1.7 - Stationary points and optimisation | Cambridge A-Level Mathematics

A zero gradient identifies a stationary point, but it does not tell you whether the point is a maximum or a minimum. You will locate stationary points, classify them with second derivatives or gradient signs, and use that information in sketches and optimisation. Domains and boundary values will decide whether a local result solves the whole problem.

Finding stationary points

Can the equation $f'(x)=0$ tell us where a curve turns? It finds the inputs with horizontal tangents. Finding those inputs is the first step; you must also recover their heights and investigate the behaviour around them.

Stationary point

A stationary point of $y=f(x)$ is a point where the derivative exists and $f'(x)=0$ . Its tangent is horizontal. A stationary point need not be a maximum or a minimum.

The equation gives inputs, not complete coordinates. Substitute each permitted solution into the original function, not into its derivative, to find the corresponding height.

Worked example 1: two stationary points

Question. Find the stationary points of $y=x^3-3x^2+2$ .

Plan. Differentiate, solve the zero-gradient equation, then calculate the original heights.

\begin{aligned} \frac{dy}{dx}&=3x^2-6x=3x(x-2)\\ 3x(x-2)&=0\quad\Longrightarrow\quad x=0\text{ or }x=2\\ y(0)&=2\\ y(2)&=8-12+2=-2. \end{aligned}

Answer. The stationary points are $(0,2)$ and $(2,-2)$ .

Check. Both inputs make $3x(x-2)$ zero. Direct substitution into the original cubic gives the stated heights.

Change only the constant term. A vertical translation changes heights but leaves every gradient unchanged.

Worked example 2: changing heights without changing inputs

Question. Find the stationary points of $y=x^3-3x^2+5$ .

Plan. Use the unchanged derivative and recalculate the heights from the new function.

\begin{aligned} \frac{dy}{dx}&=3x(x-2)\\ \frac{dy}{dx}=0&\quad\Longrightarrow\quad x=0\text{ or }x=2\\ y(0)&=5\\ y(2)&=8-12+5=1. \end{aligned}

Answer. The stationary points are $(0,5)$ and $(2,1)$ .

Check. The whole graph is $3$ units above the first graph. Both stationary heights have increased by $3$ , while their inputs are unchanged.

Check the domain before accepting a solution. An excluded input cannot be a point on the curve. Also, a vertical tangent does not have derivative zero: its gradient is not finite, so it is not a stationary point under this definition.

Watch out

Solving $f(x)=0$ finds axis intersections, not stationary points. Solve $f'(x)=0$ , then use $f(x)$ to recover each height.

Find both coordinates in a fresh cubic.

Classifying an extremum

At a local maximum, nearby points have lower heights; at a local minimum, nearby points have higher heights. “Local” means compared with points sufficiently close to the contact input, rather than necessarily every point in the full domain.

Local maximum and local minimum

A local maximum at an interior input $a$ has $f(a)\ge f(x)$ for all sufficiently nearby permitted $x$ . A local minimum has $f(a)\le f(x)$ for all sufficiently nearby permitted $x$ .

A positive second derivative means the first derivative is increasing locally. At a stationary point, that produces a change from negative gradient to positive gradient, giving a minimum. A negative second derivative produces the reverse change, giving a maximum.

Second-derivative test

At a stationary input $a$ , with the function twice differentiable in a neighbourhood,

\begin{aligned} f''(a)>0&\quad\Longrightarrow\quad\text{local minimum}\\ f''(a)<0&\quad\Longrightarrow\quad\text{local maximum}. \end{aligned}

If $f''(a)=0$ , this test is inconclusive. Investigate the sign of $f'$ on either side instead.

Not in the formula booklet: learn it.

The sign test works directly: $+$ then $-$ gives a maximum, and $-$ then $+$ gives a minimum. If the sign stays positive on both sides, or negative on both sides, there is no local maximum or minimum there.

Worked example 3: a maximum and a minimum

Question. Determine the nature of the stationary points $(0,2)$ and $(2,-2)$ on $y=x^3-3x^2+2$ .

Plan. Differentiate again and evaluate the second derivative at each stationary input.

\begin{aligned} y'&=3x^2-6x\\ y''&=6x-6\\ y''(0)&=-6<0\\ y''(2)&=6>0. \end{aligned}

Answer. $(0,2)$ is a local maximum; $(2,-2)$ is a local minimum.

Check. From $y'=3x(x-2)$ , the derivative signs are positive for $x<0$ , negative for $0<x<2$ , then positive for $x>2$ . These are exactly the changes required.

Now consider a flatter stationary point where the second derivative is zero. This changes the classification method, not the need to give a reason.

Worked example 4: a zero second derivative

Question. Find and classify the stationary point of $y=(x-1)^4$ .

Plan. Find the stationary input, try the second derivative, then use gradient signs if necessary.

\begin{aligned} y'&=4(x-1)^3=0\quad\Longrightarrow\quad x=1\\ y(1)&=0\\ y''&=12(x-1)^2,\qquad y''(1)=0. \end{aligned}

The second-derivative test is inconclusive. For $x<1$ , $y'<0$ ; for $x>1$ , $y'>0$ . The graph falls into the point and rises away from it.

Answer. $(1,0)$ is a minimum.

Check. The original function is a fourth power, so $(x-1)^4\ge0$ , with equality only at $x=1$ . This also proves the minimum over the whole real domain.

Zero second derivative does not imply a minimum either. For $y=x^3$ at $0$ , $y'=3x^2$ is positive on both sides. Although $y'(0)=y''(0)=0$ , the curve continues increasing through the point, so there is no maximum or minimum there. No additional classification is needed to reach that conclusion.

Watch out

“ $f''(a)=0$ , therefore neither maximum nor minimum” is invalid. The test has not decided. Use the first derivative's signs or another valid argument from the function.

Use the sign test when the second derivative cannot decide.

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1.7 - Stationary points and optimisation

Finding stationary points

Classifying an extremum

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