G3 - Stationary points and optimisation

G3 - Stationary points and optimisation

Classify stationary points and inflections using evidence from derivatives. Distinguish local from global extrema, compare boundaries and construct a constrained volume model whose optimum is physically possible.

Classifying stationary points

How large a box can you make from an 18 cm18\,\mathrm{cm} by 12 cm12\,\mathrm{cm} card by cutting equal corner squares and folding up the sides? A cut of 2 cm2\,\mathrm{cm} gives volume 224 cm3224\,\mathrm{cm^3}, while 3 cm3\,\mathrm{cm} gives 216 cm3216\,\mathrm{cm^3}. The best cut lies somewhere between: derivatives can locate it, but an equation alone will not prove the answer is the largest possible box.

You already know

A horizontal tangent has f′=0f'=0. Positive derivative means increasing and negative derivative decreasing. The second derivative measures change of gradient. These criteria are not supplied as formulae in the booklet.

A stationary point has derivative zero. A local maximum has no larger values in a sufficiently small neighbourhood; a local minimum has no smaller ones. A turning point reverses the increasing/decreasing direction. A stationary point need not turn.

Stationary-point tests

At a stationary input aa, with f′(a)=0f'(a)=0:

  • A change in f′f' from positive to negative gives a local maximum.
  • A change from negative to positive gives a local minimum.
  • For the smooth functions here, f′′(a)<0f''(a)<0 gives a local maximum and f′′(a)>0f''(a)>0 a local minimum.
  • If f′′(a)=0f''(a)=0, the second-derivative test is inconclusive; inspect nearby signs.

Not in the formula booklet: learn it.

The second-derivative test only classifies an input already shown to be stationary. Coordinates then need substitution into the original function, rather than into either derivative.

Worked example 1: two stationary points

Question. Find and classify the stationary points of f(x)=x3−3xf(x)=x^3-3x. Fully justify your answer.

Plan. Solve f′=0f'=0, find the two heights and use the second derivative at each input.

f′(x)=3x2−3=3(x−1)(x+1)differentiate and factorf′(x)=0  ⟺  x=−1 or x=1stationary inputsf(−1)=2f(1)=−2original heightsf′′(x)=6xf′′(−1)=−6f′′(1)=6test the candidates\begin{aligned} f'(x)&=3x^2-3=3(x-1)(x+1) && \text{differentiate and factor}\\ f'(x)=0&\iff x=-1\text{ or }x=1 && \text{stationary inputs}\\ f(-1)&=2\qquad f(1)=-2 && \text{original heights}\\ f''(x)&=6x\qquad f''(-1)=-6\qquad f''(1)=6 && \text{test the candidates} \end{aligned}

Why this step? The sign of the second derivative classifies the zero first derivative; it does not supply the point's height.

Answer. (−1,2)(-1,2) is a local maximum and (1,−2)(1,-2) a local minimum.

Check. The factored first derivative has signs +,−,++,-,+, confirming both reversals.

Change only the constant term, adding 44. This changes stationary heights but cannot change their inputs or nature.

Worked example 2: stationary points after a vertical shift

Question. Find and classify the stationary points of g(x)=x3−3x+4g(x)=x^3-3x+4, with justification.

Plan. Reuse the derivative calculations, then evaluate the shifted original function.

g′(x)=3(x−1)(x+1)constant has zero derivativeg′(x)=0  ⟺  x=−1 or x=1solveg(−1)=6g(1)=2new heightsg′′(−1)=−6<0g′′(1)=6>0classify\begin{aligned} g'(x)&=3(x-1)(x+1) && \text{constant has zero derivative}\\ g'(x)=0&\iff x=-1\text{ or }x=1 && \text{solve}\\ g(-1)&=6\qquad g(1)=2 && \text{new heights}\\ g''(-1)&=-6<0\qquad g''(1)=6>0 && \text{classify} \end{aligned}

Why this step? A vertical translation preserves slopes and changes only the coordinates' second entries.

Answer. Local maximum (−1,6)(-1,6); local minimum (1,2)(1,2).

Check. Each height is 44 greater than in Worked example 1, while the derivative sign pattern is unchanged.

At zero, both x3x^3 and x4x^4 have first and second derivatives zero. Yet 3x23x^2 stays positive on both sides, giving no extremum for x3x^3, while 4x34x^3 changes from negative to positive, giving a minimum for x4x^4. For −x4-x^4, the signs reverse and give a maximum. A zero second derivative leaves all these possibilities open.

Watch out

Wrong: “f′′(a)=0f''(a)=0, so there is no maximum or minimum.” Zero means this test cannot decide. Right: examine f′f' either side; x4x^4 has a minimum at zero despite f′′(0)=0f''(0)=0.

Change the constant again. Keep coordinates and classification as separate steps.

Testing points of inflection

An inflection changes curvature between convex and concave. A turning point instead changes between increasing and decreasing. These are different tests, so an inflection can have a sloping tangent.

For the smooth functions here, solve f′′=0f''=0 to find candidates, then check the second derivative's sign on both sides. A stationary inflection also has f′=0f'=0; a non-stationary inflection has f′≠0f'\ne0. These criteria are not in the formula booklet.

Cubic stationary inflection, quartic minimum and a non-stationary cubic inflection

In particular, x4x^4 has f′′=12x2≥0f''=12x^2\ge0 on both sides of zero: its zero second derivative is not an inflection. The neighbouring signs, not equality alone, decide.

Worked example 3: a stationary inflection

Question. Find the inflection of f(x)=x3+2f(x)=x^3+2, justify it and decide whether it is stationary.

Plan. Check curvature across the zero of f′′f'', then evaluate the separate height and tangent gradient.

f′(x)=3x2f′′(x)=6xdifferentiate twicef′′(x)<0 (x<0),f′′(x)>0 (x>0)curvature changes at 0f(0)=2f′(0)=0height and stationarity\begin{aligned} f'(x)&=3x^2\qquad f''(x)=6x && \text{differentiate twice}\\ f''(x)&<0\ (x<0),\quad f''(x)>0\ (x>0) && \text{curvature changes at }0\\ f(0)&=2\qquad f'(0)=0 && \text{height and stationarity} \end{aligned}

Why this step? The second derivative establishes a curvature change; the first derivative establishes whether the tangent is horizontal.

Answer. (0,2)(0,2) is a stationary inflection, from concave to convex.

Check. f′′(−1)=−6f''(-1)=-6, f′′(1)=6f''(1)=6, but f′f' is positive on both sides: no turning point.

Change only one feature by adding the linear term xx. It changes the tangent gradient without changing the second derivative.

Worked example 4: an inflection with a sloping tangent

Question. Find the inflection of g(x)=x3+x+2g(x)=x^3+x+2, justify it and decide whether it is stationary.

Plan. Repeat the curvature test, then check the new first derivative at the candidate.

g′(x)=3x2+1g′′(x)=6xdifferentiate twiceg′′(x)<0 (x<0),g′′(x)>0 (x>0)same curvature changeg(0)=2g′(0)=1≠0height and slope\begin{aligned} g'(x)&=3x^2+1\qquad g''(x)=6x && \text{differentiate twice}\\ g''(x)&<0\ (x<0),\quad g''(x)>0\ (x>0) && \text{same curvature change}\\ g(0)&=2\qquad g'(0)=1\ne0 && \text{height and slope} \end{aligned}

Why this step? Adding a linear term changes every slope by a constant, whose derivative is zero, so the curvature stays the same.

Answer. (0,2)(0,2) is a non-stationary inflection with tangent gradient 11.

Check. g′(x)=3x2+1>0g'(x)=3x^2+1>0 everywhere, so the curve keeps increasing through its change of curvature.

A striking consequence is that changing a curve's tilt by adding kxkx can turn a stationary inflection into a non-stationary one without moving its curvature change at all. The derivative calculation, rather than a special picture, explains it.

Watch out

Wrong: “There is no inflection because g′(0)=1g'(0)=1.” A nonzero first derivative excludes a stationary point, not an inflection. Right: g′′g'' changes sign, so there is a non-stationary inflection.

Change only the linear coefficient to −3-3; the height and second derivative are unchanged.

Global extrema and boundaries

A local optimum compares nearby inputs; a global maximum or global minimum compares every allowed input. A stationary point can be locally smallest while an endpoint is globally largest.

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