Solutions — Second Derivative, Concavity and Inflection Points
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f(x) = 3x4 − 5x³ + 2x − 7
Step 1 — First derivative:
f′(x) = 12x³ − 15x² + 2
Step 2 — Second derivative:
f″(x) = 36x² − 30x
f″(x) = 36x² − 30x -
f(x) = 2ex + 5 sin x − 3x²
Step 1 — First derivative:
f′(x) = 2ex + 5 cos x − 6x
Step 2 — Second derivative:
d/dx(2ex) = 2ex
d/dx(5 cos x) = −5 sin x
d/dx(−6x) = −6
f″(x) = 2ex − 5 sin x − 6 -
f(x) = x³ − 6x
f′(x) = 3x² − 6
f″(x) = 6x
At x = 2: f″(2) = 12 > 0 ⇒ concave up
At x = −1: f″(−1) = −6 < 0 ⇒ concave down -
f(x) = 4 cos x
f′(x) = −4 sin x
f″(x) = −4 cos x
On the interval 0 < x < π:
• For 0 < x < π/2: cos x > 0, so f″(x) = −4 cos x < 0 ⇒ concave down
• For π/2 < x < π: cos x < 0, so f″(x) = −4 cos x > 0 ⇒ concave up
Concavity changes at x = π/2 (where f″ = 0). So f(x) = 4 cos x has an inflection point at (π/2, 0). -
f(x) = x³ − 3x² − 9x + 2
f′(x) = 3x² − 6x − 9
f″(x) = 6x − 6
Step 1 — Set f″(x) = 0:
6x − 6 = 0 ⇒ x = 1
Step 2 — Sign test:
f″(0) = 6(0) − 6 = −6 < 0 (concave down to the left)
f″(2) = 6(2) − 6 = 6 > 0 (concave up to the right)
Sign changes ⇒ x = 1 is a genuine inflection point.
Step 3 — Coordinates:
f(1) = 1 − 3 − 9 + 2 = −9
Inflection point: (1, −9) -
f(x) = x4 − 4x³ + 6x²
f′(x) = 4x³ − 12x² + 12x
f″(x) = 12x² − 24x + 12 = 12(x² − 2x + 1) = 12(x − 1)²
Set f″(x) = 0: 12(x − 1)² = 0 ⇒ x = 1
Sign test:
f″(0) = 12(0−1)² = 12 > 0
f″(2) = 12(2−1)² = 12 > 0
The sign does NOT change at x = 1 ⇒ the curve remains concave up on both sides.
Conclusion: No inflection points. The function is concave up everywhere. -
f(x) = 2x³ + 3x² − 12x
f′(x) = 6x² + 6x − 12
f″(x) = 12x + 6
Set f″(x) = 0: 12x + 6 = 0 ⇒ x = −1/2
Sign test:
f″(−1) = 12(−1) + 6 = −6 < 0 ⇒ concave down
f″(0) = 6 > 0 ⇒ concave up
Sign changes at x = −1/2 ⇒ inflection point exists.
Coordinates:
f(−1/2) = 2(−1/8) + 3(1/4) − 12(−1/2) = −1/4 + 3/4 + 6 = 1/2 + 6 = 13/2
Concave down on (−∞, −1/2), concave up on (−1/2, +∞)
Inflection point: (−1/2, 13/2) -
f(x) = x4
f′(x) = 4x³
f″(x) = 12x²
The student correctly identifies that f″(0) = 12(0)² = 0, so x = 0 is a candidate.
Sign test:
f″(−1) = 12(−1)² = 12 > 0 (concave up)
f″(1) = 12(1)² = 12 > 0 (concave up)
The sign of f″ does NOT change at x = 0. The concavity is upward on both sides.
The student is incorrect. While f″(0) = 0 is a necessary condition for an inflection point, a sign change of f″ is required to confirm one. Since no sign change occurs at x = 0, there is no inflection point. The curve f(x) = x4 is concave up everywhere. -
s(t) = t4 − 8t³ + 18t²
v(t) = s′(t) = 4t³ − 24t² + 36t
a(t) = s″(t) = 12t² − 48t + 36 = 12(t² − 4t + 3) = 12(t − 1)(t − 3)
(a) Acceleration zero when:
12(t − 1)(t − 3) = 0 ⇒ t = 1 s and t = 3 s
(b) Sign test for a(t):
t = 0.5: a = 12(0.5−1)(0.5−3) = 12(−0.5)(−2.5) = 15 > 0
t = 2: a = 12(2−1)(2−3) = 12(1)(−1) = −12 < 0
t = 4: a = 12(4−1)(4−3) = 12(3)(1) = 36 > 0
Sign changes at both t = 1 and t = 3 ⇒ both are inflection points of s(t).
At t = 1, acceleration changes from positive to negative (object's velocity was increasing, now decreasing).
At t = 3, acceleration changes from negative to positive (velocity was decreasing, now increasing). -
f(x) = ex − 4x
f′(x) = ex − 4
f″(x) = ex
(a) No inflection points:
f″(x) = ex. Since ex > 0 for all real x, f″(x) > 0 everywhere. The sign never changes, so there are no inflection points.
(b) Interval of concavity:
Since f″(x) = ex > 0 for all x, the function is concave up on (−∞, +∞).
(c) Interpretation:
Because f is always concave up, its graph has the shape of a bowl — it curves upward continuously with no inflection or S-bend. Any tangent line to the curve lies entirely below the graph. The function has exactly one global minimum (found by solving f′(x) = 0, i.e. ex = 4, x = ln 4) and is decreasing to the left of that minimum and increasing to the right.