One Variable Optimisation

Increasing and Decreasing Functions

The behaviour of a function across an interval is governed by the sign of its first derivative. When $f'(x) > 0$ on an interval I the function is increasing: larger inputs produce larger outputs. When $f'(x) < 0$ it is decreasing. This follows from the linear approximation:

When b > a, the term (b − a) is positive, so the sign of f'(a) determines whether f(b) exceeds f(a) or falls below it.

To locate intervals of increase and decrease, find where f'(x) = 0, which divides the real line into regions where the sign of f’ is constant, then test f’ in each region. For $f(x) = x³ − 2x² − 15x$, the first derivative factorises as:

Setting $f'(x) = 0$ gives $x = −5/3$ and x = 3. For $x < −5/3$ both factors are negative so $f'(x) > 0$. For $−5/3 < x < 3$ the first factor is positive and the second negative, so $f'(x) < 0. For x > 3$ both factors are positive and f'(x) > 0. The function is increasing on $(−∞, −5/3)$ and $(3, ∞)$, and decreasing on $(−5/3, 3)$.

Stationary Points

A stationary point occurs where f'(a) = 0: the tangent line is horizontal and the function is momentarily neither increasing nor decreasing. Every local maximum and local minimum must be a stationary point, though not every stationary point is a local extremum. Some are points of inflection.

For $f(x) = 2x³ − 9x² + 12x − 4$ the derivative is $f'(x) = 6x² − 18x + 12$. Setting this to zero and simplifying:

The stationary points occur at x = 1 and x = 2. Substituting back: $f(1) = 2 − 9 + 12 − 4 = 1$ and $f(2) = 16 − 36 + 24 − 4 = 0$, giving stationary points at (1, 1) and (2, 0).

Classifying Stationary Points: The First-Order Test

The nature of a stationary point is determined by how the sign of f’ changes as x passes through it. If f’ goes from positive to negative the function rises then falls, making it a local maximum. If f’ goes from negative to positive the function falls then rises, making it a local minimum. If the sign of f’ is the same on both sides, the function continues in the same direction and the point is a point of inflection.

For $f(x) = 2x³ − 9x² + 12x − 4$, with $f'(x) = 6(x − 1)(x − 2)$, the sign of f’ is positive for x < 1, negative for 1 < x < 2, and positive again for x > 2. At x = 1 the sign changes from + to −, giving a local maximum at (1, 1). At x = 2 the sign changes from − to +, giving a local minimum at (2, 0).

Elasticity and Revenue

The tools of differentiation apply directly to demand analysis. The elasticity of demand measures the proportional sensitivity of quantity demanded to a change in price:

A monopolist’s revenue is $R(p) = pq$ and its derivative with respect to price is:

When ε > 1, demand is elastic and R'(p) < 0: a price increase reduces revenue. When ε < 1, demand is inelastic and R'(p) > 0: a price increase raises revenue. Revenue is maximised where ε = 1 and R'(p) = 0.

For the demand function $q(p) = 20 − 2p$, the derivative is q'(p) = −2, giving:

Setting ε > 1 requires p/(10 − p) > 1. Since 10 − p > 0 for p < 10, this simplifies to p > 5. Demand is elastic when 5 < p ≤ 10 and inelastic when 0 ≤ p < 5. Revenue is maximised at p = 5, the boundary between the two regions. This can be confirmed directly: $R(p) = p(20 − 2p) = 20p − 2p²$, so:

At this price, $R(5) = 5 × 10 = 50$.

The Second-Order Derivative Test

At a stationary point where f'(a) = 0, the quadratic approximation reduces to:

Since (x − a)² > 0 for x ≠ a, the sign of f”(a) alone determines the shape of the function near the stationary point. If f”(a) > 0 the function curves upward and the point is a local minimum. If f”(a) < 0 it curves downward and the point is a local maximum. If f”(a) = 0 the test is inconclusive and the first-order test must be used instead.

For f(x) = x³ − 2x² − 15x, the stationary points are at x = −5/3 and x = 3, and the second derivative is f”(x) = 6x − 4. Evaluating at each stationary point:

The function values give a local maximum at (−5/3, 400/27) and a local minimum at (3, −36).

Convexity and Concavity

While the first derivative describes the direction of the function, the second derivative describes its curvature across an interval. A function is convex on an interval when f”(x) > 0 there: it curves upward and lies above its tangent lines. It is concave when f”(x) < 0: it curves downward and lies below its tangent lines. Local minima are found in convex regions; local maxima are found in concave regions.

For $f(x) = x³ − 2x² − 15x$, the second derivative is $f”(x) = 6x − 4$, which equals zero at x = 2/3. For x < 2/3, f”(x) < 0 and the function is concave. For x > 2/3, f”(x) > 0 and the function is convex.

Points of Inflection

A point of inflection is where a function switches between convex and concave. Two conditions must both hold: f”(a) = 0 and f” changes sign at x = a. The first condition alone is not sufficient. For f(x) = x⁴ − 1, f”(0) = 0 but f”(x) = 12x² ≥ 0 everywhere, so there is no sign change and x = 0 is a local minimum, not an inflection point.

For f(x) = 2x⁴ − 4x³ + 2x², differentiating twice gives:

Setting f”(x) = 0 and dividing through by 4 leads to 6x² − 6x + 1 = 0, which gives:

Label these a ≈ 0.211 and b ≈ 0.789. Writing f”(x) = 24(x − a)(x − b), the second derivative is positive for x < a, negative for a < x < b, and positive for x > b. The sign changes at both a and b, confirming points of inflection at:

Curve Sketching

A rigorous sketch identifies all key features analytically rather than plotting individual points. The standard procedure is to find the x-intercepts by solving f(x) = 0; find the y-intercept by evaluating f(0); locate and classify all stationary points; determine convexity and concavity from the sign of f”; locate any inflection points; and examine limiting behaviour as x → ±∞.

For polynomials, limiting behaviour is determined by the leading term ax^n. If a > 0 and n is even, f(x) → +∞ in both directions. If a > 0 and n is odd, f(x) → +∞ as x → +∞ and −∞ as x → −∞. Signs reverse when a < 0. For products of a polynomial and an exponential, the exponential always dominates: x³e^{−x} → 0 as x → +∞ even though x³ → ∞.

The function f(x) = x³e^{−x} illustrates the full procedure. Since e^{−x} is never zero, the only x-intercept is at x = 0, which is also the y-intercept. Differentiating by the product rule:

Setting f'(x) = 0 gives x = 0 and x = 3. At x = 0 the second-order test fails because f”(0) = 0. Applying the first-order test, f'(x) > 0 on both sides of x = 0 (since x² ≥ 0 and (3 − x) > 0 near zero), so there is no sign change and x = 0 is a point of inflection. At x = 3, differentiating f'(x) gives:

Evaluating at $x = 3: f”(3) = (18 − 54 + 27)e^{−3} = −9e^{−3} < 0$, confirming a local maximum at (3, 27e^{−3}). As x → +∞ the exponential dominates and f(x) → 0. As x → −∞, x³ → −∞ and e^{−x} → +∞, so f(x) → −∞. The curve starts at −∞ for large negative x, passes through the origin as an inflection point, rises to the local maximum at (3, 27e^{−3}), then decays back toward zero.

Asymptotes and Cusps

When a function or its derivatives are undefined at some point x = a, the curve requires additional analysis. A vertical asymptote occurs when f(x) → ±∞ as x → a. A horizontal asymptote occurs when f(x) → L as x → ±∞. A cusp occurs when the function is defined at x = a but the derivative is not.

For f(x) = 1/(x − 1)², the function is undefined at x = 1. The derivatives are:

Since (x − 1)⁴ > 0 wherever the function is defined, f”(x) > 0 everywhere: both branches are convex. For x < 1, the denominator of f'(x) is negative, making f'(x) > 0, so the function increases toward the asymptote from the left. For x > 1, f'(x) < 0 and the function decreases away from the asymptote to the right. The y-intercept is f(0) = 1 and there are no x-intercepts. As x → 1 from either side f(x) → +∞, giving a vertical asymptote at x = 1, and as x → ±∞, f(x) → 0, giving a horizontal asymptote at y = 0.

Global Maxima and Minima

A global maximum is the largest value a function attains over its entire domain. Identifying it requires comparing the values at all local maxima with the function’s limiting behaviour as x → ±∞. If the function tends to +∞ in either direction, no global maximum exists. If the function tends to −∞ in either direction, no global minimum exists.

For $f(x) = x³e^{−x}$, the only local maximum is at x = 3 with value $27e^{−3}$. As x → +∞, f(x) → 0, which is always below $27e^{−3}$. As $x → −∞, f(x) → −∞$. The global maximum is therefore $27e^{−3}$ at x = 3, and a global minimum does not exist because the function is unbounded below.

Constrained Optimisation on a Closed Interval

When x is restricted to a closed interval [a, b], extrema can occur at interior stationary points or at the endpoints. The procedure is to find all stationary points inside the interval, evaluate f at each of these and at both endpoints, and select the largest value as the maximum and the smallest as the minimum.

For $f(x) = x³ − 2x² − 15x$ on [−3, 5], the stationary points at x = −5/3 and x = 3 both lie within the interval. Evaluating f at all four candidates: $f(−3) = −27 − 18 + 45 = 0$, $f(−5/3) = 400/27 ≈ 14.8$, $f(3) = 27 − 18 − 45 = −36$, and $f(5) = 125 − 50 − 75 = 0$. The maximum value is 400/27 at x = −5/3 and the minimum value is −36 at x = 3.

Economic Applications

Profit Maximisation

A monopolist’s profit is π(q) = R(q) − C(q), where $R(q) = q · p^D(q)$. Profit maximisation is treated as a constrained optimisation over the feasible output range. For $C(q) = q³ − 10q² + 25q + 10$ and $p^D(q) = 10 − q$ on [0, 10], the profit function is:

Differentiating and setting to zero:

The second derivative is π”(q) = −6q + 18. At q = 1: π”(1) = 12 > 0, a local minimum. At q = 5: π”(5) = −12 < 0, a local maximum. Evaluating π at all candidates: π(0) = −10, π(1) = −17, π(5) = 15, π(10) = −260. The maximum profit is 15, achieved at q = 5.

Maximising Tax Revenue

When a government imposes an excise tax T per unit, equilibrium quantity adjusts to q*(T) and tax revenue is $R(T) = q*(T) · T$. For $q*(T) = 2 − T/2$ on [0, 4], the revenue function is:

Differentiating: $R'(T) = 2 − T = 0$ gives T = 2. Since $R”(T) = −1 < 0$ this is a local maximum. Evaluating at the endpoints confirms R(0) = 0 and R(4) = 0, so the revenue-maximising tax is T* = 2.

This illustrates a general pattern for linear demand: the optimal excise tax is always exactly halfway between zero and the prohibitive tax T_m. Here T* = 2 = T_m/2 = 4/2.

The Classification Workflow

The full analysis of any function f(x) follows a fixed sequence. Begin by solving f(x) = 0 for x-intercepts and evaluating f(0) for the y-intercept. Then solve f'(x) = 0 to find stationary point candidates, apply the second-order test at each — falling back to the first-order sign analysis when f”(a) = 0 — and record the classification and function value at each stationary point. Next solve f”(x) = 0 and check for a sign change to locate any inflection points. Finally examine the behaviour as $x → ±∞$ to identify asymptotes or unbounded growth, and for a constrained problem on [a, b], evaluate f at every stationary point inside the interval and at both endpoints, selecting the largest and smallest values.

See you soon.