Topic Review — Further Applications of Differentiation — Solutions

This review covers all three lessons in Further Applications of Differentiation: second derivative, concavity and inflection points; second derivative test for stationary points; and optimisation problems.

Review Questions

Find f″(x) for f(x) = 4x⁵ − 3x³ + 7x − 2.
f′(x) = 20x⁴ − 9x² + 7
f″(x) = 80x³ − 18x
Determine the concavity of f(x) = 3x² − 12x + 1 for all x. Does the function have any inflection points?
f′(x) = 6x − 12; f″(x) = 6 > 0 for all x.
The function is concave up everywhere and has no inflection points.
Find all inflection points of f(x) = 2x³ − 9x² + 12x − 4.
f′(x) = 6x² − 18x + 12; f″(x) = 12x − 18
f″(x) = 0 ⇒ x = 3/2
f″(1) = −6 < 0 (concave down); f″(2) = 6 > 0 (concave up) ⇒ sign changes
f(3/2) = 2(27/8) − 9(9/4) + 12(3/2) − 4 = 27/4 − 81/4 + 18 − 4 = −54/4 + 14 = −13.5 + 14 = 1/2
Inflection point: (3/2, 1/2)
Find the stationary points of f(x) = x³ − 3x² − 9x + 5 and classify each using the second derivative test.
f′(x) = 3x² − 6x − 9 = 3(x² − 2x − 3) = 3(x − 3)(x + 1) = 0
Stationary points at x = 3 and x = −1.
f″(x) = 6x − 6
f″(3) = 12 > 0 ⇒ local minimum. f(3) = 27 − 27 − 27 + 5 = −22. Local min at (3, −22).
f″(−1) = −12 < 0 ⇒ local maximum. f(−1) = −1 − 3 + 9 + 5 = 10. Local max at (−1, 10).
Use the second derivative test to classify all stationary points of f(x) = x⁴ − 8x² + 3.
f′(x) = 4x³ − 16x = 4x(x² − 4) = 4x(x−2)(x+2) = 0 ⇒ x = 0, 2, −2
f″(x) = 12x² − 16
f″(0) = −16 < 0 ⇒ local maximum at (0, 3)
f″(2) = 48−16 = 32 > 0 ⇒ local minimum at (2, 16−32+3) = (2, −13)
f″(−2) = 32 > 0 ⇒ local minimum at (−2, −13)
A function has f′(x) = 3(x−2)². Show that x = 2 is a stationary point but is NOT a local maximum or minimum.
f′(2) = 3(0)² = 0 ⇒ x = 2 is a stationary point.
f″(x) = 6(x−2); f″(2) = 0. The second derivative test is inconclusive.
Sign of f′: for x < 2, f′(x) = 3(x−2)² ≥ 0; for x > 2, f′(x) = 3(x−2)² ≥ 0.
f′ does NOT change sign at x = 2 ⇒ x = 2 is a stationary point of inflection, not a local max or min.
Find the minimum value of f(x) = x² − 4x + 7 for x ≥ 0 using calculus.
f′(x) = 2x − 4 = 0 ⇒ x = 2
f″(x) = 2 > 0 ⇒ minimum at x = 2
f(2) = 4 − 8 + 7 = 3
Minimum value is 3 at x = 2.
A wire of length 24 cm is bent to form a rectangle. Find the dimensions that maximise the area.
Let width = x, length = y. Perimeter: 2x + 2y = 24 ⇒ y = 12 − x.
A(x) = x(12 − x) = 12x − x²; domain: 0 < x < 12
A′(x) = 12 − 2x = 0 ⇒ x = 6; y = 6
A″ = −2 < 0 ⇒ maximum
Dimensions: 6 cm × 6 cm (a square). Maximum area = 36 cm².
A particle moves with position s(t) = 2t³ − 9t² + 12t − 3 metres for t ≥ 0. Find all local maxima and minima of the position function.
s′(t) = 6t² − 18t + 12 = 6(t² − 3t + 2) = 6(t−1)(t−2) = 0 ⇒ t = 1 or t = 2
s″(t) = 12t − 18
s″(1) = −6 < 0 ⇒ local maximum: s(1) = 2−9+12−3 = 2 m
s″(2) = 6 > 0 ⇒ local minimum: s(2) = 16−36+24−3 = 1 m
Find the two positive numbers whose sum is 16 and whose sum of squares is minimum.
Let the numbers be x and 16 − x.
S = x² + (16−x)² = x² + 256 − 32x + x² = 2x² − 32x + 256
S′(x) = 4x − 32 = 0 ⇒ x = 8
S″ = 4 > 0 ⇒ minimum
The two numbers are 8 and 8. Minimum sum of squares = 64 + 64 = 128.
Find all inflection points of f(x) = x⁴ − 6x² and state the intervals of concavity.
f′(x) = 4x³ − 12x; f″(x) = 12x² − 12 = 12(x²−1) = 12(x−1)(x+1)
f″(x) = 0 ⇒ x = 1 or x = −1
Sign of f″: −∞ to −1: f″(−2) = 12(3) = 36 > 0 (concave up)
−1 to 1: f″(0) = −12 < 0 (concave down)
1 to ∞: f″(2) = 36 > 0 (concave up)
f(−1) = 1−6 = −5; f(1) = 1−6 = −5
Inflection points: (−1, −5) and (1, −5).
Concave up on (−∞,−1) and (1,+∞); concave down on (−1, 1).
A rectangular area is to be enclosed using 80 m of fencing. Two parallel internal dividers (parallel to the width) are also included. Find the dimensions that maximise the total enclosed area.
Let width = x m and length = y m. The fencing includes two lengths and four widths (2 outer + 2 dividers).
Constraint: 4x + 2y = 80 ⇒ y = 40 − 2x
A(x) = xy = x(40 − 2x) = 40x − 2x²; domain: 0 < x < 20
A′(x) = 40 − 4x = 0 ⇒ x = 10; y = 40 − 20 = 20
A″ = −4 < 0 ⇒ maximum
Width = 10 m, length = 20 m. Maximum area = 200 m².
For f(x) = e^x − 2x, find and classify all stationary points using the second derivative test. State the concavity for all x.
f′(x) = e^x − 2 = 0 ⇒ e^x = 2 ⇒ x = ln 2
f″(x) = e^x; f″(ln 2) = e^ln2 = 2 > 0 ⇒ local minimum
f(ln 2) = 2 − 2 ln 2
Since f″(x) = e^x > 0 for all x, the function is concave up everywhere with no inflection points.
Local (and global) minimum at (ln 2, 2 − 2 ln 2) ≈ (0.693, 0.614).
An open rectangular box has a square base and is to be constructed from 48 cm² of material (no lid). Find the dimensions that maximise the volume.
Let base side = x cm, height = h cm.
Surface area constraint: x² + 4xh = 48 ⇒ h = (48 − x²)/(4x)
Volume: V = x²h = x² · (48 − x²)/(4x) = x(48 − x²)/4 = (48x − x³)/4
Domain: x > 0 and 48 − x² > 0 ⇒ 0 < x < 4√3
V′(x) = (48 − 3x²)/4 = 0 ⇒ x² = 16 ⇒ x = 4
V″(x) = −6x/4 = −3x/2; V″(4) = −6 < 0 ⇒ maximum
h = (48 − 16)/16 = 32/16 = 2 cm
V(4) = (48×4 − 64)/4 = (192 − 64)/4 = 128/4 = 32 cm³
Base side = 4 cm, height = 2 cm. Maximum volume = 32 cm³.
A profit function is P(x) = −x³ + 9x² − 15x − 20 (dollars, x = hundreds of items). Use all three tools — stationary points, second derivative test, and inflection point — to give a complete analysis of P.
P′(x) = −3x² + 18x − 15 = −3(x² − 6x + 5) = −3(x−1)(x−5) = 0 ⇒ x = 1 or x = 5
P″(x) = −6x + 18
P″(1) = 12 > 0 ⇒ local minimum at x = 1: P(1) = −1+9−15−20 = −27
P″(5) = −12 < 0 ⇒ local maximum at x = 5: P(5) = −125+225−75−20 = 5
Inflection: P″(x) = 0 ⇒ −6x+18 = 0 ⇒ x = 3
P″(2) = 6 > 0; P″(4) = −6 < 0 ⇒ sign changes ⇒ inflection point at x = 3
P(3) = −27+81−45−20 = −11
Local minimum at (1, −27); local maximum at (5, 5); inflection point at (3, −11).
Profit is maximised (locally) when 500 items are sold, giving $5 profit.