Stationary values and optimisation

Use the change of sign

A stationary point has a horizontal tangent, so its derivative is zero. To determine its nature, ask how the original function behaves immediately before and after it. A local maximum is higher than nearby points on both sides; a local minimum is lower than nearby points on both sides. Local comparisons do not yet identify the greatest or least value over an entire restricted interval.

For f(x) = x³ − 3x, solve f'(x) = 3x²-3 = 3(x-1)(x + 1) = 0. The stationary inputs are −1 and 1. The factors show the derivative’s signs:

At −1 the sign changes from positive to negative, so the curve rises into a local maximum and falls away afterwards. Substitute into the original function to obtain (−1,2). At 1 the sign changes from negative to positive, giving local minimum (1,−2). Solving f′ = 0 gave inputs; the original f supplies the corresponding heights.

A zero second derivative leaves a decision open

At a stationary input a, f″(a) < 0 gives a local maximum and f″(a) > 0 gives a local minimum. For the cubic above, f″(x) = 6x gives −6 at x = −1 and 6 at x = 1, agreeing with the first-derivative sign tests.

If f″(a) = 0, this test is inconclusive. It does not establish a point of inflexion. Compare two simple cases at zero:

For g(x) = x³, g′(x) = 3x² is positive on both sides of zero, so the function increases through its horizontal tangent. The second derivative 6x changes from negative to positive: the curve’s bending changes there. Thus (0,0) is a stationary point of inflexion, not an extremum.

For h(x) = x⁴, h′(x) = 4x³ changes from negative to positive. Thus (0,0) is a local minimum, even though h″(0) = 0 too. Its second derivative 12x² is positive on both sides, so it does not change concavity there.

Build the objective from the constraint

Squares of side x cm are cut from the corners of a 12 cm square sheet, and the sides fold into an open box. Neglect sheet thickness. The height is x and each base side is 12 − 2x: a strip is lost from each end of a base side. Positive height and base dimensions require 0 < x < 6.

Multiply the height by the base area to form the volume, then expand so the permitted termwise derivative rules apply:

The stationary algebraic candidates are 2 and 6. Only x = 2 lies inside the allowed interval; x = 6 would have a zero-width base, so it is not a second usable box.

For 0 < x < 2, both factors in V′ are negative and their product is positive. For 2 < x < 6, they have opposite signs and V′ is negative. The volume increases up to x = 2 and decreases afterwards, so this is the greatest volume throughout the model’s allowed interval. The second derivative V″(2) = 24(2) − 96 = −48 also confirms the local maximum.

At x = 2 the box has dimensions 2 cm × 8 cm × 8 cm and volume 128 cm³. This checks the answer against the original geometry. Near either excluded boundary the volume approaches zero. Zero itself is not an attained minimum volume among the positive-dimension boxes: the boundaries describe degenerate shapes outside the model.

A restricted interval may select an endpoint

If manufacture restricts x to [1,1.5], the stationary x = 2 is unavailable. V′ is positive throughout that restricted interval, so its greatest volume occurs at the included endpoint x = 1.5. The volume is 1.5 × 9² = 121.5 cm³, despite V′ not being zero there.

For a continuous function on a closed interval, compare its values at included endpoints and all stationary inputs in the interior; also inspect any interior point where the derivative does not exist. For the differentiable examples here, endpoints and stationary candidates suffice. Keep ties: two different inputs may give the same greatest or least value.

A numerical extremum needs a stated region

For k(x) = x + e^(-x) on [−1,2], enter the whole function and choose the minimum operation over that interval. A result near (0,1) agrees with k'(x) = 1-e^(-x) = 0 at x = 0. Since k″(x) = e⁻ˣ > 0, it is a local minimum; the derivative is negative before zero and positive afterwards, establishing the least value 1 over the stated interval.

The greatest value over [−1,2] is not another stationary point. Check the endpoints: k(−1) = e − 1 ≈ 1.71828 and k(2) = 2 + e⁻² ≈ 2.13534. The greatest value is 2 + e⁻² at x = 2. A tool searching only for an interior turning point could miss that answer.

Record the original function, search interval, returned coordinates and final precision. Classify a reported extremum and check feasibility before interpreting it. A visually flat patch is not proof of an exact stationary input, and enlarging a plotting window changes what is visible without changing the task’s allowed interval.