1.07f, 1.07p - Concavity and points of inflection | OCR A-Level Mathematics A

A curve can flatten without turning, and it can change its bending without flattening. You will use second-derivative signs to describe convex and concave sections (1.07f), then locate and classify stationary and non-stationary points of inflection (1.07p).

The sign of the second derivative describes bending

How can a curve rise, become horizontal for a moment, then keep rising? The curve $y=(x-1)^3+2$ does exactly that at $(1,2)$ . Its changing bending will explain the flat point.

Start with a simpler distinction. On $y=x^2-4x$ , the gradients at $0,1,2$ are $-4,-2,0$ . The curve initially falls, but its gradients increase.

You already know

$\mathrm f'$ measures the curve's gradient; $\mathrm f''$ measures change in that gradient. A stationary point has $\mathrm f'=0$ , but need not be a turning point.

The second derivative describes bending, independently of the direction in which the height changes.

Convex and concave sections

In OCR Mathematics A, a smooth section with $\mathrm f''(x)>0$ is convex: gradients increase and it bends upwards. A section with $\mathrm f''(x)<0$ is concave: gradients decrease and it bends downwards. For example, $x^2$ is convex; $-x^2$ is concave. Neither word means increasing height.

To find the intervals, factor the second derivative and establish its sign throughout each interval. These sign rules are not in the formula sheet: learn them.

Worked example 1: bending while the curve falls

Question. Describe the bending of $y=x^2-4x$ , and compare it with the direction of the curve for $x<2$ .

Plan. Use the second derivative for bending and the first for direction.

\begin{aligned} y'&=2x-4 &&\text{first derivative}\\ y''&=2>0 &&\text{second derivative}\\ y'&<0\quad(x<2) &&\text{height decreases there} \end{aligned}

Why this step? A falling curve can have increasing gradient, so the two derivative signs answer separate questions.

Answer. Convex everywhere, although decreasing for $x<2$ .

Check. Its slopes $-4,-2,0$ increase through the falling part to its minimum.

Change only the sign of the whole function. This reverses both its direction and its bending.

Worked example 2: the reflected bending

Question. Describe the bending of $y=-x^2+4x$ , and its direction for $x<2$ .

Plan. Differentiate twice and interpret each sign separately.

\begin{aligned} y'&=-2x+4 &&\text{gradient}\\ y''&=-2<0 &&\text{gradients decrease}\\ y'&>0\quad(x<2) &&\text{height increases there} \end{aligned}

Why this step? Reflection reverses gradient change as well as gradient itself.

Answer. Concave everywhere, while increasing for $x<2$ .

Check. At $0,1,2$ , the slopes $4,2,0$ decrease despite the rising height.

The reflected examples separate direction from bending.

Watch out

Wrong: $\mathrm f'>0$ , therefore the curve is convex. Increasing height uses $\mathrm f'$ ; convexity uses $\mathrm f''>0$ . A rising curve can be concave, as the second example shows before its maximum.

For a cubic, bending can differ on the two sides of one point. For instance, $y=x^3-3x^2$ has $y''=6(x-1)$ , so it is concave for $x<1$ and convex for $x>1$ .

A zero second derivative is only a candidate

A zero second derivative tells you where to investigate; a change of bending tells you what you have found.

Point of inflection

A point of inflection is a point on a curve where its bending changes from concave to convex, or vice versa. For the smooth curves here, find candidates from $\mathrm f''=0$ and establish a sign change across the point. The origin on $x^3$ is an example; the origin on $x^4$ is not.

The sign-change criterion is not in the formula sheet: learn it.

The two cubics below have identical second derivatives, but different tangent gradients at zero. Adding a straight-line term can change the slope without changing the bending at all.

Two inflections and a zero second derivative without an inflection

This is a useful surprise: the same bending need not mean the same direction or the same stationary points.

Worked example 3: proving a change of bending

Question. Find the point of inflection of $\mathrm f(x)=x^3+x$ , giving a reason.

Plan. Find a second-derivative zero, prove its sign change and calculate the height.

\begin{aligned} \mathrm f''(x)&=6x &&\text{differentiate twice}\\ \mathrm f''(x)=0&\Rightarrow x=0 &&\text{candidate}\\ 6x&<0\ (x<0),\quad6x>0\ (x>0) &&\text{concave to convex}\\ \mathrm f(0)&=0 &&\text{point on the curve} \end{aligned}

Why this step? The sign change proves an inflection; solving $\mathrm f''=0$ alone only locates a candidate.

Answer. Point of inflection $(0,0)$ .

Check. The second derivatives at $-1$ and 1 are $-6$ and 6, consistent with the interval signs.

Change only the leading power from 3 to 4. A repeated zero of the second derivative behaves differently.

Worked example 4: rejecting a candidate

Question. Decide whether $(0,0)$ is a point of inflection of $\mathrm g(x)=x^4+x$ .

Plan. Compare second-derivative signs on both sides of zero.

\begin{aligned} \mathrm g''(x)&=12x^2 &&\text{differentiate twice}\\ \mathrm g''(0)&=0 &&\text{candidate}\\ 12x^2&>0\quad(x\ne0) &&\text{positive on both sides} \end{aligned}

Why this step? An even power stays positive across its zero, so the curve does not change its bending.

Answer. No inflection at the origin: the curve is convex on both sides.

Check. At $x=-1$ and $x=1$ , the second derivative is 12 in both cases.

A repeated second-derivative zero is a particularly useful warning against stopping too soon.

Watch out

Wrong: $\mathrm g''(0)=0$ , therefore the origin is an inflection. The quartic example has no change in sign. Establish a change of bending before calling a candidate an inflection.

Domain gaps also matter. For $y=1/x$ , $y''=2/x^3$ changes sign between the two branches, but there is no point at zero. The gap is not an inflection. Use the zero-second-derivative method on the smooth curves for which it is stated; do not assume it describes every possible non-smooth curve.

Stationary and non-stationary inflections

Once an inflection is proved, one first-derivative value tells you whether its tangent is horizontal.

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1.07f, 1.07p - Concavity and points of inflection

The sign of the second derivative describes bending

A zero second derivative is only a candidate

Stationary and non-stationary inflections

Continue this lesson.

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