Optimisation 1 · Stationary points: where the gradient is zero

Module 35 of 47Lesson 1 of 6 · Optimisation 1 · Stationary points: where the gradient is zero

Stationary points & optimisation · Essential understanding 1

Stationary points: where the gradient is zero

IB AI SL 5.6with Prof. Kitso

💻 Using the TI-Nspire CX Student Software on a laptop

The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
enter = Enter · tab = Tab (next box) · esc = Esc · del = Backspace.
Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
Fraction: type / · √: type sqrt( · π: type pi · eˣ: type exp( · store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
New page: Ctrl+I (or the Insert menu) · change page: Ctrl+← / Ctrl+→ · graph entry line: Ctrl+G · var: Ctrl+L.
Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
If you type nDeriv(, the software shows it as centralDiff( — it is the same command.

The big idea: The gradient of a curve tells you how steeply it climbs or falls. At the top of a hill or the bottom of a valley the ground is flat: the gradient is zero. Points like these are called stationary points, and you find them by solving f′(x) = 0.

By the end of this lesson you can:

explain why the tangent is horizontal where the gradient is zero
differentiate and solve $f'(x) = 0$ by hand and with the GDC
give a stationary point as a pair of coordinates (x, y)
use $f'(a) = 0$ to find an unknown number in a function
say what a stationary point means in a real situation

🧑🏾 Imagine this…

Thato walks along a path over a small hill near Kanye. At the start the path climbs: the gradient is positive. At the very top the ground is level for a moment: the gradient is zero. Then the path drops: the gradient is negative. If you put a ruler on the curve at the top, the ruler lies flat. That flat ruler is the horizontal tangent.

Gradient function f′(x)

the derivative. It gives the gradient of the curve at every x. Also written

\dfrac{dy}{dx}

Tangent

a straight line that just touches the curve at one point and has the same direction as the curve there.

Horizontal tangent

a flat tangent. A flat line has gradient 0.

Stationary point

a point on the curve where

f'(x) = 0

. The curve is neither rising nor falling there.

★ Finding stationary points: four steps

1. Differentiate. Power rule: $x^n \to n\,x^{n-1}$ . Rewrite first if needed: $\dfrac{1}{x} = x^{-1}$ and $\dfrac{5}{x^2} = 5x^{-2}$ .
2. Set $f'(x) = 0$ and solve (factorise, quadratic formula, or the GDC).
3. Find the y-value. Put each x back into the original f(x), not into f′(x).
4. Write the point as coordinates $(x,\,y)$ .

⚠️ Watch out

Substitute into f(x) to get y. Putting x into f′(x) just gives 0 again, which is no help.
“Find the stationary point” wants two numbers. “Find the value of x” wants one. Read the command carefully.
A tangent that is horizontal does not always mean a hill top or a valley bottom. Lesson 2 shows the odd cases.

💡 Three GDC routes to f′(x) = 0

(a) Differentiate by hand, then polyRoots(f′(x),x) or menu💻 click menu on keypad▸3: Algebra▸1: Numerical Solve on the result. (b) On a Graphs page enter f1(x) = your function, f2(x) = the derivative template of f1(x) (from the templates palette), then menu💻 click menu on keypad▸6: Analyze Graph▸1: Zero on f2. (c) Graph the function itself and use menu💻 click menu on keypad▸6: Analyze Graph▸2: Minimum or menu💻 click menu on keypad▸6: Analyze Graph▸3: Maximum. Route (c) also shows y at once, but only works for turning points.

Prof. Kitso says

Say it out loud: “derivative equals zero, flat tangent.” Whenever a question talks about a flat tangent, a turning point, a stationary point, or “the rate of change is zero”, your first move is always the same: write f′(x) = 0.

🧪 Try it yourselfSlide the tangent

This is f(x) = x³ − 9x² + 24x + 5 (Example 2). Drag the tangent along the curve and watch the gradient. Where does it read 0? At which two x-values is the tangent flat?

Worked examples

Example 1.1A flat tangent with an unknown number

The curve

y = ax^2 - 12x

has a horizontal tangent at the point where x = 3. Find the value of a.

🧠 Think it through

Horizontal tangent means gradient = 0, so $\dfrac{dy}{dx} = 0$ when x = 3.
Differentiate: $\dfrac{dy}{dx} = 2ax - 12$ .
Substitute x = 3 and set equal to zero, then solve for a.

✍️ By hand

\dfrac{dy}{dx} = 2ax - 12

x = 3

2a(3) - 12 = 0 \;\Rightarrow\; 6a = 12 \;\Rightarrow\; a = 2

🔢 On your TI-Nspire CX II

Check with a solver. Type nSolve(2·a·3−12=0,a) enter ↵💻 Enter.
📍 Where to find it on the TI-Nspire
nSolve( is in menu💻 click menu on keypad▸3: Algebra▸1: Numerical Solve

✅ Answer a = 2.

Check: with a = 2 the gradient is 4x − 12, which is 0 when x = 3 ✓.

Example 1.2Stationary points of a cubic

Find the coordinates of the stationary points of

f(x) = x^3 - 9x^2 + 24x + 5

🧠 Think it through

Differentiate term by term.
Solve f′(x) = 0: it is a quadratic, so factorise or use polyRoots.
Put each x-value into f(x) to get the y-values.

✍️ By hand

f'(x) = 3x^2 - 18x + 24 = 3(x^2 - 6x + 8) = 3(x-2)(x-4)

f'(x) = 0 \;\Rightarrow\; x = 2 \text{ or } x = 4

f(2) = 8 - 36 + 48 + 5 = 25

and

f(4) = 64 - 144 + 96 + 5 = 21

🔢 On your TI-Nspire CX II

Type polyRoots(3x^2−18x+24,x) enter ↵💻 Enter to solve f′(x) = 0.
📍 Where to find it on the TI-Nspire
polyRoots( is in menu💻 click menu on keypad▸3: Algebra▸3: Polynomial Tools▸2: Real Roots of Polynomial
Define the function and evaluate it: Define f(x)=x^3−9x^2+24x+5 then f(2) and f(4).
Or on a Graphs page: f1 = the function, f2 = derivative template of f1. menu💻 click menu on keypad▸6: Analyze Graph▸1: Zero on f2 gives x = 2 and x = 4.

✅ Answer The stationary points are (2, 25) and (4, 21).

The zeros of the derivative graph give the x-values only. The y-values come from the original function.

Example 1.3A function with

\dfrac{1}{x}

in it

Find the stationary points of

f(x) = 3x + \dfrac{12}{x}

, for

x \neq 0

🧠 Think it through

Rewrite $\dfrac{12}{x}$ as 12x⁻¹ so the power rule works.
f′(x) = 3 − 12x⁻² = 3 − $\dfrac{12}{x^{2}}$ . Set it equal to zero.
There are two solutions (±). Find the y-value for each.

✍️ By hand

f(x) = 3x + 12x^{-1} \;\Rightarrow\; f'(x) = 3 - 12x^{-2} = 3 - \dfrac{12}{x^2}

3 - \dfrac{12}{x^2} = 0 \;\Rightarrow\; x^2 = 4 \;\Rightarrow\; x = 2 \text{ or } x = -2

f(2) = 6 + 6 = 12

and

f(-2) = -6 - 6 = -12

🔢 On your TI-Nspire CX II

Graph 3x+12/x with window x from −6 to 6, y from −30 to 30. Use menu💻 click menu on keypad▸6: Analyze Graph▸2: Minimum and menu💻 click menu on keypad▸6: Analyze Graph▸3: Maximum on each branch.

✅ Answer The stationary points are (2, 12) and (−2, −12).

The curve has two separate branches (it cannot cross x = 0). Notice that the point on the left, (−2, −12), is lower than the one on the right. That is fine: each point is only a hill top or valley bottom for its own neighbourhood. More in Lesson 2.

Example 1.4Finding two unknowns from a stationary point

The curve

y = x^3 + ax^2 + bx

has a stationary point at

(2,\,-8)

(a) Find the values of a and b.
(b) Find the other stationary point.

🧠 Think it through

Two facts about the point (2, −8): the gradient is 0 there, AND the point is on the curve.
Fact 1: $\dfrac{dy}{dx} = 3x^2 + 2ax + b = 0$ at x = 2. Fact 2: put x = 2, y = −8 into the curve.
That gives two equations in a and b. Solve them.
Then write the gradient function with the numbers, solve it again and take the other root.

✍️ By hand
; at : , so   …(1)
On the curve: , so   …(2)
(2) − (1): . Then , so .
Now  and , so  or .

🔢 On your TI-Nspire CX II

Solve the pair at once: menu💻 click menu on keypad▸3: Algebra▸2: Solve System of Linear Equations or type linSolve({4a+b=−12,4a+2b=−16},{a,b}).
For part (b): polyRoots(3x^2−4x−4,x) gives the x-values, then evaluate y.
📍 Where to find it on the TI-Nspire
polyRoots( is in menu💻 click menu on keypad▸3: Algebra▸3: Polynomial Tools▸2: Real Roots of Polynomial

✅ Answer (a) a = −2, b = −4. (b) The other stationary point is (−0.667, 1.48), that is

\left(-\dfrac{2}{3},\,\dfrac{40}{27}\right)

Two unknowns need two equations. Using “gradient = 0” alone gives only one.

Example 1.5A drone that stops climbing

Kagiso flies a drone over a farm near Palapye. For the first 4 seconds its height is

h(t) = t^3 - 6t^2 + 9t + 4

metres,

0 \le t \le 4

(a) Find the rate of change of height, h′(t).
(b) Find when the drone is momentarily neither climbing nor falling, and its height then.
(c) Is the drone climbing or falling at t = 0.5 s and at t = 2 s?

🧠 Think it through

h′(t) is the gradient of the height graph: the vertical speed in m s⁻¹.
“Neither climbing nor falling” means h′(t) = 0.
For (c), substitute the time into h′(t). Positive gradient = climbing, negative = falling.

✍️ By hand

(a)

h'(t) = 3t^2 - 12t + 9

(b)

3t^2 - 12t + 9 = 3(t-1)(t-3) = 0 \Rightarrow t = 1 \text{ or } t = 3

h(1) = 1 - 6 + 9 + 4 = 8

and

h(3) = 27 - 54 + 27 + 4 = 4

(c)

h'(0.5) = 0.75 - 6 + 9 = 3.75 > 0

(climbing);

h'(2) = 12 - 24 + 9 = -3 < 0

(falling)

🔢 On your TI-Nspire CX II

Graph t^3−6t^2+9t+4 with window x from 0 to 4, y from 0 to 12 and find the turning points with the Maximum and Minimum tools.
Check the gradients: define h′ and evaluate.

✅ Answer (a) h′(t) = 3t² − 12t + 9 m s⁻¹. (b) At t = 1 s (height 8 m) and t = 3 s (height 4 m). (c) Climbing at 0.5 s (3.75 m s⁻¹), falling at 2 s (−3 m s⁻¹).

So the drone climbs to 8 m, falls to 4 m, then climbs again. The stationary moments are where it turns round.

Challenge questions

⭐ Challenge 1.6Flat at x = 3Paper 1 style

The curve

y = x^3 - 3x^2 + kx

has a horizontal tangent at the point where

x = 3

(a) Find the value of k.
(b) Find the gradient of the curve at x = 0.

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

$\dfrac{dy}{dx} = 3x^2 - 6x + k$ , and it equals 0 when x = 3.
27 − 18 + k = 0.
Put x = 0 into the gradient function with your k.

✍️ By hand

\dfrac{dy}{dx} = 3x^2 - 6x + k

; at

x=3

27 - 18 + k = 0 \Rightarrow k = -9

x = 0

\dfrac{dy}{dx} = k = -9

🔢 On your TI-Nspire CX II

Check: nSolve(3·9−6·3+k=0,k).
📍 Where to find it on the TI-Nspire
nSolve( is in menu💻 click menu on keypad▸3: Algebra▸1: Numerical Solve

✅ Answer (a) k = −9. (b) The gradient at x = 0 is −9.

⭐ Challenge 1.7A cubic with four questionsPaper 2 style

Let

f(x) = 2x^3 - 3x^2 - 12x + 1

(a) Find f′(x).
(b) Find the coordinates of the two stationary points of the graph of f.
(c) Find the equation of the tangent to the curve at x = 0.
(d) The line y = k meets the graph of f at three different points. Use the stationary points to find the possible values of k.

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

f′(x) = 6x² − 6x − 12 = 6(x² − x − 2).
Solve f′(x) = 0, then find f at each x.
At x = 0 the gradient is f′(0) and the point is (0, f(0)). Use y = mx + c.
A horizontal line crosses the curve three times only if it passes between the lowest turning value and the highest turning value.

✍️ By hand

(a)

f'(x) = 6x^2 - 6x - 12

(b)

6(x-2)(x+1) = 0 \Rightarrow x = 2,\,-1

f(2) = 16 - 12 - 24 + 1 = -19

;

f(-1) = -2 - 3 + 12 + 1 = 8

(c)

f'(0) = -12

and

f(0) = 1

, so

y = -12x + 1

(d) The turning values are 8 (hill top) and −19 (valley), so

-19 < k < 8

🔢 On your TI-Nspire CX II

Graph the function and mark the two turning points; draw a horizontal line such as y = 0 to see three crossings.

✅ Answer (a) f′(x) = 6x² − 6x − 12. (b) (−1, 8) and (2, −19). (c) y = −12x + 1. (d) −19 < k < 8.

In (d) k = 8 or k = −19 would touch the curve at a turning point, giving only two meeting points.

Next: Lesson 2 · Local maximum and minimum points

IB DP Mathematics: Applications and Interpretation (AI) — SL and HL Optimisation 1 · Stationary points: where the gradient is zero

IB DP Mathematics: Applications and Interpretation (AI) — SL and HL

Stationary points: where the gradient is zero

Worked examples

Challenge questions