Mc6–Mc8, MC5 - Second derivatives, stationary points and monotonicity | OCR A-Level Mathematics B (MEI)

Use the second derivative to track changing gradient (Mc6), locate and classify stationary points (Mc7), and find increasing or decreasing intervals (Mc8). You will combine these decisions in curve sketches (MC5), including cases where a horizontal tangent does not mean a turn.

The rate of change of gradient

Does a horizontal tangent always mark a turning point? At zero, each of $x^4$ , $-x^4$ and $x^3$ has a horizontal tangent, yet their shapes are different. We need more than one number to read a curve.

For a first concrete case, $y=x^2-4x$ has gradient $2x-4$ . At $x=0$ the curve falls, but the gradient is increasing: it becomes less negative as you move right.

You already know

The derivative $\mathrm f'(x)$ is the gradient of $y=\mathrm f(x)$ . Use the power rule and differentiate sums, differences and constant multiples term by term.

Apply the same process to the gradient function itself.

Second derivative

The second derivative is the rate of change of the gradient:

\mathrm f''(x)=\frac{\mathrm d^2y}{\mathrm dx^2}=\frac{\mathrm d}{\mathrm dx}\left(\frac{\mathrm dy}{\mathrm dx}\right).

For $y=x^2-4x$ , it is 2. It is not $(\mathrm f'(x))^2$ .

If $y$ is a depth in centimetres and $x$ a time in seconds, the first derivative has units $\mathrm{cm\,s^{-1}}$ , and the second has units $\mathrm{cm\,s^{-2}}$ . Positive second derivative means increasing gradient, even if the depth is currently decreasing. The second-derivative interpretation is not in the formula sheet: learn it.

Worked example 1: a decreasing height with increasing gradient

Question. For $y=x^3-3x^2-9x+4$ , find $y''$ and interpret its value at $x=2$ .

Plan. Differentiate twice and distinguish the signs of the first and second derivatives.

\begin{aligned} y'&=3x^2-6x-9 &&\text{gradient function}\\ y''&=6x-6 &&\text{rate of change of gradient}\\ y'(2)&=-9,\quad y''(2)=6 &&\text{evaluate both} \end{aligned}

Why this step? A negative first derivative and positive second derivative describe different features, so they can occur together.

Answer. $y''=6x-6$ . At $x=2$ , the curve falls but its gradient increases locally.

Check. At $1.9$ and $2.1$ , the gradients are $-9.57$ and $-8.37$ , becoming less negative.

Change only the overall sign of the function. Both derivative signs reverse.

Worked example 2: an increasing height with decreasing gradient

Question. For $y=-x^3+3x^2+9x-4$ , find $y''$ and interpret its value at $x=2$ .

Plan. Differentiate the reflected function twice and compare its rates.

\begin{aligned} y'&=-3x^2+6x+9 &&\text{differentiate}\\ y''&=-6x+6 &&\text{differentiate again}\\ y'(2)&=9,\quad y''(2)=-6 &&\text{substitute} \end{aligned}

Why this step? Reflection in the horizontal axis reverses heights and gradients, including how those gradients change.

Answer. $y''=-6x+6$ . The curve rises at $x=2$ , with a decreasing gradient.

Check. Its gradients at $1.9$ and $2.1$ are $9.57$ and $8.37$ , both positive but decreasing.

Reading each derivative's sign separately avoids a common mixed interpretation.

Watch out

Wrong: $y''<0$ , so the curve must be falling. A negative second derivative means a decreasing gradient. The curve can still rise when its first derivative is positive.

A vertical shift leaves every gradient unchanged. Use that fact to check your next calculation.

Locating and classifying stationary points

A stationary point has a zero gradient; a maximum or minimum also requires a decision about the neighbouring curve.

Stationary point

A stationary point is a point where the derivative exists and is zero. A maximum turning point is locally higher than nearby points; a minimum turning point is locally lower. The origin on $y=x^3$ is stationary but is neither type of turning point.

Use the second derivative for a quick classification when its value is nonzero.

Classifying a stationary point

For a smooth curve with $\mathrm f'(a)=0$ :

\begin{aligned} \mathrm f''(a)>0&\Rightarrow\text{minimum turning point}\\ \mathrm f''(a)<0&\Rightarrow\text{maximum turning point}. \end{aligned}

If $\mathrm f''(a)=0$ , use another argument.

Not in the formula sheet: learn it.

The derivative locates the horizontal tangents, but the function supplies their heights.

Worked example 3: finding both turning points

Question. Determine the stationary points of $y=x^3-3x^2-9x+4$ and their nature.

Plan. Factor the derivative, evaluate the curve and use the second-derivative signs.

\begin{aligned} y'&=3(x+1)(x-3)=0 &&\text{stationary equation}\\ x&=-1,3 &&\text{solve the factors}\\ y(-1)&=9,\quad y(3)=-23 &&\text{find heights}\\ y''&=6x-6 &&\text{second derivative}\\ y''(-1)&=-12<0,\quad y''(3)=12>0 &&\text{classify} \end{aligned}

Why this step? “Determine” needs the reasoning that establishes the points and their nature, rather than a list from a graphing calculator.

Answer. Maximum turning point $(-1,9)$ ; minimum turning point $(3,-23)$ .

Check. $y'$ has signs positive, negative, positive across the two roots.

Change only the sign of the whole function. Predict the new nature before calculating it.

Worked example 4: exchanging maximum and minimum

Question. Determine the stationary points of $y=-x^3+3x^2+9x-4$ and their nature.

Plan. Solve the reflected derivative equation and provide sign evidence.

\begin{aligned} y'&=-3(x+1)(x-3)=0\Rightarrow x=-1,3 &&\text{stationarity}\\ y(-1)&=-9,\quad y(3)=23 &&\text{coordinates}\\ y''(-1)&=12>0,\quad y''(3)=-12<0 &&\text{nature} \end{aligned}

Why this step? The stationary horizontal coordinates survive reflection, but the turning directions reverse.

Answer. Minimum turning point $(-1,-9)$ ; maximum turning point $(3,23)$ .

Check. The first-derivative signs are negative, positive, negative, agreeing with both labels.

The coordinates and the nature each need their own evidence.

Exam tip

For “Determine”, show the zero-gradient equation, calculate the heights and justify the nature with derivative signs. A calculator picture or unsupported maximum/minimum labels do not establish those claims.

Changing the constant term now moves both points vertically without changing their nature.

Increasing and decreasing intervals

A sign table explains the journey between turning points and also identifies horizontal tangents that are not turns.

Direction from the first derivative

A derivative that is positive throughout an interval gives an increasing function there; a derivative that is negative throughout gives a decreasing function there.

Not in the formula sheet: learn it.

Split the domain at derivative zeros and any gaps. For a continuous derivative, its sign cannot change between successive roots without passing through another zero.

Worked example 5: three sign intervals

Question. Find the intervals where $y=x^3-3x^2-9x+4$ has positive and negative gradient.

Plan. Analyse $y'=3(x+1)(x-3)$ on the intervals separated by its roots.

Interval	$x+1$	$x-3$	$y'$
$x<-1$	negative	negative	positive
$-1<x<3$	positive	negative	negative
$x>3$	positive	positive	positive

Why this step? The product's sign must be established across each whole interval, not just at a stationary point.

Answer. Positive gradient for $x<-1$ or $x>3$ ; negative gradient for $-1<x<3$ .

Check. Gradients at $-2,0,4$ are $15,-9,15$ .

Change only the separation of the derivative roots: let them coincide. Integrating is not needed; the function is supplied.

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Mc6–Mc8, MC5 - Second derivatives, stationary points and monotonicity

The rate of change of gradient

Locating and classifying stationary points

Increasing and decreasing intervals

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