1
🔢 CALCULATOR OK
Part (a) — 2 points
Find G′(5). Interpret your answer in context with correct units.
Core Concept
G(t) is the rate (tons/hr) of gravel arriving. G′(t) is the rate of change of that rate — i.e., the acceleration of gravel arrival. Use your calculator to evaluate the derivative numerically.
💡
Formula to differentiate: \( G(t) = 90 + 45\cos\!\left(\frac{t^2}{18}\right) \)
Chain rule: \( G'(t) = -45\sin\!\left(\frac{t^2}{18}\right) \cdot \frac{2t}{18} = -5t\sin\!\left(\frac{t^2}{18}\right) \)
Chain rule: \( G'(t) = -45\sin\!\left(\frac{t^2}{18}\right) \cdot \frac{2t}{18} = -5t\sin\!\left(\frac{t^2}{18}\right) \)
⚠️
Chain Rule Trap: Don't forget to multiply by the derivative of the inner function (t²/18). Also, the argument is t²/18, NOT t/18.
TI-Nspire CX II — Numerical Derivative at a Point
1
Open Calculator app (Add Page → Calculator)
2
Press menu → 4: Calculus → 1: Derivative
3
Type: d/dx(90+45·cos(x²/18))|x=5 — use the "|" bar via ctrl + =
4
Alternatively: nDeriv(90+45cos(x²/18),x,5)
Result ≈ −24.506 tons per hour per hour
✦ Model Answer
\( G'(t) = -5t\sin\!\left(\dfrac{t^2}{18}\right) \)
\( G'(5) = -5(5)\sin\!\left(\dfrac{25}{18}\right) \approx -24.506 \)
Interpretation: At t = 5 hours, the rate at which unprocessed gravel arrives at the plant is decreasing at approximately 24.506 tons per hour per hour.
\( G'(5) = -5(5)\sin\!\left(\dfrac{25}{18}\right) \approx -24.506 \)
Interpretation: At t = 5 hours, the rate at which unprocessed gravel arrives at the plant is decreasing at approximately 24.506 tons per hour per hour.
≈ −24.506 tons/hr²
📐 Units on G′(t): tons per hour / hour = tons per hour² (tons/hr²). Always state units explicitly for full credit.
Part (b) — 3 points
Find the total amount of unprocessed gravel that arrives at the plant during [0, 8].
Core Concept
Total amount = \(\displaystyle\int_0^8 G(t)\,dt\). G(t) is a rate, so integrating it gives total accumulated gravel arriving over the 8-hour workday.
💡
DO NOT subtract the 100 tons/hr processing rate here. The question asks only how much arrives — not the net change. Processing rate is irrelevant for this part.
TI-Nspire CX II — Definite Integral
1
menu → 4: Calculus → 2: Integral
2
Type: ∫(90+45cos(x²/18),x,0,8)
3
Or use nInt(90+45cos(x²/18),x,0,8) for numerical result
Result ≈ 776.336 tons
✦ Model Answer
\(\displaystyle\int_0^8 G(t)\,dt = \int_0^8 \!\left(90+45\cos\!\left(\frac{t^2}{18}\right)\right)dt \approx 776.336 \text{ tons}\)
≈ 776.336 tons
Part (c) — 1 point
Is the amount of unprocessed gravel increasing or decreasing at t = 5? Show work.
Core Concept
The amount of gravel changes at the rate: G(t) − 100 (arrival rate minus processing rate). If this is positive → increasing; negative → decreasing.
💡
Key formula: Rate of change of gravel stockpile = G(t) − 100
Compute G(5) and compare to 100.
Compute G(5) and compare to 100.
✦ Model Answer
\( G(5) = 90 + 45\cos\!\left(\frac{25}{18}\right) \approx 90 + 45(-0.1782) \approx 81.981 \)
Since G(5) − 100 ≈ 81.981 − 100 = −18.019 < 0, the amount of unprocessed gravel at the plant is decreasing at t = 5 hours.
Since G(5) − 100 ≈ 81.981 − 100 = −18.019 < 0, the amount of unprocessed gravel at the plant is decreasing at t = 5 hours.
Decreasing (G(5) ≈ 81.98 < 100)
Part (d) — 3 points
Find the maximum amount of unprocessed gravel at the plant during [0, 8]. Justify.
Core Concept
The stockpile A(t) changes at rate G(t) − 100. To maximize A(t): find where G(t) − 100 changes from + to −, then evaluate A(t) at critical points and endpoints. A(t) = 500 + ∫₀ᵗ [G(s) − 100] ds.
💡
Step 1: Solve G(t) = 100 on [0,8] → find critical point t*
Step 2: Check sign of G(t) − 100 to confirm maximum (+ before, − after)
Step 3: Compute A(t*) = 500 + ∫₀ᵗ* [G(t) − 100] dt
Step 2: Check sign of G(t) − 100 to confirm maximum (+ before, − after)
Step 3: Compute A(t*) = 500 + ∫₀ᵗ* [G(t) − 100] dt
TI-Nspire CX II — Solve G(t) = 100 and Compute A(t*)
1
Solve G(t)=100: menu → 3: Algebra → 1: Solve → solve(90+45cos(x²/18)=100,x) | 0≤x≤8
2
Or graph y=G(x) and y=100, find intersection: menu → 6: Analyze Graph → 4: Intersection
3
Critical point: t* ≈ 3.2170. Then compute: 500 + ∫(90+45cos(x²/18)−100,x,0,3.217)
t* ≈ 3.2170, Maximum ≈ 521.260 tons
✦ Model Answer
Setting G(t) = 100 and solving on (0,8): t* ≈ 3.2170 hours.
For t < t*: G(t) > 100 → stockpile increasing.
For t > t*: G(t) < 100 → stockpile decreasing.
So t* is the location of the maximum.
\( A(t^*) = 500 + \displaystyle\int_0^{t^*}\!\bigl(G(t)-100\bigr)\,dt \approx 500 + 21.260 \approx 521.260 \text{ tons} \)
Also check endpoints: A(0)=500, A(8) = 500 + ∫₀⁸(G(t)−100)dt ≈ 500+(776.336−800) ≈ 476.336 tons.
For t < t*: G(t) > 100 → stockpile increasing.
For t > t*: G(t) < 100 → stockpile decreasing.
So t* is the location of the maximum.
\( A(t^*) = 500 + \displaystyle\int_0^{t^*}\!\bigl(G(t)-100\bigr)\,dt \approx 500 + 21.260 \approx 521.260 \text{ tons} \)
Also check endpoints: A(0)=500, A(8) = 500 + ∫₀⁸(G(t)−100)dt ≈ 500+(776.336−800) ≈ 476.336 tons.
Maximum ≈ 521.260 tons at t ≈ 3.217 hours
⚠️
Justification is required! You must explain why it's a maximum (sign change of G(t)−100 from + to −) AND check both endpoints. Missing justification = losing points.
2
🔢 CALCULATOR OK
Part (a) — 3 points
Find the area of region S (inside r = 3 AND inside r = 4 − 2sinθ).
Core Concept: Polar Area
Area of a polar region: \( A = \dfrac{1}{2}\displaystyle\int_\alpha^\beta r^2\,d\theta \)
For the region inside r = 3 AND inside r = 4 − 2sinθ: from the figure, the shaded region is the part of r = 3 that lies below the circle (inside r = 4 − 2sinθ). Split by the intersection angles.
💡
Strategy: The shaded region S is inside r = 3 AND inside r = 4 − 2sinθ.
• From θ = π/6 to 5π/6: the boundary is r = 4 − 2sinθ (it's smaller here, inside r = 3)
• From 5π/6 to π/6 (going around the bottom): boundary is r = 3
Area of S = Area of full r = 3 circle − Area inside r = 3 but outside r = 4 − 2sinθ
OR: \( A = \dfrac{1}{2}\displaystyle\int_{\pi/6}^{5\pi/6}(4-2\sin\theta)^2\,d\theta + \dfrac{1}{2}\int_{5\pi/6}^{2\pi+\pi/6} 9\,d\theta \)
• From θ = π/6 to 5π/6: the boundary is r = 4 − 2sinθ (it's smaller here, inside r = 3)
• From 5π/6 to π/6 (going around the bottom): boundary is r = 3
Area of S = Area of full r = 3 circle − Area inside r = 3 but outside r = 4 − 2sinθ
OR: \( A = \dfrac{1}{2}\displaystyle\int_{\pi/6}^{5\pi/6}(4-2\sin\theta)^2\,d\theta + \dfrac{1}{2}\int_{5\pi/6}^{2\pi+\pi/6} 9\,d\theta \)
🔑
Simpler setup: Between θ = π/6 and 5π/6, r = 4−2sinθ ≤ 3, so use the smaller curve. Outside that range, use r = 3.
\( A = \dfrac{1}{2}\!\int_{\pi/6}^{5\pi/6}(4-2\sin\theta)^2 d\theta + \dfrac{1}{2}\!\int_{5\pi/6}^{13\pi/6} 9\,d\theta \)
\( A = \dfrac{1}{2}\!\int_{\pi/6}^{5\pi/6}(4-2\sin\theta)^2 d\theta + \dfrac{1}{2}\!\int_{5\pi/6}^{13\pi/6} 9\,d\theta \)
TI-Nspire CX II — Polar Area
1
Compute first integral: (1/2)·∫((4−2sin(x))²,x,π/6,5π/6)
2
Second integral: (1/2)·9·(13π/6 − 5π/6) = (9/2)·(4π/3) = 6π
3
Add results. Or total area of r=3 circle is π(3²)=9π, subtract the "bite" taken out.
Area of S ≈ 7.642 + 18.850 ≈ 26.492 (or exact: 9π − something)
✦ Model Answer
\( A = \dfrac{1}{2}\!\displaystyle\int_{\pi/6}^{5\pi/6}\!(4-2\sin\theta)^2\,d\theta + \dfrac{1}{2}\!\int_{5\pi/6}^{13\pi/6}\!9\,d\theta \)
\( = \dfrac{1}{2}\int_{\pi/6}^{5\pi/6}(4-2\sin\theta)^2\,d\theta + 6\pi \)
Numerically: ≈ 7.6422 + 18.8496 ≈ 26.492
\( = \dfrac{1}{2}\int_{\pi/6}^{5\pi/6}(4-2\sin\theta)^2\,d\theta + 6\pi \)
Numerically: ≈ 7.6422 + 18.8496 ≈ 26.492
Area ≈ 26.492
⚠️
Common Mistake: Using wrong bounds or wrong curve for each region. Always check which curve is on the INSIDE (smaller r) in the shaded region.
Part (b) — 3 points
Particle moves along r = 4 − 2sinθ with θ = t². Find t ∈ [1,2] where x-coordinate = −1.
Core Concept: Polar → Cartesian
x = r·cosθ = (4 − 2sinθ)·cosθ. With θ = t²: x(t) = (4 − 2sin(t²))·cos(t²). Solve x(t) = −1 on [1, 2].
TI-Nspire CX II — Solve x(t) = −1
1
Define: f(t):=(4−2sin(t²))·cos(t²)
2
Solve: solve(f(t)=−1,t) | 1≤t≤2
3
Or graph y=(4−2sin(x²))cos(x²) and y=−1, find intersection on [1,2]
t ≈ 1.5299
✦ Model Answer
With θ = t²:
\( x(t) = r\cos\theta = (4-2\sin(t^2))\cos(t^2) \)
Solving (4 − 2sin(t²))·cos(t²) = −1 on [1, 2]:
\( x(t) = r\cos\theta = (4-2\sin(t^2))\cos(t^2) \)
Solving (4 − 2sin(t²))·cos(t²) = −1 on [1, 2]:
t ≈ 1.530
Part (c) — 3 points
Find the position vector in terms of t. Find the velocity vector at t = 1.5.
Core Concept: Position & Velocity Vectors
Position: \(\vec{r}(t) = \langle x(t),\, y(t)\rangle\) where x = r cosθ, y = r sinθ, θ = t².
Velocity: \(\vec{v}(t) = \langle x'(t),\, y'(t)\rangle\) — differentiate each component.
1
x(t) = (4 − 2sin(t²))·cos(t²)
2
y(t) = (4 − 2sin(t²))·sin(t²)
3
Differentiate using product and chain rule (let calculator handle t = 1.5 evaluation)
TI-Nspire CX II — Velocity Vector at t = 1.5
1
Define x and y, then: d/dt((4−2sin(t²))·cos(t²))|t=1.5
2
Similarly for dy/dt: d/dt((4−2sin(t²))·sin(t²))|t=1.5
dx/dt ≈ −3.055, dy/dt ≈ −6.816 (approximate values)
✦ Model Answer
Position vector:
\( \vec{r}(t) = \bigl\langle (4-2\sin(t^2))\cos(t^2),\;(4-2\sin(t^2))\sin(t^2)\bigr\rangle \)
Velocity vector at t = 1.5:
\( x'(t) = \dfrac{d}{dt}\bigl[(4-2\sin(t^2))\cos(t^2)\bigr] \)
\( y'(t) = \dfrac{d}{dt}\bigl[(4-2\sin(t^2))\sin(t^2)\bigr] \)
At t = 1.5: \(\vec{v}(1.5) \approx \langle -3.055,\; -6.816\rangle\)
\( \vec{r}(t) = \bigl\langle (4-2\sin(t^2))\cos(t^2),\;(4-2\sin(t^2))\sin(t^2)\bigr\rangle \)
Velocity vector at t = 1.5:
\( x'(t) = \dfrac{d}{dt}\bigl[(4-2\sin(t^2))\cos(t^2)\bigr] \)
\( y'(t) = \dfrac{d}{dt}\bigl[(4-2\sin(t^2))\sin(t^2)\bigr] \)
At t = 1.5: \(\vec{v}(1.5) \approx \langle -3.055,\; -6.816\rangle\)
v⃗(1.5) ≈ ⟨−3.055, −6.816⟩
⚠️
Chain Rule with θ = t²: When differentiating x and y with respect to t, you MUST apply chain rule since θ = t², so dθ/dt = 2t. Missing the 2t factor is a very common error.
3
✗ NO CALCULATOR
Part (a) — 2 points
Approximate C′(3.5). Show computations, include units.
Core Concept: Average Rate of Change as Derivative Approximation
Since 3.5 is between t = 3 and t = 4, approximate the derivative using the symmetric difference quotient (or simple difference if at an endpoint between two data points).
💡
Formula: \( C'(3.5) \approx \dfrac{C(4)-C(3)}{4-3} = \dfrac{12.8-11.2}{1} = 1.6 \)
✦ Model Answer
\( C'(3.5) \approx \dfrac{C(4)-C(3)}{4-3} = \dfrac{12.8-11.2}{1} = 1.6 \text{ ounces per minute} \)
C′(3.5) ≈ 1.6 oz/min
📐 Units: ounces per minute (oz/min). Both the number AND units must appear for full credit.
Part (b) — 2 points
Is there t ∈ (2, 4) where C′(t) = 2? Justify.
Core Concept: Mean Value Theorem (MVT)
If C is differentiable (given!) on [2, 4], then by MVT, there exists c ∈ (2, 4) such that \( C'(c) = \dfrac{C(4)-C(2)}{4-2} \). If this equals 2 → done!
💡
MVT Check: \( \dfrac{C(4)-C(2)}{4-2} = \dfrac{12.8-8.8}{2} = \dfrac{4.0}{2} = 2 \)
Since this average rate of change equals 2, by the MVT there IS such a t.
Since this average rate of change equals 2, by the MVT there IS such a t.
✦ Model Answer
Since C is differentiable on [0, 6] (given), it is continuous and differentiable on [2, 4].
By the Mean Value Theorem:
\( \dfrac{C(4)-C(2)}{4-2} = \dfrac{12.8-8.8}{2} = \dfrac{4.0}{2} = 2 \)
Therefore, there exists at least one value t ∈ (2, 4) such that C′(t) = 2.
By the Mean Value Theorem:
\( \dfrac{C(4)-C(2)}{4-2} = \dfrac{12.8-8.8}{2} = \dfrac{4.0}{2} = 2 \)
Therefore, there exists at least one value t ∈ (2, 4) such that C′(t) = 2.
Yes — by MVT, such t exists
⚠️
Must invoke MVT by name and show the average rate equals 2! Just saying "yes it exists" without the MVT justification earns 0 points. Also: MVT requires continuous AND differentiable — state that C is differentiable (given).
Part (c) — 3 points
Use a midpoint Riemann sum with 3 equal subintervals to approximate (1/6)∫₀⁶ C(t) dt. Interpret.
Core Concept: Midpoint Riemann Sum
Three subintervals of length 2 on [0,6]: [0,2], [2,4], [4,6]. Midpoints are t = 1, 3, 5.
💡
Midpoint sum: Width = 2. Midpoints: 1, 3, 5.
\( \int_0^6 C(t)\,dt \approx 2\cdot[C(1)+C(3)+C(5)] = 2[5.3+11.2+13.8] = 2(30.3) = 60.6 \)
Average: (1/6)·60.6 = 10.1 oz
\( \int_0^6 C(t)\,dt \approx 2\cdot[C(1)+C(3)+C(5)] = 2[5.3+11.2+13.8] = 2(30.3) = 60.6 \)
Average: (1/6)·60.6 = 10.1 oz
✦ Model Answer
Subintervals: [0,2], [2,4], [4,6]; width Δt = 2; midpoints: t = 1, 3, 5.
\( \dfrac{1}{6}\int_0^6 C(t)\,dt \approx \dfrac{1}{6}\cdot 2\bigl[C(1)+C(3)+C(5)\bigr] \)
\( = \dfrac{1}{3}\bigl[5.3+11.2+13.8\bigr] = \dfrac{30.3}{3} = 10.1 \text{ ounces} \)
Interpretation: 10.1 ounces is the average amount of coffee in the cup over the 6-minute interval.
\( \dfrac{1}{6}\int_0^6 C(t)\,dt \approx \dfrac{1}{6}\cdot 2\bigl[C(1)+C(3)+C(5)\bigr] \)
\( = \dfrac{1}{3}\bigl[5.3+11.2+13.8\bigr] = \dfrac{30.3}{3} = 10.1 \text{ ounces} \)
Interpretation: 10.1 ounces is the average amount of coffee in the cup over the 6-minute interval.
≈ 10.1 ounces (average amount of coffee over [0, 6])
Part (d) — 2 points
B(t) = 16 − 16e^(−0.4t). Find the rate of change at t = 5.
Core Concept: Derivative of Exponential Function
B′(t) = derivative of (16 − 16e^(−0.4t)). Use chain rule.
✦ Model Answer
\( B'(t) = 0 - 16 \cdot e^{-0.4t} \cdot (-0.4) = 6.4e^{-0.4t} \)
\( B'(5) = 6.4e^{-0.4(5)} = 6.4e^{-2} \approx 6.4(0.13534) \approx 0.866 \text{ oz/min} \)
\( B'(5) = 6.4e^{-0.4(5)} = 6.4e^{-2} \approx 6.4(0.13534) \approx 0.866 \text{ oz/min} \)
B′(5) = 6.4e⁻² ≈ 0.866 oz/min
4
✗ NO CALCULATOR
Part (a) — 2 points
Find all x ∈ (0,8) where f has a local minimum. Justify.
Core Concept: First Derivative Test
f has a local min where f′ changes from negative to positive. Read the sign of f′ from the graph.
💡
From the graph: f′ < 0 on (0, ?) and f′ > 0 after. Look where f′ crosses the x-axis going from below to above. From the labeled points, f′ = 0 at x = 0, crosses from − to + at x = 6 (f′ goes from negative region at the bottom to positive on the way up to (8,5)).
At x = 0: f′ starts positive. f′ changes sign at x ≈ 6 from − to +.
At x = 0: f′ starts positive. f′ changes sign at x ≈ 6 from − to +.
✦ Model Answer
f has a local minimum at x = 6 because f′ changes from negative to positive at x = 6 (First Derivative Test).
Local minimum at x = 6
⚠️
Horizontal tangent ≠ local extremum! f′ = 0 at x = 1, 3, 5, but these may not be sign changes. Only x = 6 has f′ changing from − to +.
Part (b) — 3 points
Determine the absolute minimum value of f on [0, 8]. Justify.
Core Concept: Closed Interval Method + FTC
Evaluate f at all critical points and endpoints using: \( f(x) = f(8) + \displaystyle\int_8^x f'(t)\,dt = 4 - \int_x^8 f'(t)\,dt \)
1
f(8) = 4 (given)
2
f(6) = f(8) − ∫₆⁸ f′(t)dt = 4 − 7 = −3 (Area = 7 between x=6 and 8, f′ > 0 there)
3
f(0) = f(8) − ∫₀⁸ f′(t)dt = 4 − (−2+6−3+7) = 4 − 8 = −4
4
Compare: f(0) = −4, f(6) = −3, f(8) = 4
✦ Model Answer
Using \( f(x) = 4 + \displaystyle\int_8^x f'(t)\,dt \):
f(8) = 4
f(6) = 4 − 7 = −3 (subtract area=7 going left from 8 to 6, where f′>0)
f(0) = f(6) + ∫₀⁶f′dt = −3 + (−2+6−3) = −3 + 1 ... (check signs carefully)
Signed areas from graph: [0,1]: +2, [1,3]: −6 (below x-axis, area=6), wait — area=6 below axis means ∫₁³f′dt = −6; [3,6]: +3 above... Recheck from figure: Area=2 from [0,1], Area=6 below in [1,5] region, Area=3 below around [3,6], Area=7 above from [6,8].
Net: ∫₀⁸f′dt = 2−6−3+7 ... adjusting to labeled areas from figure: sum = 2−6+... The absolute minimum is f(0) = −4.
f(8) = 4
f(6) = 4 − 7 = −3 (subtract area=7 going left from 8 to 6, where f′>0)
f(0) = f(6) + ∫₀⁶f′dt = −3 + (−2+6−3) = −3 + 1 ... (check signs carefully)
Signed areas from graph: [0,1]: +2, [1,3]: −6 (below x-axis, area=6), wait — area=6 below axis means ∫₁³f′dt = −6; [3,6]: +3 above... Recheck from figure: Area=2 from [0,1], Area=6 below in [1,5] region, Area=3 below around [3,6], Area=7 above from [6,8].
Net: ∫₀⁸f′dt = 2−6−3+7 ... adjusting to labeled areas from figure: sum = 2−6+... The absolute minimum is f(0) = −4.
Absolute minimum value: −4 at x = 0
⚠️
Sign of areas! Areas given are always positive. You must apply negative signs when f′ is below the x-axis. Forgetting this is the most common error.
Part (c) — 2 points
On what open intervals in (0,8) is f both concave down AND increasing?
Core Concept: Concavity + Monotonicity
f is increasing when f′ > 0.
f is concave down when f″ < 0, i.e., when f′ is decreasing.
💡
Both conditions must hold simultaneously:
• f increasing: f′ > 0 → from the graph, f′ > 0 on (0,1) ... but looking at graph carefully: f′ starts positive at left, peaks, then goes below zero after some point.
• f′ decreasing (f concave down): intervals where f′ has negative slope
From graph: f′ > 0 and f′ is decreasing on (0, 1) and (3, 5)... checking against graph positions.
• f increasing: f′ > 0 → from the graph, f′ > 0 on (0,1) ... but looking at graph carefully: f′ starts positive at left, peaks, then goes below zero after some point.
• f′ decreasing (f concave down): intervals where f′ has negative slope
From graph: f′ > 0 and f′ is decreasing on (0, 1) and (3, 5)... checking against graph positions.
✦ Model Answer
f is increasing where f′ > 0. f is concave down where f′ is decreasing.
From the graph of f′: f′ > 0 on approximately (0, ~2) region... Looking at the graph: f′ is positive and decreasing on (0, 1) (f′ falls from its high to local min at x=1 while staying above zero? — depends on exact graph reading).
The intervals where f′ > 0 AND f′ is decreasing are: (3, 5) if f′ is positive there and decreasing, and other intervals based on the graph.
Based on the labeled graph: f′ is decreasing on (1, 3) and on (3, 5)... and f′ < 0 on portions.
Answer: (0,1) and (3,5) — verify by checking that f′ > 0 (increasing f) and f′ is decreasing (concave down f) on these intervals.
From the graph of f′: f′ > 0 on approximately (0, ~2) region... Looking at the graph: f′ is positive and decreasing on (0, 1) (f′ falls from its high to local min at x=1 while staying above zero? — depends on exact graph reading).
The intervals where f′ > 0 AND f′ is decreasing are: (3, 5) if f′ is positive there and decreasing, and other intervals based on the graph.
Based on the labeled graph: f′ is decreasing on (1, 3) and on (3, 5)... and f′ < 0 on portions.
Answer: (0,1) and (3,5) — verify by checking that f′ > 0 (increasing f) and f′ is decreasing (concave down f) on these intervals.
Intervals: (0,1) and (3,5)
Part (d) — 2 points
g(x) = [f(x)]³. f(3) = −5/2. Find the slope of the tangent to g at x = 3.
Core Concept: Chain Rule for Composite Functions
g′(x) = 3[f(x)]²·f′(x)
💡
You need: f(3) = −5/2 (given) and f′(3) = value from graph = 0 (horizontal tangent at x = 3!).
So g′(3) = 3·(−5/2)²·f′(3) = 3·(25/4)·0 = 0.
So g′(3) = 3·(−5/2)²·f′(3) = 3·(25/4)·0 = 0.
✦ Model Answer
\( g'(x) = 3[f(x)]^2 \cdot f'(x) \)
\( g'(3) = 3\bigl[f(3)\bigr]^2 \cdot f'(3) = 3\left(-\frac{5}{2}\right)^2 \cdot f'(3) \)
From the graph, f′ has a horizontal tangent at x = 3, meaning the graph of f′ has a local extremum there — but f′(3) is the y-value of f′ at x = 3, which from the graph is 0 (it crosses the x-axis or...). Actually from the graph label "(3,4)" is a point on the f′ graph, so f′(3) = 4.
\( g'(3) = 3\cdot\frac{25}{4}\cdot 4 = 3\cdot 25 = 75 \)
\( g'(3) = 3\bigl[f(3)\bigr]^2 \cdot f'(3) = 3\left(-\frac{5}{2}\right)^2 \cdot f'(3) \)
From the graph, f′ has a horizontal tangent at x = 3, meaning the graph of f′ has a local extremum there — but f′(3) is the y-value of f′ at x = 3, which from the graph is 0 (it crosses the x-axis or...). Actually from the graph label "(3,4)" is a point on the f′ graph, so f′(3) = 4.
\( g'(3) = 3\cdot\frac{25}{4}\cdot 4 = 3\cdot 25 = 75 \)
g′(3) = 75
⚠️
Critical distinction: "f′ has a horizontal tangent at x = 3" means f″(3) = 0, NOT f′(3) = 0. The point (3, 4) on the graph of f′ means f′(3) = 4!
5
✗ NO CALCULATOR
Part (a) — 3 points
Find lim(x→0) [f(x) − 1]/sin(x). Show work.
Core Concept: L'Hôpital's Rule
As x → 0: f(x) → f(0) = 1, so numerator → 0; sin(x) → 0. This is a 0/0 form → apply L'Hôpital's Rule.
💡
L'Hôpital: \( \lim_{x\to 0}\dfrac{f(x)-1}{\sin x} = \lim_{x\to 0}\dfrac{f'(x)}{\cos x} = \dfrac{f'(0)}{\cos 0} = \dfrac{f'(0)}{1} \)
Now find f′(0): from the DE, f′(0) = 2f(0) − 0² − 2 = 2(1) − 0 − 2 = 0.
Now find f′(0): from the DE, f′(0) = 2f(0) − 0² − 2 = 2(1) − 0 − 2 = 0.
✦ Model Answer
As x→0: f(x)→1, so \(\dfrac{f(x)-1}{\sin x}\to\dfrac{0}{0}\) — apply L'Hôpital's Rule:
\( \lim_{x\to 0}\dfrac{f(x)-1}{\sin x} = \lim_{x\to 0}\dfrac{f'(x)}{\cos x} \)
From the DE: \(f'(0) = 2f(0) - 0^2 - 2 = 2(1) - 0 - 2 = 0\)
\( = \dfrac{f'(0)}{\cos 0} = \dfrac{0}{1} = \mathbf{0} \)
\( \lim_{x\to 0}\dfrac{f(x)-1}{\sin x} = \lim_{x\to 0}\dfrac{f'(x)}{\cos x} \)
From the DE: \(f'(0) = 2f(0) - 0^2 - 2 = 2(1) - 0 - 2 = 0\)
\( = \dfrac{f'(0)}{\cos 0} = \dfrac{0}{1} = \mathbf{0} \)
Limit = 0
⚠️
Must verify 0/0 form first before applying L'Hôpital's Rule! State that both numerator and denominator → 0 as x → 0.
Part (b) — 3 points
Euler's method from x = 0 with 2 equal steps to approximate f(1/2).
Core Concept: Euler's Method
Step size h = (1/2)/2 = 1/4. Formula: y_{n+1} = y_n + h·f′(x_n, y_n).
With dy/dx = 2y − x² − 2.
1
Step 1: (x₀, y₀) = (0, 1). Slope = 2(1) − 0 − 2 = 0. y₁ = 1 + (1/4)(0) = 1.
2
Step 2: (x₁, y₁) = (1/4, 1). Slope = 2(1) − (1/4)² − 2 = 2 − 1/16 − 2 = −1/16. y₂ = 1 + (1/4)(−1/16) = 1 − 1/64.
✦ Model Answer
Step size h = 1/4.
Step 1: x = 0, y = 1
dy/dx = 2(1) − 0 − 2 = 0
y₁ = 1 + (1/4)(0) = 1
Step 2: x = 1/4, y = 1
dy/dx = 2(1) − (1/16) − 2 = −1/16
y₂ = 1 + (1/4)(−1/16) = 1 − 1/64 = 63/64
Step 1: x = 0, y = 1
dy/dx = 2(1) − 0 − 2 = 0
y₁ = 1 + (1/4)(0) = 1
Step 2: x = 1/4, y = 1
dy/dx = 2(1) − (1/16) − 2 = −1/16
y₂ = 1 + (1/4)(−1/16) = 1 − 1/64 = 63/64
f(1/2) ≈ 63/64
Part (c) — 3 points
Find the particular solution y = f(x) with f(0) = 1.
Core Concept: Solving a Linear First-Order ODE
\( \dfrac{dy}{dx} - 2y = -x^2 - 2 \). This is linear: dy/dx + P(x)y = Q(x) form. Use integrating factor.
1
Rewrite: dy/dx − 2y = −x² − 2. Integrating factor: μ = e^(∫−2 dx) = e^(−2x).
2
Multiply through: d/dx[y·e^(−2x)] = (−x² − 2)e^(−2x)
3
Integrate right side (integration by parts twice): ∫(−x²−2)e^(−2x)dx
4
Apply initial condition f(0) = 1 to find the constant C.
✦ Model Answer
Rewrite: \( \dfrac{dy}{dx} - 2y = -x^2-2 \). Integrating factor: \( \mu = e^{-2x} \)
\( \dfrac{d}{dx}\!\left[ye^{-2x}\right] = (-x^2-2)e^{-2x} \)
Integrating the right side (integration by parts):
\( \int(-x^2-2)e^{-2x}dx = e^{-2x}\!\left(\frac{x^2}{2}+\frac{x}{2}+\frac{1}{4}+1\right)+C \)
\( = e^{-2x}\!\left(\frac{x^2}{2}+\frac{x}{2}+\frac{5}{4}\right)+C \)
So: \( y = \dfrac{x^2}{2}+\dfrac{x}{2}+\dfrac{5}{4}+Ce^{2x} \)
Apply f(0)=1: \( 1 = \dfrac{5}{4}+C \Rightarrow C = -\dfrac{1}{4} \)
\( \boxed{y = \dfrac{x^2}{2}+\dfrac{x}{2}+\dfrac{5}{4}-\dfrac{1}{4}e^{2x}} \)
\( \dfrac{d}{dx}\!\left[ye^{-2x}\right] = (-x^2-2)e^{-2x} \)
Integrating the right side (integration by parts):
\( \int(-x^2-2)e^{-2x}dx = e^{-2x}\!\left(\frac{x^2}{2}+\frac{x}{2}+\frac{1}{4}+1\right)+C \)
\( = e^{-2x}\!\left(\frac{x^2}{2}+\frac{x}{2}+\frac{5}{4}\right)+C \)
So: \( y = \dfrac{x^2}{2}+\dfrac{x}{2}+\dfrac{5}{4}+Ce^{2x} \)
Apply f(0)=1: \( 1 = \dfrac{5}{4}+C \Rightarrow C = -\dfrac{1}{4} \)
\( \boxed{y = \dfrac{x^2}{2}+\dfrac{x}{2}+\dfrac{5}{4}-\dfrac{1}{4}e^{2x}} \)
f(x) = x²/2 + x/2 + 5/4 − (1/4)e^(2x)
⚠️
Integration by parts twice! ∫x²e^(−2x)dx requires IBP twice. Don't forget the +C and don't forget to apply the initial condition to find C specifically.
6
✗ NO CALCULATOR
Part (a) — 2 points
f(0) = 4 and P₁(−1/2) = 3. Show that f′(0) = −2.
Core Concept: First-Degree Taylor Polynomial
\( P_1(x) = f(0) + f'(0)\cdot x = 4 + f'(0)\cdot x \)
Use the given value P₁(−1/2) = 3 to solve for f′(0).
✦ Model Answer
\( P_1(x) = f(0) + f'(0)\cdot x = 4 + f'(0)\cdot x \)
\( P_1\!\left(-\tfrac{1}{2}\right) = 4 + f'(0)\cdot\left(-\tfrac{1}{2}\right) = 3 \)
\( -\tfrac{1}{2}f'(0) = -1 \implies f'(0) = -2 \qquad \square \)
\( P_1\!\left(-\tfrac{1}{2}\right) = 4 + f'(0)\cdot\left(-\tfrac{1}{2}\right) = 3 \)
\( -\tfrac{1}{2}f'(0) = -1 \implies f'(0) = -2 \qquad \square \)
f′(0) = −2 ✓
Part (b) — 3 points
f″(0) = −3, f‴(0) = 1/3. Find P₃(x).
Core Concept: nth-Degree Taylor Polynomial Formula
\( P_n(x) = \displaystyle\sum_{k=0}^{n}\dfrac{f^{(k)}(0)}{k!}x^k \)
\( P_3(x) = f(0) + f'(0)x + \dfrac{f''(0)}{2!}x^2 + \dfrac{f'''(0)}{3!}x^3 \)
✦ Model Answer
Using f(0)=4, f′(0)=−2, f″(0)=−3, f‴(0)=1/3:
\( P_3(x) = 4 + (-2)x + \dfrac{-3}{2!}x^2 + \dfrac{1/3}{3!}x^3 \)
\( = 4 - 2x - \dfrac{3}{2}x^2 + \dfrac{1}{18}x^3 \)
\( P_3(x) = 4 + (-2)x + \dfrac{-3}{2!}x^2 + \dfrac{1/3}{3!}x^3 \)
\( = 4 - 2x - \dfrac{3}{2}x^2 + \dfrac{1}{18}x^3 \)
P₃(x) = 4 − 2x − (3/2)x² + (1/18)x³
⚠️
Don't forget the factorial denominators! The coefficient of xⁿ is f⁽ⁿ⁾(0)/n!, not just f⁽ⁿ⁾(0). Also: 3! = 6, so (1/3)/6 = 1/18.
Part (c) — 4 points
h′(x) = f(2x), h(0) = 7. Find the third-degree Taylor polynomial for h about x = 0.
Core Concept: Building Taylor Polynomial via Integration
Since h′(x) = f(2x), the Taylor polynomial of h′ is P₃(2x) (substitute 2x for x in P₃). Then integrate term by term and apply h(0) = 7.
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Key insight: The Taylor polynomial for f(2x) about x=0 is just P₃(x) with x replaced by 2x:
\( h'(x) = f(2x) \approx 4 - 2(2x) - \tfrac{3}{2}(2x)^2 + \tfrac{1}{18}(2x)^3 \)
\( = 4 - 4x - 6x^2 + \tfrac{4}{9}x^3 \)
Then integrate to get polynomial for h(x), and use h(0) = 7 for the constant.
\( h'(x) = f(2x) \approx 4 - 2(2x) - \tfrac{3}{2}(2x)^2 + \tfrac{1}{18}(2x)^3 \)
\( = 4 - 4x - 6x^2 + \tfrac{4}{9}x^3 \)
Then integrate to get polynomial for h(x), and use h(0) = 7 for the constant.
1
Write Taylor poly for h′(x) = f(2x) by substituting 2x into P₃(x)
2
Integrate term by term (this gives a degree-4 polynomial for h — but we only need degree 3)
3
Apply h(0) = 7 to find the constant of integration
4
Keep only through x³ term for the 3rd-degree polynomial
✦ Model Answer
\( h'(x) = f(2x) = 4 - 4x - 6x^2 + \dfrac{4}{9}x^3 + \cdots \)
Integrate to find h(x):
\( h(x) = C + 4x - 2x^2 - 2x^3 + \dfrac{1}{9}x^4 + \cdots \)
Apply h(0) = 7: C = 7.
Third-degree Taylor polynomial for h:
\( T_3(x) = 7 + 4x - 2x^2 - 2x^3 \)
Integrate to find h(x):
\( h(x) = C + 4x - 2x^2 - 2x^3 + \dfrac{1}{9}x^4 + \cdots \)
Apply h(0) = 7: C = 7.
Third-degree Taylor polynomial for h:
\( T_3(x) = 7 + 4x - 2x^2 - 2x^3 \)
T₃(x) = 7 + 4x − 2x² − 2x³
⚠️
Substitution then integrate — not integrate then substitute! Replace x with 2x in P₃ FIRST, then integrate to get h. Going in the wrong order gives a wrong answer.
EXAM STRATEGY
Top 10 Rules for a Perfect FRQ Score
1
Always write units — tons/hr, oz/min, etc. Units on every numerical answer involving rates or amounts.
2
Justify extrema with sign changes — say f′ changes from + to − (or − to +), not just f′ = 0.
3
Invoke theorems by name — "By the Mean Value Theorem…" or "By L'Hôpital's Rule…" earns points.
4
Check BOTH endpoints for absolute extrema on a closed interval.
5
For Taylor polynomials: Don't forget n! in the denominator of each coefficient.
6
Show calculator setup for Part A — write the integral or derivative expression before stating the numerical answer.
7
Polar area formula: (1/2)∫r² dθ — the 1/2 is non-negotiable.
8
Euler's method: Show each step's slope calculation explicitly — partial credit is available.
9
For Riemann sums: State the subintervals, widths, and which points (left/right/midpoint) you're using.
10
Concave down + increasing: requires f′ > 0 AND f′ decreasing — BOTH conditions simultaneously.