5.8 Stationary points, optimisation and inflexion (IB Maths: Analysis and Approaches HL): Revision notes

Section 1

Finding and classifying stationary points

A stationary point is where $f'(x)=0$ . It is a local maximum if the graph turns from increasing to decreasing, and a local minimum if it turns from decreasing to increasing.

First derivative test: check the sign of $f'$ just either side. $+\to-$ is a maximum; $-\to+$ is a minimum; no change of sign means neither.

Second derivative test: if $f'(a)=0$ and

$f''(a)<0$ : local maximum;
$f''(a)>0$ : local minimum;
$f''(a)=0$ : the test fails, so use the first derivative test instead.

Example: $f(x)=x^{3}-6x^{2}+9x+1$ has $f'(x)=3(x-1)(x-3)$ and $f''(x)=6x-12$ : maximum $(1,5)$ , minimum $(3,1)$ .

Key termsstationary pointlocal maximumlocal minimum

Common mistake

Concluding "not a max or min" because $f''(a)=0$ . For $y=x^{4}$ , $f''(0)=0$ but $(0,0)$ is a minimum. Use the sign of $f'$ instead.

Exam tip

Always give both coordinates of a stationary point unless the question asks only for $x$ .

Section 2

Concavity

A graph is concave-up where $f''(x)>0$ (it bends upwards, like a cup, and its gradient is increasing) and concave-down where $f''(x)<0$ (it bends downwards and its gradient is decreasing).

This is why the second derivative test works: a stationary point on a concave-down section must be a maximum.

Key termsconcave-upconcave-down

Common mistake

Confusing concavity with increasing/decreasing. $f(x)=x^{3}-6x^{2}+9x+1$ is concave-down for all $x<2$ , but it is increasing for $x<1$ and decreasing for $1<x<2$ .

Section 3

Points of inflexion

A point of inflexion is where the concavity changes: $f''(x)=0$ and $f''$ changes sign.

Zero gradient (stationary inflexion): also $f'=0$ , e.g. $g(x)=x^{4}-4x^{3}+5$ at $x=0$ , where $g'=4x^{2}(x-3)$ is negative on both sides.
Non-zero gradient: $f'\neq0$ , e.g. the same $g$ at $x=2$ , where $g'(2)=-16$ .

$f''(x)=0$ on its own is not sufficient. For $y=x^{4}$ , $y''=12x^{2}=0$ at $x=0$ , but $y''>0$ on both sides, so there is no inflexion.

Key termspoint of inflexionstationary point of inflexion

Exam tip

To show a change of sign, evaluate $f''$ at a value just below and just above, or factorise $f''$ and reason about each factor.

Section 4

Optimisation

To optimise a real-world quantity:

Write the quantity in terms of one variable, using any constraint (e.g. a fixed volume) to eliminate the other.
Differentiate and set the derivative to zero.
Justify the nature (second derivative or sign change).
Check any end points of the domain, then answer the question in context with units.

Example: an open box with square base $x$ and volume $32\,000$ cm $^{3}$ has $h=\frac{32\,000}{x^{2}}$ and card area $S=x^{2}+\frac{128\,000}{x}$ . $S'=0$ gives $x=40$ , $h=20$ , and $S''(40)=6>0$ confirms a minimum of 4800 cm $^{2}$ .

Key termsoptimisationconstraintend points

Common mistake

Stopping at a local maximum without checking the ends of the domain. For $P(x)=-x^{3}+9x^{2}-15x-10$ on $[0,8]$ the local maximum $P(5)=15$ must be compared with $P(0)$ and $P(8)$ .

Section 5

Interpreting inflexion in context

At a non-stationary point of inflexion, $f'$ has a maximum or minimum. In a profit model this is where profit is growing (or falling) fastest. For $P(x)=-x^{3}+9x^{2}-15x-10$ the inflexion is at $(3,-1)$ with $P'(3)=12$ : at 300 units, profit is rising at its greatest rate, 12 thousand dollars per extra hundred units.

Key termspoint of diminishing returns

Must know

Stationary points: solve $f'(x)=0$ ; classify with the sign of $f'$ or of $f''$ .
$f''>0$ minimum and concave-up; $f''<0$ maximum and concave-down.
Inflexion: $f''=0$ and changes sign. The gradient there may be zero or non-zero.
$y=x^{4}$ at $(0,0)$ shows $f''=0$ is not enough.
Optimisation: one variable, differentiate, justify, check end points, answer in context.

That's the notes covered.

Carry on to the next subtopic.