1.07PureStretch

Second Derivatives

Q: What does the second derivative measure?

How fast the gradient is changing. Physically, if the function is displacement, it is acceleration.

Q: What if the second derivative is zero at a stationary point?

The test fails. Use the sign of the first derivative either side to classify the point.

Q: Is every point with f^ (x) = 0 a point of inflection?

No. The second derivative must change sign there.

Q: How do I know a local optimum is the overall best?

Check the ends of the domain as well. If there is a single stationary point and the function falls away from it on both sides, it is the global optimum.

Differentiate a function and you get its gradient. Differentiate again and you get the rate at which the gradient itself is changing — the $\textbf{second derivative}$ . It tells you whether a curve is bending upwards or downwards, gives a quick test for maxima and minima, and turns real-world "largest" and "smallest" questions into routine calculus.

45 min Differentiation

The big picture

If the first derivative is velocity, the second is acceleration: it measures how the gradient responds as you move along the curve. A positive second derivative means the gradient is increasing, so the curve is $\textbf{convex}$ — shaped like a cup — and a stationary point there must be a minimum. A negative second derivative means $\textbf{concave}$ , and a maximum. Where the curve switches between the two is a $\textbf{point of inflection}$ , whether or not it is also stationary. Put together with the first derivative, this lets you model a situation, find the best value of a quantity, and prove that it really is the best — which is the whole of optimisation.

What you'll be able to do

Find second derivatives, written $\mathrm{f}^{\prime\prime}(x)$ or $\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}$
Use the second derivative to classify stationary points
Identify convex and concave sections of a curve and points of inflection
Solve optimisation problems and justify the nature of the optimum

The second derivative test

At a stationary point, where $\mathrm{f}^{\prime}(x) = 0$ : if $\mathrm{f}^{\prime\prime}(x) > 0$ it is a $\textbf{minimum}$ ; if $\mathrm{f}^{\prime\prime}(x) < 0$ it is a $\textbf{maximum}$ .

If $\mathrm{f}^{\prime\prime}(x) = 0$ the test is $\textbf{inconclusive}$ . The point could be a maximum, a minimum or a point of inflection — use the sign test from the previous lesson.

The logic: at a minimum the gradient passes from negative through zero to positive, so it is increasing — the rate of change of the gradient is positive.

\mathrm{f}^{\prime}(a) = 0,\ \mathrm{f}^{\prime\prime}(a) > 0 \Rightarrow \text{minimum}

And

\mathrm{f}^{\prime\prime}(a) < 0

gives a maximum.

Example 1Find and classify the stationary points of

y = 2x^{3} - 3x^{2} - 12x + 5

\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6x^{2} - 6x - 12 = 6(x - 2)(x + 1) = 0

, so

x = 2

x = -1

\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = 12x - 6

3At

x = 2

18 > 0

, minimum;

y = 16 - 12 - 24 + 5 = -15

4At

x = -1

-18 < 0

, maximum;

y = -2 - 3 + 12 + 5 = 12

AnswerMaximum

(-1, 12)

; minimum

(2, -15)

Tip — Write the value of the second derivative and its sign explicitly — " $\mathrm{f}^{\prime\prime}(2) = 18 > 0$ , so minimum". A bare "minimum" gets no justification mark.

Convex, concave and points of inflection

A curve is $\textbf{convex}$ on an interval where $\mathrm{f}^{\prime\prime}(x) > 0$ (it lies above its tangents, like a cup) and $\textbf{concave}$ where $\mathrm{f}^{\prime\prime}(x) < 0$ (below its tangents, like a cap).

A $\textbf{point of inflection}$ is where the curve changes between convex and concave. At such a point $\mathrm{f}^{\prime\prime}(x) = 0$ — but you must also check that $\mathrm{f}^{\prime\prime}$ genuinely $\textit{changes sign}$ there.

A point of inflection need not be stationary. On $y = x^{3} - 3x$ , the origin is a point of inflection where the gradient is $-3$ , not zero.

Example 1Find the point of inflection of

y = x^{3} - 6x^{2} + 4

and state where the curve is convex.

\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = 6x - 12

, zero at

x = 2

2For

x < 2

it is negative; for

x > 2

positive: a sign change, so an inflection.

3At

x = 2

y = 8 - 24 + 4 = -12

AnswerPoint of inflection

(2, -12)

; convex for

x > 2

Why this works

$\mathrm{f}^{\prime\prime}(x) = 0$ is necessary but not sufficient. For $y = x^{4}$ , the second derivative $12x^{2}$ is zero at the origin but positive on both sides, so the curve is convex throughout and the origin is a minimum, not an inflection.

Optimisation

To find the largest or smallest value of a quantity: write the quantity as a function of a single variable (using any constraint to eliminate others), differentiate, set equal to zero, and solve.

Then $\textbf{justify}$ the nature of the stationary point with the second derivative or a sign test, and answer the question actually asked — often the optimal value, not just the variable.

Check the solution makes sense in context: lengths positive, within any stated limits.

Example 1An open-topped box with a square base of side

x

cm has volume

32\,000

cm³. Find the dimensions that minimise the surface area.

1Height

h = \dfrac{32\,000}{x^{2}}

. Surface area

A = x^{2} + 4xh = x^{2} + \dfrac{128\,000}{x}

\dfrac{\mathrm{d}A}{\mathrm{d}x} = 2x - \dfrac{128\,000}{x^{2}} = 0

, so

x^{3} = 64\,000

and

x = 40

\dfrac{\mathrm{d}^{2}A}{\mathrm{d}x^{2}} = 2 + \dfrac{256\,000}{x^{3}} = 6 > 0

x = 40

, so a minimum.

4Then

h = \dfrac{32\,000}{1600} = 20

AnswerBase

40

cm by

40

cm, height

20

cm (minimum area

4800

cm²).

Tip — The first line — expressing the quantity in one variable — is where most marks are lost. Draw a diagram, label everything, and write the constraint equation before differentiating.

Second derivatives in context

In kinematics, if $s$ is displacement then $\dfrac{\mathrm{d}s}{\mathrm{d}t}$ is velocity and $\dfrac{\mathrm{d}^{2}s}{\mathrm{d}t^{2}}$ is acceleration.

In other models the second derivative describes whether growth is speeding up or slowing down. A population with positive rate of growth but negative second derivative is still increasing, but ever more slowly — a point of inflection marks where growth was fastest.

Example 1A particle has displacement

s = t^{3} - 9t^{2} + 24t

metres. Find its acceleration when it is momentarily at rest for the first time.

v = 3t^{2} - 18t + 24 = 3(t - 2)(t - 4)

, zero first at

t = 2

a = \dfrac{\mathrm{d}v}{\mathrm{d}t} = 6t - 18

3At

t = 2

a = 12 - 18

Answer

-6

m s⁻²

Think like an examiner

State the value and sign of $\mathrm{f}^{\prime\prime}$ when classifying.
For points of inflection, show that $\mathrm{f}^{\prime\prime}$ changes sign.
In optimisation, express the quantity in one variable before differentiating.
Answer the question asked, including units and the optimal value if requested.

Common misconceptions

Concluding a point of inflection because

\mathrm{f}^{\prime\prime}(x) = 0

Check for a sign change;

y = x^{4}

has

\mathrm{f}^{\prime\prime}(0) = 0

at a minimum.

Reading

\mathrm{f}^{\prime\prime} > 0

as a maximum.

Positive second derivative: convex, minimum.

Differentiating a quantity that still depends on two variables.

Use the constraint to eliminate one variable first.

Stopping at

x = 40

when asked for the minimum area.

Substitute back to find the optimal value.

Second derivatives

\mathrm{f}^{\prime\prime}(x) = \dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}

Notation.

\mathrm{f}^{\prime\prime} > 0: \text{convex, minimum}

Positive.

\mathrm{f}^{\prime\prime} < 0: \text{concave, maximum}

Negative.

Stretch yourself

A rectangle is inscribed under the curve $y = 12 - x^{2}$ , with its base on the $x$ -axis and two upper corners on the curve, symmetric about the $y$ -axis. Find the maximum possible area of the rectangle, justifying that it is a maximum.

Hint — If the upper right corner is $(x, 12 - x^{2})$ , the width is $2x$ . Write the area as a function of $x$ with $0 < x < \sqrt{12}$ .

Reveal solution

SolutionArea

A = 2x(12 - x^{2}) = 24x - 2x^{3}

for

0 < x < 2\sqrt3

\dfrac{\mathrm{d}A}{\mathrm{d}x} = 24 - 6x^{2} = 0

, so

x^{2} = 4

and

x = 2

(taking the positive root in the domain).

\dfrac{\mathrm{d}^{2}A}{\mathrm{d}x^{2}} = -12x = -24 < 0

x = 2

, so this is a maximum. Then

A = 2(2)(12 - 4) = 32

. Since

A \to 0

at both ends of the domain and there is only one stationary point inside it, this local maximum is the greatest possible area:

32

square units.

\blacksquare

Questions students ask

What does the second derivative measure?

How fast the gradient is changing. Physically, if the function is displacement, it is acceleration.

What if the second derivative is zero at a stationary point?

The test fails. Use the sign of the first derivative either side to classify the point.

Is every point with

\mathrm{f}^{\prime\prime}(x) = 0

a point of inflection?

No. The second derivative must change sign there.

How do I know a local optimum is the overall best?

Check the ends of the domain as well. If there is a single stationary point and the function falls away from it on both sides, it is the global optimum.

Key takeaways

The second derivative is the rate of change of the gradient.
At a stationary point, $\mathrm{f}^{\prime\prime} > 0$ gives a minimum and $\mathrm{f}^{\prime\prime} < 0$ a maximum; zero is inconclusive.
Convex where $\mathrm{f}^{\prime\prime} > 0$ , concave where $\mathrm{f}^{\prime\prime} < 0$ .
Points of inflection need $\mathrm{f}^{\prime\prime}$ to change sign.
Optimisation: one-variable model, differentiate, solve, justify, answer in context.