AP Unit 5 Solutions
Worked solutions to all thirty Unit 5 problems. Each solution gives the mathematical steps and the reason for using them. Compare your setup and justification as well as the final answer.
Problem 1
Verify MVT for f(x) = x² on [1,3], and find every c in the open interval satisfying its conclusion.
A polynomial is continuous on [1,3] and differentiable on (1,3).
Why: These are exactly the MVT hypotheses.
Average rate = (9 − 1)/(3 − 1) = 4.
Why: Use endpoint function values for the secant slope.
Solve 2c = 4 to get c = 2, inside (1,3).
Why: The instantaneous derivative must match the secant slope at an interior point.
Final answer
The hypotheses hold; c = 2.
Review the teaching: Topic 5.1
Problem 2
Can MVT be applied to f(x) = |x| on [−1,1]? Does any interior derivative equal the endpoint secant slope?
f is continuous but has a corner at zero.
Why: MVT requires differentiability at every interior point.
Endpoint secant slope is [1 − 1]/2 = 0.
Why: The equal endpoint values give a horizontal secant.
f′ = −1 for x < 0 and +1 for x > 0, and f′(0) does not exist.
Why: There is no matching derivative here; the failed hypothesis prevented the guarantee.
Final answer
No: f is not differentiable at 0. The secant slope is 0, but no existing derivative in (−1,1) equals 0.
Review the teaching: Topic 5.1
Problem 3
A temperature function is continuous on [2,6] and differentiable on (2,6). T(2) = 30°C and T(6) = 20°C, with t in minutes. What instantaneous rate is guaranteed, and where? Is the time unique?
The stated smoothness satisfies the MVT hypotheses.
Why: Do not infer them merely from the two data values.
Compute (20 − 30)/(6 − 2) = −2.5.
Why: This is the average temperature change per minute.
MVT guarantees at least one interior c with this derivative value.
Why: It proves existence, not a particular time or uniqueness.
Final answer
T′(c) = −2.5 °C/min for at least one c in (2,6); uniqueness is not guaranteed.
Review the teaching: Topic 5.1
Problem 4
Compare f(x) = x on [0,1] and on (0,1). Give their absolute extrema, and explain why the domain changes the answer.
On [0,1], continuity and a closed bounded interval satisfy EVT.
Why: The theorem guarantees attained extrema.
The increasing function reaches its smallest and largest outputs at the endpoints.
Why: The endpoints belong to the closed domain.
On (0,1), every input has smaller and larger allowed neighbors.
Why: The limiting endpoint values are approached but never attained.
Final answer
On [0,1], minimum 0 at 0 and maximum 1 at 1. On (0,1), neither is attained, so there is no absolute minimum or maximum.
Review the teaching: Topic 5.2
Problem 5
For f(x) = x^(2/3) and g(x) = x³ on the real line, classify the critical point at zero for each. Explain why “critical” does not mean “extremum.”
For x ≠ 0, f′(x) = (2/3)x^(−1/3); it is undefined at zero, while f(0) exists.
Why: A defined point with nonexistent derivative is critical.
f(x) ≥ 0 and f(0) = 0.
Why: Direct value comparison proves a minimum.
g′(0) = 0, but g has negative values to the left and positive to the right.
Why: The cubic increases through zero, demonstrating a critical point without an extremum.
Final answer
f has a nondifferentiable local/absolute minimum at zero; g has a stationary critical point at zero with neither local extremum.
Review the teaching: Topic 5.2
Problem 6
A differentiable function has f′(x) = (x + 1)(x − 2). Determine all increasing and decreasing intervals.
Zeros of f′ are −1 and 2.
Why: They split the real line into possible sign intervals.
The factor signs give f′ signs +, −, + on the three intervals.
Why: No other zeros or derivative breaks occur in this polynomial.
Translate positive to increasing and negative to decreasing.
Why: The first derivative controls the direction of function change.
Final answer
Increasing on (−∞,−1) and (2,∞); decreasing on (−1,2).
Review the teaching: Topic 5.3
Problem 7
A differentiable function has exact derivative values in the table. What can you conclude at each listed input, and can the table prove increasing behavior throughout [0,4]?
| x | 0 | 2 | 4 |
|---|---|---|---|
| f′(x) | 3 | 1 | 2 |
Read all three derivatives as positive instantaneous slopes.
Why: These are derivative values, not function heights.
A derivative could be negative at an unsampled input.
Why: A finite table does not control every value between its entries.
Request interval information before making an interval conclusion.
Why: A proof of increase requires suitable derivative-sign or direct comparison evidence throughout the interval.
Final answer
The slopes are positive at all three inputs; the table alone does not prove that f increases throughout [0,4].
Review the teaching: Topic 5.3
Problem 8
For a differentiable function with f′(x) = (x − 1)²(x + 2), classify the critical numbers by the first derivative test.
The critical numbers are −2 and 1.
Why: Both make the derivative zero.
f′ is negative before −2 and positive between −2 and 1 and after 1.
Why: The squared factor is positive away from 1 and does not change sign there.
A negative-to-positive change gives a minimum at −2; no sign change gives neither at 1.
Why: The first derivative test classifies behavior on both sides.
Final answer
Local minimum at x = −2; neither local maximum nor minimum at x = 1.
Review the teaching: Topic 5.4
Problem 9
A continuous derivative graph passes linearly through (−1,2), (0,0), (1,−2), (2,0), and (3,2). Classify the relative extrema of f in (−1,3).
f′ is positive on (−1,0), negative on (0,2), positive on (2,3).
Why: Read derivative height relative to the horizontal axis.
At zero, the sign changes positive to negative.
Why: The function rises then falls, giving a local maximum.
At two, the sign changes negative to positive.
Why: The function falls then rises, giving a local minimum.
Final answer
Local maximum at x = 0; local minimum at x = 2.
Review the teaching: Topic 5.4
Problem 10
Find all absolute extrema of f(x) = x³ − 3x on [−2,3]. Show a complete candidates comparison.
f is continuous; f′ = 3(x² − 1) gives interior critical numbers −1 and 1.
Why: EVT and the candidate method apply on the closed interval.
Evaluate endpoints and critical numbers: f(−2) = −2, f(−1) = 2, f(1) = −2, f(3) = 18.
Why: Every candidate must be checked in the original function.
Compare the four outputs and report both locations of the minimum.
Why: An absolute extremum may be attained more than once.
Final answer
Absolute maximum 18 at x = 3; absolute minimum −2 at x = −2 and x = 1.
Review the teaching: Topic 5.5
Problem 11
Find all absolute extrema of f(x) = |x − 1| on [−1,3]. Include the nondifferentiable candidate.
The function is continuous, with an interior corner at x = 1.
Why: The corner is a candidate even though f′ is undefined.
Evaluate f(−1) = 2, f(1) = 0, f(3) = 2.
Why: Include both endpoints and the interior critical number.
Compare values to identify the minimum and tied maxima.
Why: The candidates test concerns function outputs.
Final answer
Minimum 0 at x = 1; maximum 2 at x = −1 and x = 3.
Review the teaching: Topic 5.5
Problem 12
A twice-differentiable function on the real line has f″(x) = x(x − 2). Determine concavity intervals, where f′ increases/decreases, and the x-coordinates of inflection points.
Second-derivative zeros are 0 and 2; signs are +, −, +.
Why: The factored expression supplies complete sign intervals.
Positive f″ means f′ increases and f is concave up; negative means the opposite.
Why: Concavity is the change in first-derivative slopes.
At both zero and two the concavity changes, and f is continuous there.
Why: These satisfy the inflection-point definition; f″ = 0 alone would not suffice.
Final answer
Concave up and f′ increasing on (−∞,0) and (2,∞); concave down and f′ decreasing on (0,2). Inflections at x = 0 and x = 2.
Review the teaching: Topic 5.6
Problem 13
Compare the origin for f(x) = x⁴ and g(x) = x^(1/3). Which has an inflection there? Explain using concavity, including any undefined derivatives.
f″(x) = 12x² is positive on both sides of zero.
Why: Its zero at the origin does not create a concavity change.
For x ≠ 0, g″(x) = −2/[9x^(5/3)], positive left of zero and negative right of zero.
Why: Odd-root power signs determine the concavity on each side.
g is continuous at zero, so the concavity change gives an inflection point there.
Why: An inflection can exist even if the first and second derivatives do not exist at that point.
Final answer
Only g has an inflection at the origin. f stays concave up; g changes from concave up to concave down despite undefined derivatives at zero.
Review the teaching: Topic 5.6
Problem 14
Use the second derivative test to classify critical points of f(x) = x³ − 3x, including their function values.
f′ = 3x² − 3 gives c = −1 and 1.
Why: A stationary-point condition is required before the second derivative test.
f″ = 6x gives f″(−1) = −6 and f″(1) = 6.
Why: Negative/positive values at stationary points give maximum/minimum.
Evaluate f(−1) = 2 and f(1) = −2.
Why: Ordered pairs need heights from the original function.
Final answer
Local maximum (−1,2); local minimum (1,−2).
Review the teaching: Topic 5.7
Problem 15
Both f(x) = x⁴ and g(x) = x³ have f′(0) = g′(0) = 0 and f″(0) = g″(0) = 0. What does the second derivative test say, and what actually happens?
The second derivative is zero at both stationary points.
Why: A zero second derivative supplies no classification by this test.
f′ = 4x³ changes from negative to positive.
Why: The first derivative test establishes a minimum of f.
g′ = 3x² stays positive on both sides.
Why: The function g increases through zero without an extremum.
Final answer
The test is inconclusive for both. f has a local minimum; g has neither local maximum nor minimum.
Review the teaching: Topic 5.7
Problem 16
For f(x) = x + 9/x on (0,∞), find and justify the absolute minimum. State why a local conclusion extends to the entire interval.
f′ = 1 − 9/x² has exactly one critical number, x = 3, on the positive domain.
Why: The negative root is outside the domain and the function is continuous for x > 0.
f″ = 18/x³, so f″(3) > 0 and there is a local minimum.
Why: The second derivative test applies at the stationary point.
The sole critical point is an absolute minimum: f′ is negative on (0,3) and positive on (3,∞).
Why: The unique-critical-point criterion is corroborated by complete interval sign evidence.
f(3) = 3 + 3 = 6.
Why: Evaluate the original objective for its minimum value.
Final answer
Absolute minimum 6 at x = 3.
Review the teaching: Topic 5.7
Problem 17
For f(x) = x³ − 3x, give the labeled features needed for a calculus sketch: intercepts, extrema, concavity, and end behavior.
Factor f = x(x² − 3) to find intercepts; f′ = 3(x² − 1) gives stationary numbers ±1.
Why: Function zeros and derivative zeros describe different graph features.
Derivative signs +, −, + give the two extrema; evaluate their heights 2 and −2.
Why: A sketch needs both locations and output values.
f″ = 6x changes sign at zero, giving the inflection and concavity intervals.
Why: Second derivative controls the bending.
The positive leading cubic term controls opposite infinite ends.
Why: End behavior completes the sketch beyond the local landmarks.
Final answer
Intercepts −√3, 0, √3; local max (−1,2), local min (1,−2); concave down for x < 0 and up for x > 0; inflection (0,0); f → −∞ left and +∞ right.
Review the teaching: Topic 5.8
Problem 18
The derivative is f′(x) = |x| − 1, with the graph shown. f(0) = 2. Describe extrema, concavity, the inflection, and the anchor for a sketch of f. Is f differentiable at zero?
Derivative signs are + for x < −1, − for −1 < x < 1, and + for x > 1.
Why: Derivative height gives function increase/decrease and sign-change extrema.
The derivative graph slopes downward for x < 0 and upward for x > 0.
Why: Derivative slope gives the function’s concavity.
The continuous f changes concavity at the known point (0,2).
Why: The given anchor fixes the inflection height; a second derivative at zero is not needed for this definition.
f′(0) exists and is −1, even though the derivative graph has a corner.
Why: A corner in f′ affects f″, not the existence of f′ itself.
Final answer
Local max at x = −1; local min at x = 1; concave down for x < 0 and up for x > 0; inflection (0,2). f is differentiable at zero with f′(0) = −1, though f″(0) does not exist.
Problem 19
On an interval, f′ > 0 and f″ < 0. Explain the behavior of f and f′. Can you determine the sign of f?
f′ > 0 means positive slopes, so f increases.
Why: The first derivative’s sign controls function direction.
f″ < 0 means f′ decreases, so f is concave down.
Why: The second derivative controls slope change.
Adding any constant leaves both derivatives unchanged.
Why: Derivative information does not fix the function’s vertical height or sign.
Final answer
f increases and is concave down; f′ decreases. The sign of f is undetermined.
Review the teaching: Topic 5.9
Problem 20
These are exact instantaneous values, and f′ and f″ are continuous near each listed input. Classify the local direction and concavity near each listed x; then say whether the three rows alone establish behavior on the full intervals between rows.
| x | f′(x) | f″(x) |
|---|---|---|
| −1 | −2 | 3 |
| 0 | 4 | −1 |
| 1 | −3 | −2 |
Use f′ signs −, +, − for direction near the listed inputs.
Why: Continuity keeps the nonzero first-derivative sign fixed on a sufficiently small neighborhood.
Use f″ signs +, −, − for concavity on sufficiently small neighborhoods of those inputs.
Why: Continuity of f″ keeps each nonzero second-derivative sign unchanged on a sufficiently small neighborhood.
Do not infer signs at every unlisted input.
Why: Finite samples cannot exclude intervening zeros or sign changes.
Final answer
At −1: decreasing, concave up. At 0: increasing, concave down. At 1: decreasing, concave down. The rows alone do not establish full interval behavior.
Review the teaching: Topic 5.9
Problem 21
A rectangular garden borders a wall, so only three sides need fencing. With 96 m of fencing, let x be a perpendicular side. Build the area model, give its feasible domain, and find the maximizing dimensions and area.
The constraint is 2x + y = 96, so y = 96 − 2x.
Why: Only the three fenced sides count.
A(x) = x(96 − 2x), and positivity gives 0 < x < 48.
Why: The objective is area and both dimensions must be positive.
A′ = 96 − 4x changes from positive to negative at x = 24.
Why: The complete derivative sign change proves the maximum.
Then y = 48 and A = 24·48 = 1152.
Why: Recover all physical quantities and their units.
Final answer
A(x) = x(96 − 2x), 0 < x < 48; perpendicular sides 24 m, parallel side 48 m; maximum area 1152 m².
Review the teaching: Topic 5.10
Problem 22
An open box is made by cutting equal squares of side h from the corners of a 20-cm by 12-cm sheet and folding the sides. Write its volume objective, feasible domain, and derivative equation for candidates. You need not solve it.
The base dimensions become 20 − 2h and 12 − 2h; the height is h.
Why: Each original side loses one corner square at each end.
Multiply to obtain V = 240h − 64h² + 4h³, with 0 < h < 6.
Why: All three box dimensions must be positive.
V′ = 240 − 128h + 12h²; set it equal to zero.
Why: Stationary candidates come from the derivative of the correctly constrained objective.
Final answer
V(h) = h(20 − 2h)(12 − 2h), 0 < h < 6; candidates satisfy 12h² − 128h + 240 = 0.
Review the teaching: Topic 5.10
Problem 23
A closed cylinder must hold 54π cm³. Minimize its surface area S = 2πr² + 2πrh, with positive dimensions. Give radius, height, area, and a global justification.
Volume πr²h = 54π gives h = 54/r² and S(r) = 2πr² + 108π/r for r > 0.
Why: Use the volume constraint to form a one-variable objective.
S′ = 4πr − 108π/r² = 4π(r³ − 27)/r².
Why: Differentiate and expose a sign-friendly form.
S′ is negative for 0 < r < 3 and positive for r > 3.
Why: Complete interval sign evidence proves an absolute minimum.
At r = 3, h = 6 and S = 18π + 36π = 54π.
Why: Translate the optimum back into dimensions and the objective, with units.
Final answer
r = 3 cm, h = 6 cm, minimum surface area 54π cm².
Review the teaching: Topic 5.10 Topic 5.11
Problem 24
A rectangle must have area 100 m². Find positive side lengths that minimize its perimeter, with a global justification.
Let sides be x and y; xy = 100 gives y = 100/x, x > 0.
Why: The area constraint eliminates one variable.
P(x) = 2x + 200/x and P′ = 2 − 200/x².
Why: Perimeter is the objective, not area.
P′ is negative on (0,10) and positive on (10,∞).
Why: The perimeter decreases then increases across the whole feasible domain.
x = 10 gives y = 10 and P = 40.
Why: Report both dimensions and the minimum output with length units.
Final answer
10 m by 10 m; minimum perimeter 40 m.
Review the teaching: Topic 5.10 Topic 5.11
Problem 25
For the box model V(h) = h(20 − 2h)(12 − 2h), 0 < h < 6, find the maximizing h numerically to three decimals and the volume to three decimals. Justify that the chosen root is feasible and globally maximal.
V′ = 12h² − 128h + 240 gives roots h = (16 ± 2√19)/3.
Why: The derivative equation can be solved exactly or by a recorded calculator root search.
Only h = (16 − 2√19)/3 ≈ 2.427400704 lies in (0,6); the other root exceeds 6.
Why: Physical dimensions restrict candidate validity.
V′ changes from positive to negative at the feasible root and the boundary volume limits are zero.
Why: This proves the global maximum, not just a graph-window turn.
Use the unrounded h to evaluate V ≈ 262.682263531... and round as requested.
Why: Guard digits prevent volume error from early input rounding.
Final answer
h ≈ 2.427 cm; maximum volume ≈ 262.682 cm³.
Review the teaching: Topic 5.10 Topic 5.11
Problem 26
On the upper branch of x² + y² = 25, with −5 < x < 5, find the interior critical point and classify it using implicit derivative evidence.
Implicit differentiation gives y′ = −x/y, and y > 0 on this branch.
Why: The branch restriction ensures a finite slope except at excluded endpoints.
y′ = 0 only at x = 0, where the relation gives y = 5.
Why: The numerator zero and nonzero denominator identify the stationary point.
A second differentiation gives y″ = −[1 + (y′)²]/y < 0.
Why: The product and chain rules yield negative concavity on the upper branch.
The derivative changes positive to negative at zero, and the whole branch lies at y ≤ 5.
Why: Either complete sign evidence or the relation proves the absolute maximum.
Final answer
The interior critical point is (0,5), a local and absolute maximum on the upper branch.
Review the teaching: Topic 5.12
Problem 27
The implicit relation y² + xy = 6 passes through (1,2). Find y′ and y″ there and describe the local direction and concavity.
Differentiate to get (2y + x)y′ + y = 0, so y′ = −y/(2y + x).
Why: Product and chain rules apply, and the denominator is nonzero at (1,2).
At (1,2), y′ = −2/5.
Why: The point satisfies the original relation and supplies a valid local branch.
Differentiate again: (2y + x)y″ + 2(y′)² + 2y′ = 0.
Why: Differentiating the coefficient and final y produces both slope terms.
y″ = −[2(4/25) − 4/5]/5 = 12/125 > 0.
Why: Insert the first derivative only after forming the second-derivative equation; the signs distinguish direction from concavity.
Final answer
y′ = −2/5, y″ = 12/125; decreasing and concave up locally.
Review the teaching: Topic 5.12
Problem 28
Near (0,1), the relation y² = x³ + 1 uses its positive branch. Show that the point is critical, explain why the second derivative test is inconclusive, and classify extrema using first-derivative signs.
Implicit differentiation gives 2yy′ = 3x², hence y′ = 3x²/(2y).
Why: The positive branch has y > 0 near the point.
At (0,1), y′ = 0. Differentiate again: 2(y′)² + 2yy″ = 6x, giving y″ = 0 there.
Why: The second derivative equation retains the square of the first derivative.
The second derivative test supplies no classification when y″ = 0.
Why: Zero second derivative is inconclusive, not a proof of neither.
For small nonzero x on both sides, y′ > 0, so there is no sign change and no local extremum.
Why: The first derivative evidence makes the actual classification.
Final answer
y′(0) = 0 and y″(0) = 0. The second derivative test is inconclusive; there is neither local maximum nor minimum because the branch increases through the point.
Review the teaching: Topic 5.12
Problem 29
A differentiable function has f′(x) = sin x on [0,2π]. Determine increasing/decreasing intervals, interior relative extrema, concavity intervals, and inflection x-coordinates. Can you determine the inflection heights without an f-value?
sin x is positive on (0,π) and negative on (π,2π).
Why: The sign of f′ determines direction and its positive-to-negative crossing gives the interior maximum.
f″ = cos x, positive on the outer subintervals and negative on the middle one.
Why: The derivative of f′ supplies concavity.
Concavity changes at π/2 and 3π/2, where f is continuous.
Why: Sign changes justify inflections; zeros alone are not the proof.
Any vertical shift of f preserves f′ and f″.
Why: Derivative information alone cannot determine function heights.
Final answer
Increasing (0,π), decreasing (π,2π); local maximum at π. Concave up (0,π/2) and (3π/2,2π), down (π/2,3π/2); inflections at π/2 and 3π/2. Heights need a function anchor.
Review the teaching: Topic 5.3 Topic 5.4 Topic 5.6 Topic 5.8 Topic 5.9
Problem 30
Find all points on y = x² closest to (0,3). Minimize squared distance and give the actual minimum distance, with a global justification.
Squared distance is D² = x² + (x² − 3)² = x⁴ − 5x² + 9.
Why: The distance is nonnegative, so minimizing its square gives the same inputs.
Its derivative is 4x³ − 10x = 2x(2x² − 5), with candidates 0 and ±√(5/2).
Why: Differentiate the one-variable objective and solve every critical value.
At zero D² = 9; at both nonzero candidates D² = 11/4. As |x| → ∞, D² → ∞.
Why: Comparison of every critical value plus unbounded endpoint behavior proves the global minimum.
The curve gives y = 5/2 at both minimizers, and D = √(11/4) = √11/2.
Why: Recover the points and take the square root to report distance rather than squared distance.
Final answer
Closest points (−√(5/2), 5/2) and (√(5/2), 5/2); minimum distance √11/2.
Review the teaching: Topic 5.10 Topic 5.11