SL 5.6 Find and classify stationary points | International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Calculus

IB Mathematics: Analysis and Approaches SL — Study topic SL 5.6

A curve can rise, fall, or flatten as its input changes. A stationary point is a point where the curve has a horizontal tangent: its gradient is zero. Finding such a point is not enough to describe its behaviour. You must also decide whether the curve changes from rising to falling, falling to rising, or continues in the same direction. For a stationary point of inflection, the curve must also change concavity: its shape changes from bending one way to bending the other. This lesson develops these ideas using derivatives, sign checks, graphs, and contextual interpretation.

What you will learn

Explain what makes a point on a curve stationary.
Find stationary points by solving an equation involving the first derivative.
Classify stationary points as local maxima, local minima, or stationary points of inflection.
Use derivative signs and, where appropriate, the second derivative to justify a classification.
Check analytical results with a graphing calculator and interpret points in context.

1. Prior knowledge: gradients and derivatives

For a differentiable function

y=f(x)

, the derivative

f'(x)

gives the gradient of the curve at each input

x

. A positive derivative means the curve is increasing, and a negative derivative means it is decreasing. A zero derivative means the tangent is horizontal.

A stationary point occurs at an input

x=a

where

f'(a)=0

. The point on the graph is

(a,f(a))

, so finding the input alone is not a complete answer. Solve

f'(x)=0

, then substitute each solution into the original function.

A stationary point is not necessarily a maximum or minimum. A curve may flatten briefly while continuing to increase. Classification depends on what the curve does on either side of the point. Work within the stated domain: an input outside the domain cannot be a stationary point of the given function.

f'(a)=0

Differentiate the function, then solve $f'(x)=0$ .
Use the original function to find the point's $y$ -coordinate.
A horizontal tangent alone does not determine the classification.

2. Classifying a stationary point

Check the sign of

f'(x)

just to the left and right of a stationary input. If the sign changes from positive to negative, the curve changes from increasing to decreasing, so the point is a local maximum. If the sign changes from negative to positive, the curve changes from decreasing to increasing, so the point is a local minimum.

If the derivative has the same sign on both sides, the curve continues in the same direction rather than turning. This alone does not establish an inflection point. To classify a stationary point of inflection, also check that the curve changes concavity. One way is to examine the sign of the second derivative

f''(x)

on either side: a change in its sign shows that the curve changes concavity.

The second derivative describes how the gradient changes. At a stationary input

a

, a negative value of

f''(a)

indicates a local maximum, while a positive value indicates a local minimum. If

f''(a)=0

, this test does not decide the classification. Check the signs of

f'(x)

on either side; if there is no change in direction, also check for a change in concavity before calling the point a stationary point of inflection.

These classifications are local: they describe behaviour near the stationary point. They do not, by themselves, show that it is the highest or lowest value over an entire domain.

Positive to negative derivative: local maximum.
Negative to positive derivative: local minimum.
Same derivative sign means no change in direction; a stationary point of inflection additionally requires a change in concavity.
A zero second derivative is inconclusive; check derivative signs and, for an inflection classification, concavity on either side.

3. Representations, context, and technology

The algebraic method identifies candidate inputs by solving

f'(x)=0

. A sign check then shows whether the function increases or decreases around each candidate. Concavity can be checked from the sign of

f''(x)

on either side. On a graph, a local maximum looks like a nearby peak and a local minimum like a nearby valley. A stationary point of inflection has a horizontal tangent, continues in the same direction, and changes concavity.

Graphing technology is useful for checking the approximate location and shape of a stationary point, especially when an equation is difficult to solve exactly. Enter the function and inspect its graph near each candidate. A numerical maximum or minimum feature can estimate coordinates, but the derivative equation and a classification argument explain why the point qualifies. A graph window can hide behaviour, so choose a suitable scale and confirm the result analytically where possible.

In a context, the input and output may have units. If

t

is time in seconds and

s(t)

is position in metres, then

s'(t)

is measured in metres per second. A stationary point of

s

has zero instantaneous velocity. Check whether position changes from increasing to decreasing or vice versa, and interpret the point only within the stated time interval and model.

f'(x)>0,\quad f'(x)<0

Use derivative signs to connect algebra with increasing and decreasing behaviour.
For a stationary point of inflection, establish both no change in direction and a change in concavity.
Use a graph as a check, not as a replacement for reasoning.
Include coordinates, domain restrictions, and units when the context requires them.

4. A reliable solution structure

State the function and its relevant domain. Differentiate carefully, solve the stationary condition, and find the corresponding function values. For each candidate, classify it using the second derivative when its value is non-zero, or a first-derivative sign check when needed.

If the derivative has the same sign on both sides of a stationary input, check whether the second derivative changes sign on either side before classifying the point as a stationary point of inflection. A zero value of the second derivative at the point itself is not enough to prove that concavity changes.

Write a conclusion that includes the point and its classification. If an answer is numerical, give a suitable accuracy and retain enough precision during intermediate calculations. In an exam-style response, show the equation used to find the candidate and the evidence supporting the classification.

For example, a candidate input

a

gives a stationary point only when

f'(a)=0

and the point lies in the function's domain. Its coordinates are then

(a,f(a))

.

Find candidates, coordinates, and classifications in that order.
Do not classify a point from its coordinates alone.
Give exact values when they are available and appropriate.

Worked example

A local maximum and a local minimum

Find and classify all stationary points of

f(x)=x^3-3x^2+2

.

Differentiate
Apply the power rule to find the gradient function.
$f'(x)=3x^2-6x$
Find candidate inputs
Set the derivative equal to zero and factor. Both solutions are stationary inputs.
$3x(x-2)=0\Rightarrow x=0,\ 2$
Find the coordinates
Substitute each input into the original function, rather than the derivative, to obtain the corresponding output values.
$f(0)=2,\quad f(2)=-2$
Classify the points
The second derivative is $f''(x)=6x-6$ . At $x=0$ it is negative, so the point is a local maximum. At $x=2$ it is positive, so the point is a local minimum.
$f''(0)=-6,\quad f''(2)=6$

Answer: The stationary points are

(0,2)

, a local maximum, and

(2,-2)

, a local minimum.

Check: The derivative changes from positive to negative at

x=0

, and from negative to positive at

x=2

, agreeing with the classifications.

Worked example

A stationary point of inflection

Find and classify the stationary point of

g(x)=(x-1)^3+4

.

Differentiate
Differentiate the cubic expression to obtain the gradient.
$g'(x)=3(x-1)^2$
Solve for the stationary input
The derivative is zero only when $x=1$ . Substituting into the original function gives the point's output.
$g'(x)=0\Rightarrow x=1,\quad g(1)=4$
Check direction and concavity
For inputs on either side of $1$ , $g'(x)=3(x-1)^2$ is positive, so the function keeps increasing and does not turn. The second derivative is $g''(x)=6(x-1)$ ; it is negative to the left of $1$ and positive to the right. Thus the curve changes concavity at the stationary point.
$g'(x)>0\ (x\ne1),\quad g''(x)<0\ (x<1),\quad g''(x)>0\ (x>1)$
Classify
The point has a horizontal tangent, no change in direction, and a change in concavity. It is therefore a stationary point of inflection.
$g''(1)=0$

Answer: The stationary point is

(1,4)

, a stationary point of inflection.

Check: A graph shows the curve flattening at

(1,4)

while continuing to rise and changing concavity there.

Worked example

Using a graphing calculator to check a result

For

h(x)=x^3-6x

, find and classify the stationary points. Use a graphing calculator to check the result.

Find candidates analytically
Differentiate, factor, and solve the stationary condition.
$h'(x)=3x^2-6=3(x^2-2)=0\Rightarrow x=\pm\sqrt{2}$
Find the coordinates
Substitute the candidate inputs into $h$ . The outputs have opposite signs.
$h(-\sqrt{2})=4\sqrt{2},\quad h(\sqrt{2})=-4\sqrt{2}$
Classify analytically
Since $h''(x)=6x$ , its value is negative at $-\sqrt{2}$ and positive at $\sqrt{2}$ . Therefore the first point is a local maximum and the second is a local minimum.
$h''(-\sqrt{2})<0,\quad h''(\sqrt{2})>0$
Check with technology
Graph $h(x)$ in a window that includes both candidate inputs. A calculator's maximum and minimum features should give approximate coordinates near $(-1.414,5.657)$ and $(1.414,-5.657)$ . These decimals check the exact results; they do not replace the derivative-based justification.
$\sqrt{2}\approx1.414,\quad4\sqrt{2}\approx5.657$

Answer: The local maximum is

(-\sqrt{2},4\sqrt{2})

and the local minimum is

(\sqrt{2},-4\sqrt{2})

.

Check: The derivative is positive for inputs outside the interval between the two stationary inputs and negative between them, matching a maximum followed by a minimum.

Common mistakes and how to avoid them

Calling every point where

f'(x)=0

a maximum or minimum.

Correction: Check the derivative signs on both sides, or use the second derivative when its value is non-zero. A stationary point can have another classification.

Calling a stationary point of inflection whenever the derivative has the same sign on either side.

Correction: The same derivative sign shows no change in direction, but does not prove an inflection. Also check that concavity changes, for example by checking whether

f''(x)

changes sign across the point.

Treating

f''(a)=0

as proof of a stationary point of inflection.

Correction: A zero second derivative at the point does not decide the classification. Check the first-derivative signs for direction and the second-derivative signs on either side for a change in concavity.

Giving only the input value where the derivative is zero.

Correction: Substitute into the original function to give the full point

(a,f(a))

, then state its classification.

Using a calculator graph as the entire justification.

Correction: Use the graph to check shape and approximate coordinates, and show the derivative equation and classification reasoning.

Lesson summary

A stationary point occurs where the derivative is zero.
Find its coordinates by substituting the stationary input into the original function.
A positive-to-negative derivative sign change gives a local maximum; a negative-to-positive change gives a local minimum.
A stationary point of inflection requires no change in direction and a change in concavity.
The second derivative can classify a stationary point when it is non-zero; if it is zero, check derivative signs and concavity on either side.

Check your understanding

Question 1

For

p(x)=x^3-3x

, which classification applies at the stationary input

x=1

?

Local maximum
Local minimum
Stationary point of inflection
The input is not stationary

Show answer and explanation

Local minimum

Here

p'(x)=3x^2-3

, so

p'(1)=0

. Also,

p''(1)=6>0

, so the stationary point is a local minimum.

Question 2

A function has a stationary input

x=a

. Its derivative is positive just to the left and negative just to the right. What is the classification?

Local maximum
Local minimum
Stationary point of inflection
The classification cannot be determined from these signs

Show answer and explanation

Local maximum

The function changes from increasing to decreasing, so it has a local maximum.

Question 3

At a stationary input, the second derivative is zero. What should you do next to classify a possible stationary point of inflection?

Conclude it is a local maximum
Conclude it is a local minimum
Check the first-derivative signs for direction and second-derivative signs on either side for a concavity change
Conclude it is not stationary

Show answer and explanation

Check the first-derivative signs for direction and second-derivative signs on either side for a concavity change

A zero second derivative is inconclusive. Check whether direction changes using the first derivative. To identify a stationary point of inflection, also establish a change in concavity, for example from a sign change in the second derivative.

Key terms

Derivative: A function that gives the gradient of a curve at each input.
Stationary point: A point on a differentiable curve where the gradient is zero.
Local maximum: A point whose function value is greater than nearby function values.
Local minimum: A point whose function value is less than nearby function values.
Concavity: The way a curve bends over an interval; a change in concavity means it bends in opposite ways on either side.
Stationary point of inflection: A stationary point where the curve changes concavity and does not change from increasing to decreasing or from decreasing to increasing.

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SL 5.6 · Find and classify stationary points

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

IB Mathematics: Analysis and Approaches SL — Study topic SL 5.6

What you will learn

1. Prior knowledge: gradients and derivatives

2. Classifying a stationary point

3. Representations, context, and technology

4. A reliable solution structure

A local maximum and a local minimum

A stationary point of inflection

Using a graphing calculator to check a result

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through IB AA SL

About this lesson and its review