1.07d, 1.07e, 1.07n, 1.07o - Second derivatives, stationary points and monotonicity | OCR A-Level Mathematics A

A horizontal tangent can mark a peak, a trough or neither. You will find second derivatives (1.07d), interpret changes of gradient (1.07e), classify stationary points (1.07n) and use derivative signs to describe increasing and decreasing intervals (1.07o).

The rate of change of gradient

Can a curve have a local maximum of 6 but no largest value? The curve $y=x^3-6x^2+9x+2$ does. We will find that peak, then explain why it does not put a ceiling on the curve.

First separate a curve's height from its slope. For $y=x^2$ , the gradient is $2x$ : it increases by 2 whenever $x$ increases by 1, even on the falling left-hand branch.

You already know

The derivative $\mathrm f'(x)$ gives the tangent gradient. Differentiate a polynomial term by term using $\frac{\mathrm d}{\mathrm dx}x^n=nx^{n-1}$ .

Differentiating that gradient function answers a new question: how fast is the gradient itself changing?

Second derivative

The second derivative is the derivative of the first derivative:

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

For example, $y=x^2$ has second derivative 2. The notation does not mean $(\mathrm f'(x))^2$ .

A positive second derivative means increasing gradient; it does not necessarily mean increasing height. If volume is in litres and time in minutes, the second derivative has units $\mathrm{L\,min^{-2}}$ : change in filling rate per minute. This interpretation is not in the formula sheet: learn it.

Worked example 1: differentiating the gradient

Question. Find the second derivative of $y=2x^3-3x^2+5$ and interpret its value at $x=1$ .

Plan. Differentiate twice, then use the sign of the second derivative.

\begin{aligned} y'&=6x^2-6x &&\text{differentiate each term}\\ y''&=12x-6 &&\text{differentiate the gradient}\\ y''(1)&=6>0 &&\text{substitute }x=1 \end{aligned}

Why this step? The second differentiation measures change in gradient, rather than giving another height.

Answer. $y''=12x-6$ ; the gradient is increasing near $x=1$ .

Check. The gradients at $0.9$ and $1.1$ are $-0.54$ and $0.66$ : they increase through zero.

Now change only the leading power from a cube to a fourth power. Keep both differentiations visible.

Worked example 2: a fourth-power term

Question. Find the second derivative of $y=2x^4-3x^2+5$ and interpret its value at $x=1$ .

Plan. Apply the power rule twice, then evaluate.

\begin{aligned} y'&=8x^3-6x &&\text{first derivative}\\ y''&=24x^2-6 &&\text{second derivative}\\ y''(1)&=18>0 &&\text{evaluate} \end{aligned}

Why this step? The first derivative is now cubic, so its derivative is quadratic.

Answer. $y''=24x^2-6$ ; the gradient increases locally at $x=1$ .

Check. $y'(0.9)=0.432$ and $y'(1.1)=4.048$ , consistent with increasing gradient.

The two differentiations answer different questions, even when their values happen to agree.

Watch out

Wrong: $y''=(y')^2$ . Squaring a gradient is not differentiating it. For $y=x^2$ , the second derivative is 2, whereas the square of the first derivative is $4x^2$ .

The next check changes the leading power again; the meaning of the second derivative stays the same.

Locating and classifying stationary points

A peak or trough has a horizontal tangent when the derivative exists, but a horizontal tangent alone does not tell you which shape occurs.

Stationary point

A stationary point is a point where the derivative exists and equals zero. For example, the origin is stationary on $y=x^3$ , although that curve does not turn there. A root of $\mathrm f(x)=0$ need not be stationary.

Near a minimum, the gradient changes from negative to positive; near a maximum it changes from positive to negative. The second derivative can detect this at a stationary point.

Second derivative test

At a stationary point $x=a$ of a smooth curve:

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

If $\mathrm f''(a)=0$ , this test is inconclusive.

Not in the formula sheet: learn it.

Find horizontal coordinates using the derivative, then use the curve itself for the heights. That distinction keeps the two jobs tidy.

Worked example 3: two stationary points

Question. Find and classify the stationary points of $\mathrm f(x)=x^3-6x^2+9x+2$ .

Plan. Solve $\mathrm f'=0$ , find the heights and test the second derivative.

\begin{aligned} \mathrm f'(x)&=3(x-1)(x-3) &&\text{differentiate and factor}\\ \mathrm f'(x)=0&\Rightarrow x=1,3 &&\text{horizontal tangents}\\ \mathrm f(1)&=6,\quad\mathrm f(3)=2 &&\text{heights on the curve}\\ \mathrm f''(x)&=6x-12 &&\text{differentiate again}\\ \mathrm f''(1)&=-6<0,\quad\mathrm f''(3)=6>0 &&\text{classify} \end{aligned}

Why this step? Substituting into $\mathrm f'$ again would give zero, not the stationary heights.

Answer. Local maximum $(1,6)$ ; local minimum $(3,2)$ .

Check. The derivative has signs positive, negative, positive across 1 and 3, matching both classifications.

Change only the overall sign of the function. Reflection in the horizontal axis should interchange peaks and troughs.

Worked example 4: reflecting the curve

Question. Find and classify the stationary points of $\mathrm g(x)=-x^3+6x^2-9x-2$ .

Plan. Repeat the derivative test and compare with the reflection.

\begin{aligned} \mathrm g'(x)&=-3(x-1)(x-3)=0 &&\text{stationary equation}\\ x&=1,3 &&\text{solve the factors}\\ \mathrm g(1)&=-6,\quad\mathrm g(3)=-2 &&\text{find coordinates}\\ \mathrm g''(1)&=6>0,\quad\mathrm g''(3)=-6<0 &&\text{test their nature} \end{aligned}

Why this step? Negating the function negates both derivatives, so the signs used for classification reverse.

Answer. Local minimum $(1,-6)$ ; local maximum $(3,-2)$ .

Check. Both heights and classifications are the reflections of the preceding example.

Locating the point and finding its height use different functions.

Watch out

Solving $\mathrm f(x)=0$ finds axis crossings, not stationary points. Solve $\mathrm f'(x)=0$ , then substitute those locations into $\mathrm f$ for the heights.

A vertical translation changes the heights but leaves both derivatives unchanged.

Increasing and decreasing intervals

Knowing the turning points is useful; knowing what happens between them finishes the story.

Derivative signs and direction

On an interval where the derivative exists, $\mathrm f'(x)>0$ throughout implies that $\mathrm f$ is increasing; $\mathrm f'(x)<0$ throughout implies that it is decreasing.

Not in the formula sheet: learn it.

Split a sign analysis at derivative zeros and excluded domain values. A continuous derivative cannot change sign between successive zeros without another zero. For the opening cubic, its derivative is $3(x-1)(x-3)$ .

Worked example 5: a derivative sign table

Question. Find where $\mathrm f(x)=x^3-6x^2+9x+2$ has positive and negative derivative.

Plan. Use the signs of both factors on the intervals split by 1 and 3.

Interval	$x-1$	$x-3$	$\mathrm f'$
$x<1$	negative	negative	positive
$1<x<3$	positive	negative	negative
$x>3$	positive	positive	positive

Why this step? Testing only the zeros gives no information about whether the curve rises or falls between them.

Answer. $\mathrm f'>0$ for $x<1$ or $x>3$ ; $\mathrm f'<0$ for $1<x<3$ .

Check. At $0,2,4$ , the gradients are respectively $9,-3,9$ .

Change one structural feature: let the derivative have a repeated root. A zero gradient may then leave the direction unchanged.

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1.07d, 1.07e, 1.07n, 1.07o - 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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