Curve Behaviour: Maxima, Minima and Points of Inflection

• A function f is increasing (ascending) on an interval when f′(x) > 0 at every point of it, and decreasing (descending) when f′(x) < 0.
• Procedure: compute f′(x), solve f′(x) = 0 for the critical numbers, then test the sign of f′ in each interval bounded by them (substitute a convenient value, or use the factor-sign chart).
• f is never increasing at a point where f′ does not exist (e.g. a cusp) — always state the domain first.

Worked example

f(x) = x³ − 6x² + 9x + 2 ⟹ f′(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3). f′ = 0 at x = 1, 3.

• x < 1: f′ > 0 (increasing) · 1 < x < 3: f′ < 0 (decreasing) · x > 3: f′ > 0 (increasing).
• So f increases on (−∞, 1) ∪ (3, ∞) and decreases on (1, 3).

The strictly-increasing condition

If f(x) = x³ + 3x² + kx is increasing on ℝ, then f′(x) = 3x² + 6x + k ≥ 0 for all x, so its discriminant ≤ 0: 36 − 12k ≤ 0 ⟹ k ≥ 3.

A stationary point is a point on the curve where f′(x) = 0; the tangent there is horizontal (parallel to the x-axis).

Worked example

f(x) = x³ − 6x² + 9x + 2, stationary at x = 1 and x = 3.

• At x = 1 the sign goes + → −, so maximum at (1, f(1)) = (1, 6).
• At x = 3 the sign goes − → +, so minimum at (3, 2).

Note that a maximum value is local: the local maximum 6 is smaller than the value at x = 5, which is 27.

Compute f″(x) and substitute the x-coordinate of each stationary point:

• f″(a) < 0 ⟹ f has a local maximum at x = a.
• f″(a) > 0 ⟹ f has a local minimum at x = a.
• f″(a) = 0 ⟹ the test fails; go back to the sign of f′ (it may be a maximum, a minimum or a point of inflection).

Worked example

f(x) = 2x³ − 9x² + 12x − 3 ⟹ f′ = 6x² − 18x + 12 = 6(x − 1)(x − 2) ⟹ stationary at x = 1, 2. f″ = 12x − 18.

• f″(1) = −6 < 0 ⟹ maximum at (1, 2).
• f″(2) = 6 > 0 ⟹ minimum at (2, 1).

Finding constants from a given extremum

If f(x) = ax³ + bx² has a maximum at (1, 3), then f(1) = 3 ⟹ a + b = 3, and f′(1) = 3a + 2b = 0 ⟹ solving, a = −6, b = 9.

• The curve is concave up (∪, "holds water") where f″(x) > 0 and concave down (∩) where f″(x) < 0.
• A point of inflection is where the concavity changes: solve f″(x) = 0 and check that the sign of f″ really changes across the root.
• At a point of inflection the tangent crosses the curve, and the value of f′ there is a maximum or a minimum (the slope is extreme).
• For a cubic, the point of inflection is the midpoint of the segment joining the maximum and the minimum; its x-coordinate is the average of the two critical numbers.

Worked example

f(x) = x³ − 6x² + 9x + 2 ⟹ f″ = 6x − 12 = 0 ⟹ x = 2, f(2) = 4. Sign of f″: negative before 2, positive after ⟹ point of inflection (2, 4) — indeed the midpoint of (1, 6) and (3, 2).

When there is no inflection point

f(x) = x⁴ has f″ = 12x², which is zero at x = 0 but positive on both sides, so x = 0 is a minimum, not an inflection point. Zero second derivative alone is never enough.

Draw y = f(x) by collecting the information in this order:

1. Domain and any values excluded (denominators, even roots, logarithms).
2. Intercepts: set x = 0 for the y-intercept and solve f(x) = 0 for the x-intercepts.
3. Symmetry: even function → symmetric about the y-axis; odd → symmetric about the origin.
4. Asymptotes: vertical where the denominator vanishes, horizontal from the limit as x → ±∞ (compare degrees).
5. Stationary points with their nature, from f′ and f″.
6. Points of inflection and the concavity on each side.
7. End behaviour: the sign of the leading term as x → ±∞, then join the points smoothly.

Method: write the quantity to be optimised as a function of one variable using the constraint, state the admissible interval, differentiate, set f′ = 0, justify the nature, then answer the question in words with units.

• Fencing: a rectangular field of area 1,200 m² is to be fenced. Perimeter P = 2(x + 1200/x) ⟹ P′ = 2 − 2400/x² = 0 ⟹ x = √1200 ≈ 34.6 m, P minimum ≈ 138.6 m.
• Open box: from a 20 cm × 20 cm square, cut squares of side x and fold. V = x(20 − 2x)² ⟹ V′ = (20 − 2x)(20 − 6x) ⟹ x = 10/3 cm gives the greatest volume ≈ 592.6 cm³.
• Closest point: the point on y² = 4x nearest (3, 0): minimise D² = (x − 3)² + y² = (x − 3)² + 4x ⟹ d/dx = 2x − 2 = 0 ⟹ x = 1 ⟹ points (1, ±2), distance 2.
• Revenue: p = 100 − 0.5x, R = px = 100x − 0.5x² ⟹ R′ = 100 − x = 0 ⟹ x = 100 units, price EGP 50, R = EGP 5,000.

1. Find the intervals of increase and decrease of f(x) = 2x³ − 3x² − 12x + 5. — f′ = 6(x − 2)(x + 1): increasing on (−∞, −1) ∪ (2, ∞), decreasing on (−1, 2).
2. Locate and classify the stationary points of f(x) = x⁴ − 4x³. — f′ = 4x²(x − 3) ⟹ minimum (3, −27); x = 0 is a stationary point of inflection.
3. The curve y = ax² + bx passes through (1, 4) and has a minimum at x = 2. Find a, b. — a + b = 4 and 2a(2) + b = 0 ⟹ a = 4, b = −8.
4. Find the point of inflection of y = x³ − 3x² + 4. — y″ = 6x − 6 ⟹ (1, 2).
5. A farmer has 240 m of fencing for a rectangular plot against a river (no fence on the river side). Greatest area? — A = x(240 − 2x), A′ = 240 − 4x ⟹ x = 60, other side 120, area 7,200 m².
6. Find the maximum value of f(x) = x e^−x. — f′ = e^−x(1 − x) = 0 ⟹ x = 1, f″(1) = −e⁻¹ < 0 ⟹ maximum 1/e ≈ 0.368.
7. For what k does x³ + kx have no stationary points? — f′ = 3x² + k ≥ 0 always ⟹ k ≥ 0.