Stationary points AI HL

IB Mathematics: Applications and Interpretation HL · Unit 5: Calculus

Lesson 5.6 · Stationary Points

Solve $f'(x)=0$ , classify local maxima and minima, and check the whole domain for greatest and least values.

Share the message beyond familiar places

Paul and Barnabas understood that the good news of Jesus Christ was for people beyond their first audience. They recalled God’s purpose that his salvation should reach “unto the ends of the earth” (Acts 13:47, KJV). When the Gentiles heard the message, many received it with joy (Acts 13:46–48).

Jesus died for our sins and rose again. His invitation to repent and believe is meant to be shared with every person. Paul and Barnabas kept speaking even when their circumstances changed. As we study where a curve changes direction, we can ask God to make us faithful in every season and ready to tell others about Christ.

Learning objectives

By the end of this lesson, you should be able to:

find the values of $x$ for which a curve has zero gradient;
find the coordinates of stationary points by solving $f'(x)=0$ ;
classify a stationary point as a local maximum, local minimum or neither;
use a graphing calculator to locate and check stationary points;
distinguish local extrema from greatest and least values on a stated domain;
interpret stationary points in practical situations.

1. What is a stationary point?

A stationary point is a point on a differentiable curve where the tangent is horizontal. Its gradient is zero:

$\boxed{f'(x)=0}$

Solving this equation gives the possible $x$ -coordinates. Substitute each value into the original function to find the complete coordinates $(x,f(x))$ .

What the curve does	Classification	Sign of $f'$
Rises, then falls	Local maximum	Positive to negative
Falls, then rises	Local minimum	Negative to positive
Continues in the same direction	Neither maximum nor minimum	No sign change

The word local means that the value is greatest or least compared with nearby points. It does not automatically describe the entire domain.

2. Find and classify stationary points

Worked example · Two turning points.

Find and classify the stationary points of:

$f(x)=x^3-3x^2-9x+5$

Differentiate and solve $f'(x)=0$ :

$f'(x)=3x^2-6x-9 =3(x+1)(x-3)$

$3(x+1)(x-3)=0 \quad\Longrightarrow\quad x=-1\ \text{or}\ x=3$

Find the corresponding outputs:

$f(-1)=10,\qquad f(3)=-22$

Interval	Sign of $f'(x)$	Behaviour of $f$
$x<-1$	Positive	Increasing
$-1<x<3$	Negative	Decreasing
$x>3$	Positive	Increasing

At $x=-1$ , the derivative changes from positive to negative: $(-1,10)$ is a local maximum. At $x=3$ , it changes from negative to positive: $(3,-22)$ is a local minimum.

A zero derivative does not guarantee a turning point

For $g(x)=x^3$ , the derivative is $g'(x)=3x^2$ , so $g'(0)=0$ . Yet $g'(x)$ is positive on both sides of zero. The curve keeps rising through $(0,0)$ .

Therefore $(0,0)$ is a stationary point of inflection, not a local maximum or minimum. Always check what happens on both sides of a stationary point.

3. Local versus greatest and least values

On a closed interval $[a,b]$ , check both the stationary points inside the interval and the two endpoints when asked for the greatest or least value.

Worked example · A bounded cubic.

Consider $h(x)=x^3-3x$ on $-3\leq x\leq3$ .

$h'(x)=3x^2-3=3(x-1)(x+1)$

The stationary points are at $x=-1$ and $x=1$ . Evaluate the function at both of these values and at the endpoints:

$x$	$-3$	$-1$	$1$	$3$
$h(x)$	$-18$	$2$	$-2$	$18$

The stationary point $(-1,2)$ is a local maximum, but the greatest value on this domain is $18$ at the endpoint $x=3$ . The stationary point $(1,-2)$ is a local minimum, but the least value is $-18$ at $x=-3$ .

4. Stationary points in context

Worked example · Maximum profit.

A company’s profit from selling $q$ items is modelled by $P(q)=-0.5q^2+20q-100$ euros, for $0\leq q\leq30$ .

$P'(q)=-q+20$

Setting $P'(q)=0$ gives $q=20$ . The derivative is positive before $20$ and negative after $20$ , so this is a local maximum. Check the endpoints to find the greatest value on the stated domain:

$P(0)=-100,\qquad P(20)=100,\qquad P(30)=50$

The greatest modelled profit is €100 from selling 20 items. The model predicts a loss of €100 if no items are sold.

Using a TI-84 Plus CE

Enter $f(x)$ as Y1 and graph it in a window that includes the stated domain and relevant outputs.
To graph a numerical derivative, enter nDeriv(Y1,X,X) as Y2. Find where this graph meets the $x$ -axis.
Alternatively, use 2nd → TRACE (CALC) → maximum or minimum on the graph of Y1. Select left and right bounds around the turning point.
Check the sign of the derivative on each side. A graph or a numerical zero alone does not establish the classification.
For greatest or least values on a closed interval, also evaluate the original function at both endpoints.

The TI-84 computes and graphs numerical derivatives; it does not produce a symbolic derivative formula. Differentiate algebraically when the question calls for a formula.

IB-style Paper 1 · Skill practice

Show your method. Mark allocations are shown in square brackets.

Question 1 · Two stationary points [8 marks]

Let $f(x)=x^3-3x^2-9x+5$ .

Find $f'(x)$ . [2]
Find the coordinates of all stationary points. [4]
Classify each stationary point. [2]

Show solution

(a) $f'(x)=3x^2-6x-9=3(x+1)(x-3)$ . (b) $f'(x)=0$ at $x=-1$ and $x=3$ . Since $f(-1)=10$ and $f(3)=-22$ , the points are $(-1,10)$ and $(3,-22)$ . (c) The sign of $f'$ changes $+$ to $-$ at $-1$ , giving a local maximum. It changes $-$ to $+$ at $3$ , giving a local minimum.

Question 2 · A stationary point without a turn [5 marks]

Let $g(x)=x^3$ .

Find its stationary point. [2]
Explain why this point is neither a local maximum nor a local minimum. [3]

Show solution

(a) $g'(x)=3x^2=0$ at $x=0$ , giving $(0,0)$ . (b) For $x\neq0$ , $3x^2>0$ on both sides of zero. The function increases before and after the stationary point, so there is no change from rising to falling or falling to rising.

Question 3 · A repeated solution [7 marks]

Let $p(x)=x^4-4x^3$ .

Show that $p'(x)=4x^2(x-3)$ . [2]
Find the coordinates of its stationary points. [3]
Classify both points. [2]

Show solution

(a) $p'(x)=4x^3-12x^2=4x^2(x-3)$ . (b) $p'(x)=0$ at $x=0$ and $x=3$ , giving $(0,0)$ and $(3,-27)$ . (c) The derivative is negative on both sides of $0$ , so $(0,0)$ is a stationary point with no turn. At $3$ , the derivative changes from negative to positive, so $(3,-27)$ is a local minimum.

Question 4 · Reading a derivative [5 marks]

A differentiable function has derivative $f'(x)=3(x+2)(x-4)$ .

Find the $x$ -coordinates of its stationary points. [2]
State which one is a local maximum and which one is a local minimum. Explain using derivative signs. [3]

Show solution

(a) $x=-2$ and $x=4$ . (b) The signs of $f'$ are $+$ for $x<-2$ , $-$ for $-2<x<4$ , and $+$ for $x>4$ . Hence there is a local maximum at $x=-2$ and a local minimum at $x=4$ . The $y$ -coordinates cannot be determined from the given derivative alone.

IB-style Paper 2 · Applications

Question 5 · Calculator investigation [9 marks]

Let $f(x)=x^3-3x^2-14x+432$ , for $-5\leq x\leq6$ .

Use your GDC to find the coordinates of its stationary points. Give each coordinate to three significant figures. [4]
Classify the two stationary points. [2]
Find the greatest and least values of $f$ on the stated domain. [3]

Show solution

(a) $f'(x)=3x^2-6x-14$ . Its zeros are approximately $x=-1.38048$ and $x=3.38048$ . The stationary points are approximately $(-1.38,443)$ and $(3.38,389)$ to three significant figures. (b) The derivative signs are $+$ , $-$ , $+$ , so the first point is a local maximum and the second a local minimum. (c) Check the endpoints: $f(-5)=302$ and $f(6)=456$ . Comparing all four values, the greatest is $456$ at $x=6$ and the least is $302$ at $x=-5$ .

Question 6 · Profit model [8 marks]

Profit from selling $q$ items is modelled by $P(q)=-0.5q^2+20q-100$ euros, where $0\leq q\leq30$ .

Find $P'(q)$ and the stationary value of $q$ . [3]
Show that this stationary point is a local maximum. [2]
Find the greatest and least modelled profits on the domain, stating when they occur. [3]

Show solution

(a) $P'(q)=-q+20=0$ gives $q=20$ . (b) $P'(q)>0$ before $20$ and $P'(q)<0$ after $20$ , so profit changes from increasing to decreasing. (c) $P(0)=-100$ , $P(20)=100$ and $P(30)=50$ . The greatest profit is €100 at $q=20$ ; the least modelled profit is $-€100$ at $q=0$ , meaning a €100 loss.

Question 7 · Designing an open box [9 marks]

Equal squares of side $x$ cm are cut from the corners of a $24$ cm by $18$ cm sheet. The sides are folded up to make an open box. Its volume is $V(x)=x(24-2x)(18-2x)$ cm $^3$ , where $0<x<9$ .

Show that $V'(x)=12x^2-168x+432$ . [3]
Find the stationary value of $x$ within the physical domain. [3]
Find the maximum volume to three significant figures. [2]
Explain why the other solution of $V'(x)=0$ is rejected. [1]

Show solution

(a) Expanding gives $V(x)=4x^3-84x^2+432x$ , so $V'(x)=12x^2-168x+432$ . (b) Solving $V'(x)=0$ gives $x=7\pm\sqrt{13}$ . Only $x=7-\sqrt{13}\approx3.39$ cm lies in $0<x<9$ . (c) At this value, $V\approx\boxed{655\text{ cm}^3}$ to three significant figures. The volume increases before this value and decreases after it within the physical domain. (d) $7+\sqrt{13}\approx10.6$ cm exceeds the possible cut size, so it cannot form the box.

Question 8 · Displacement and turning points [8 marks]

A particle’s displacement is $s(t)=t^3-6t^2+9t$ metres for $0\leq t\leq4$ , where $t$ is measured in seconds.

Find its velocity $v(t)$ and the times at which it is momentarily at rest. [3]
Find and classify the stationary points of its displacement. [3]
Find the greatest and least displacement values over the stated interval. [2]

Show solution

(a) $v(t)=s'(t)=3t^2-12t+9=3(t-1)(t-3)$ , so the particle is momentarily at rest at $t=1$ s and $t=3$ s. (b) $s(1)=4$ m and $s(3)=0$ m. Velocity changes from positive to negative at $t=1$ , giving a local maximum $(1,4)$ ; it changes from negative to positive at $t=3$ , giving a local minimum $(3,0)$ . (c) The endpoint values are $s(0)=0$ and $s(4)=4$ . The greatest displacement is $4$ m at $t=1$ and $t=4$ ; the least is $0$ m at $t=0$ and $t=3$ .

Exam checklist

Solve $f'(x)=0$ , then substitute each solution into $f(x)$ to obtain coordinates. Check derivative signs on both sides before classifying a stationary point. For the greatest or least value on a closed interval, compare every stationary value and both endpoint values. In applications, state the answer within the permitted domain and include units.

Lesson 5.6 · Stationary Points

Share the message beyond familiar places

Learning objectives

1. What is a stationary point?

2. Find and classify stationary points

A zero derivative does not guarantee a turning point

3. Local versus greatest and least values

4. Stationary points in context

Using a TI-84 Plus CE

IB-style Paper 1 · Skill practice

Question 1 · Two stationary points [8 marks]

Question 2 · A stationary point without a turn [5 marks]

Question 3 · A repeated solution [7 marks]

Question 4 · Reading a derivative [5 marks]

IB-style Paper 2 · Applications

Question 5 · Calculator investigation [9 marks]

Question 6 · Profit model [8 marks]

Question 7 · Designing an open box [9 marks]

Question 8 · Displacement and turning points [8 marks]

Exam checklist

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IB Mathematics: Applications and Interpretation HL 6 Sections80 LessonsLifetime

Curriculum

Stationary points AI HL

Share the message beyond familiar places

Learning objectives

1. What is a stationary point?

2. Find and classify stationary points

A zero derivative does not guarantee a turning point

3. Local versus greatest and least values

4. Stationary points in context

Using a TI-84 Plus CE

IB-style Paper 1 · Skill practice

Question 1 · Two stationary points [8 marks]

Question 2 · A stationary point without a turn [5 marks]

Question 3 · A repeated solution [7 marks]

Question 4 · Reading a derivative [5 marks]

IB-style Paper 2 · Applications

Question 5 · Calculator investigation [9 marks]

Question 6 · Profit model [8 marks]

Question 7 · Designing an open box [9 marks]

Question 8 · Displacement and turning points [8 marks]

Exam checklist

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