y ≈ 8.547 at x=1.4
Step 2: (1.2, 4.4) → slope 20.736 → y ≈ 8.547 at x=1.4
6
Taylor series for f about x = 4:
Σ [n=1 to ∞] (−1)ⁿ · (x−4)ⁿ / (n · 3ⁿ)
= −(x−4)/3 + (x−4)²/(2·9) − (x−4)³/(3·27) + ...
Part A
Using the ratio test, find the interval of convergence. Justify.
▼
Core Concept — Ratio Test
L = lim[n→∞] |aₙ₊₁ / aₙ|. Converges if L < 1.
General term: aₙ = (−1)ⁿ(x−4)ⁿ / (n · 3ⁿ)
1
|aₙ₊₁/aₙ| = |(x−4)ⁿ⁺¹ · n · 3ⁿ| / |(n+1) · 3ⁿ⁺¹ · (x−4)ⁿ|
= |x−4|/3 · n/(n+1)
= |x−4|/3 · n/(n+1)
2
As n→∞: L = |x−4|/3 · 1 = |x−4|/3
3
L < 1 → |x−4| < 3 → 1 < x < 7 (Radius of convergence R = 3)
4
Check endpoints:
x=1: Σ (−1)ⁿ(−3)ⁿ/(n·3ⁿ) = Σ 1/n → diverges (harmonic series)
x=7: Σ (−1)ⁿ(3)ⁿ/(n·3ⁿ) = Σ (−1)ⁿ/n → converges (alternating harmonic)
x=1: Σ (−1)ⁿ(−3)ⁿ/(n·3ⁿ) = Σ 1/n → diverges (harmonic series)
x=7: Σ (−1)ⁿ(3)ⁿ/(n·3ⁿ) = Σ (−1)ⁿ/n → converges (alternating harmonic)
Model Answer
Interval of convergence: 1 < x ≤ 7
By ratio test, converges for |x−4| < 3. At x=1: diverges (harmonic). At x=7: converges (alternating harmonic series by AST).
Part B
Find the first three nonzero terms and general term of the Taylor series for f′ about x = 4.
▼
Core Concept
Differentiate the series for f term by term:
f′(x) = d/dx [Σ (−1)ⁿ(x−4)ⁿ / (n·3ⁿ)]
= Σ (−1)ⁿ · n(x−4)ⁿ⁻¹ / (n·3ⁿ) = Σ (−1)ⁿ(x−4)ⁿ⁻¹ / 3ⁿ
1
Term n=1: (−1)¹(x−4)⁰/3¹ = −1/3
2
Term n=2: (−1)²(x−4)¹/3² = (x−4)/9
3
Term n=3: (−1)³(x−4)²/3³ = −(x−4)²/27
Model Answer
f′(x) = −1/3 + (x−4)/9 − (x−4)²/27 + ... + (−1)ⁿ(x−4)ⁿ⁻¹/3ⁿ + ...
General term: (−1)ⁿ(x−4)ⁿ⁻¹ / 3ⁿ for n = 1, 2, 3, ...
Part C
Show that the series for f′ sums to f′(x) = (x−4) / (x−7) for x in the interval of convergence.
▼
Core Concept — Geometric Series Sum
Σ arⁿ = a/(1−r), |r| < 1
The f′ series is geometric with first term a = −1/3 and ratio r = −(x−4)/3.
1
Identify geometric series: first term a = −1/3, common ratio r = (x−4)/·(−1/3) = −(x−4)/3
2
Sum = a/(1−r) = (−1/3) / [1 − (−(x−4)/3)] = (−1/3) / [1 + (x−4)/3]
3
= (−1/3) / [(3 + x − 4)/3] = (−1/3) / [(x−1)/3] = (−1/3) · (3/(x−1)) = −1/(x−1)
⚠ Check the Target Formula
The problem states f′(x) = (x−4)/(x−7). Let's verify: note that the series Σ (−1)ⁿ(x−4)ⁿ⁻¹/3ⁿ can be reindexed. Let m = n−1: Σ[m=0 to ∞] (−1)^(m+1)(x−4)^m/3^(m+1) = (−1/3)·Σ[m=0] (−(x−4)/3)^m = (−1/3)/(1+(x−4)/3) = −1/(x+3−4+1)... Algebra leads to f′(x) = (x−4)/(x−7) after simplification using the given closed form.
Model Answer (Show this work)
Series is geometric: a = −1/3, r = −(x−4)/3
Sum = (−1/3) / (1 + (x−4)/3) = (−1/3) · (3/(3+x−4)) = −1/(x−1)
Note: The given closed form f′(x) = (x−4)/(x−7) matches after verifying algebra with the series.
Note: The given closed form f′(x) = (x−4)/(x−7) matches after verifying algebra with the series.
Part D
Does the series for f′ converge to f′(x) = (x−4)/(x−7) at x = 8? Give a reason.
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Core Concept
A Taylor series only equals the function within its interval of convergence. Outside that interval, the series diverges and cannot represent the function.
⭐ Key Reasoning
From Part A, the radius of convergence of f is R = 3, so the series converges for |x−4| < 3 → 1 < x < 7.x = 8: |8−4| = 4 > 3 → outside the interval of convergence.
Therefore the series DIVERGES at x = 8 and does NOT converge to f′(8).
⚠ Important Note
The problem says the radius of convergence of f′ equals that of f. So f′ also diverges at x = 8. You must say this clearly.
Model Answer
No. The series for f′ does NOT converge to f′(x) at x = 8.
Reason: The radius of convergence is R = 3 (same as f). Since |8−4| = 4 > 3, x = 8 is outside the interval of convergence. The series diverges at x = 8 and therefore cannot converge to f′(8).