IGCSE, Edexcel, IB and AP Science Courses | Dr Dabour Hub Academy

Overview

This lesson covers lessons 2-3 and 2-4 of the unit "Behaviour of the Function and Curve Sketching".
After it you should be able to:
use the second derivative for the direction of convexity, inflection points and the second-derivative test;
sketch the curve of a polynomial up to degree three from its derivatives;
model and solve maximum and minimum problems from geometry and daily life.

🌳 Concept tree

Convexity
Inflection Points
Curve Sketching and Optimisation

├── 2-3 (a) Convexity of a curve

│ ├── Direction of convexity

│ └── The second-derivative test for convexity

├── 2-3 (b) Inflection points

│ ├── Definition

│ └── Finding inflection points

├── 2-3 (c) Second-derivative test for local extrema

│ └── Rule

├── 2-3 (d) Sketching a polynomial curve

│ ├── Plan

│ ├── Quadratic functions

│ ├── Sketching from given properties

│ └── Asymptotes (supplementary)

├── 2-4 (a) Mathematical modelling

│ ├── The modelling steps

│ └── Why a critical point can give the absolute extremum

├── 2-4 (b) Geometric optimisation

└── 2-4 (c) Cost
shortest length and other applications

└── Checking the answer

Convexity, Inflection Points, Curve Sketching and Optimisation · 1 / 14

2-3 (a) Convexity of a curve

Direction of convexity

A curve is convex downward on an interval when its tangent slope $f^{'}$ is increasing there, so $f^{''} (x) > 0$ . The curve lies above its tangents and has a cup shape $\cup$ (many books call this "concave up").
A curve is convex upward on an interval when $f^{'}$ is decreasing, so $f^{''} (x) < 0$ . The curve lies below its tangents and has a cap shape $\cap$ (called "concave down").
This lesson follows the book’s names convex downward means f′′>0 and convex upward means f′′<0.

Let $f$ be twice differentiable on $] a, b [$ .
If $f^{''} (x) > 0$ for all $x \in] a, b [$ , the curve is convex downward on it.
If $f^{''} (x) < 0$ for all $x \in] a, b [$ , the curve is convex upward on it.

✍️ Worked examples

Example 1 Find the intervals of convexity of $f (x) = 2 - 3 x^{2} - x^{3}$ .
Solution:
$f^{'} (x) = - 6 x - 3 x^{2}$ and $f^{''} (x) = - 6 - 6 x = - 6 (x + 1)$ .
$f^{''} (x) = 0$ at $x = - 1$ .
For x<−1 f′′>0, so the curve is convex downward on ]−∞,−1[.
For x>−1 f′′<0, so the curve is convex upward on ]−1,∞[.
Example 2 Find the intervals of convexity of $g (x) = x^{4} - 4 x^{3}$ .
Solution:
$g^{'} (x) = 4 x^{3} - 12 x^{2}$ and $g^{''} (x) = 12 x^{2} - 24 x = 12 x (x - 2)$ .
g′′>0 for x<0 and for x>2 convex downward on ]−∞,0[ and ]2,∞[.
g′′<0 for 0<x<2 convex upward on ]0,2[.

Convexity, Inflection Points, Curve Sketching and Optimisation · 2 / 14

2-3 (b) Inflection points

Definition

A point $(c, f (c))$ is an inflection point if $f$ is continuous at $c$ , the curve has a tangent there, and the direction of convexity changes at $c$ .
At an inflection point the tangent crosses the curve the curve lies under it on one side and above it on the other.

Solve $f^{''} (x) = 0$ and find where $f^{''} (x)$ does not exist.
Test the sign of $f^{''}$ on both sides of each value; it must change.
Make sure a tangent exists at the point.
A zero of f′′ is not enough f(x)=x4 has f′′(0)=0 but f′′≥0 on both sides, so there is no inflection.

✍️ Worked examples

Example 3 Find the inflection points of $g (x) = x^{4} - 4 x^{3}$ .
Solution:
From Example 2 the sign of $g^{''}$ changes at $x = 0$ (from $+$ to $-$ ) and at $x = 2$ (from $-$ to $+$ ).
$g (0) = 0$ and $g (2) = 16 - 32 = - 16$ .
The inflection points are $(0, 0)$ and $(2, - 16)$ .
Example 4 Study the convexity and inflection of

f(x)={x2−4x<−2x3−3x+2x≥−2

Solution:
At $x = - 2$ both pieces give $0$ , so $f$ is continuous.
Left derivative 2x=−4. Right derivative: 3x2−3=9. They differ, so f is not differentiable at x=−2 and there is no tangent there.
$f^{''} (x) = 2$ for $x < - 2$ and $f^{''} (x) = 6 x$ for $x > - 2$ , which is zero at $x = 0$ .
Convex downward on $] - \infty, - 2 [$ (where $f^{''} = 2$ ) and on $] 0, \infty [$ ; convex upward on $] - 2, 0 [$ .
$(- 2, 0)$ is not an inflection point although the convexity changes, because no tangent exists.
(0,2) is an inflection point the tangent exists, with slope f′(0)=−3, so its equation is y−2=−3x.
Example 5 The point $(1, 12)$ is an inflection point of $f (x) = a x^{3} + b x^{2}$ . Find $a$ and $b$ .
Solution:
The point lies on the curve a+b=12.
$f^{''} (x) = 6 a x + 2 b$ and $f^{''} (1) = 0$ , so $6 a + 2 b = 0$ and $b = - 3 a$ .
Then $a - 3 a = 12$ , so $a = - 6$ and $b = 18$ .
Example 6 The point $(2, 2)$ is an inflection point of $f (x) = x^{3} + a x^{2} + b x$ . Find $a$ and $b$ .
Solution:
$f^{''} (x) = 6 x + 2 a$ and $f^{''} (2) = 12 + 2 a = 0$ , so $a = - 6$ .
$f (2) = 8 + 4 a + 2 b = 8 - 24 + 2 b = 2$ , so $b = 9$ .

Convexity, Inflection Points, Curve Sketching and Optimisation · 3 / 14

2-3 (c) Second-derivative test for local extrema

Rule

Let $f$ be twice differentiable near $c$ and $f^{'} (c) = 0$ .
If $f^{''} (c) < 0$ , then $f (c)$ is a local maximum (the curve is a cap at $c$ ).
If $f^{''} (c) > 0$ , then $f (c)$ is a local minimum (the curve is a cup at $c$ ).
If $f^{''} (c) = 0$ the test fails; use the first-derivative test.

✍️ Worked examples

Example 7 Use the second-derivative test on $f (x) = x^{4} - 8 x^{2} + 10$ .
Solution:
$f^{'} (x) = 4 x^{3} - 16 x = 4 x (x - 2) (x + 2)$ , so the critical points are $x = 0$ and $x = \pm 2$ .
$f^{''} (x) = 12 x^{2} - 16$ .
f′′(0)=−16<0 local maximum f(0)=10.
f′′(±2)=32>0 local minimum f(±2)=16−32+10=−6 at both points.
Example 8 Find the extrema of $f (x) = x^{3} - 3 x^{2} - 9 x$ .
Solution:
$f^{'} (x) = 3 x^{2} - 6 x - 9 = 3 (x - 3) (x + 1)$ ; critical points $x = 3$ and $x = - 1$ .
$f^{''} (x) = 6 x - 6$ .
f′′(3)=12>0 local minimum f(3)=27−27−27=−27.
f′′(−1)=−12<0 local maximum f(−1)=−1−3+9=5.

Convexity, Inflection Points, Curve Sketching and Optimisation · 4 / 14

2-3 (d) Sketching a polynomial curve

Plan

Comparison aspect	Main idea	Keywords
Domain and symmetry	an even function is symmetric about the y-axis an odd function about the origin	evenfunctionsymmetricaxis
Find f′ and solve f′=0	critical points intervals of increase and decrease local extrema	criticalpointsintervalsincrease
Find f′′ and solve f′′=0	intervals of convexity and inflection points	intervalsconvexityinflectionpoints
Collect everything in one table	sign of f′ behaviour of f sign of f′′ convexity	signbehaviourconvexity
Find intercepts	y-intercept f(0) and x-intercepts when the equation f(x)=0 is solvable	equationsolvable
•	Add a few extra points if needed plot the key points and join them smoothly following the table	extrapointsneededplot

✍️ Worked examples

Sketch $f (x) = x^{3} - 3 x^{2} + 4$ .
Solution:
Domain $R$ ; the function is neither even nor odd.
$f^{'} (x) = 3 x^{2} - 6 x = 3 x (x - 2)$ gives critical points $x = 0$ and $x = 2$ .
$f^{''} (x) = 6 x - 6 = 6 (x - 1)$ gives $x = 1$ .
Table:

$x$	$- \infty$ to $0$	$0$	$0$ to $1$	$1$	$1$ to $2$	$2$	$2$ to $\infty$
$f^{'}$	$+$	$0$	$-$	$-$	$-$	$0$	$+$
$f^{''}$	$-$	$-$	$-$	$0$	$+$	$+$	$+$
Behaviour	rising, convex upward	local max $(0, 4)$	falling, convex upward	inflection $(1, 2)$	falling, convex downward	local min $(2, 0)$	rising, convex downward

Comparison aspect	Main idea	Keywords
Intercepts	y-intercept (0 4). Since f(x)=(x+1)(x−2)2 the x-intercepts are (−1 0) and (2 the curve touches the axis at (2	Sincecurvetouchesaxis
Extra point	f(3)=27−27+4=4
•	The curve comes up from the lower left through $(- 1, 0)$ , peaks at $(0, 4)$ , falls through the inflection $(1, 2)$ to touch the axis at $(2, 0)$ , then rises through $(3, 4)$ .	curvecomeslowerleft

✍️ Worked examples

Example 10 Sketch $f (x) = 12 x - x^{3}$ .
Solution:
$f$ is odd ( $f (- x) = - f (x)$ ), so the curve is symmetric about the origin.
$f^{'} (x) = 12 - 3 x^{2} = 3 (2 - x) (2 + x)$ , so the critical points are $x = \pm 2$ .
$f^{''} (x) = - 6 x$ , zero at $x = 0$ .
$f^{'} < 0$ for $x < - 2$ , $f^{'} > 0$ on $] - 2, 2 [$ , $f^{'} < 0$ for $x > 2$ .
Local minimum $f (- 2) = - 24 + 8 = - 16$ and local maximum $f (2) = 24 - 8 = 16$ .
$f^{''} > 0$ for $x < 0$ (convex downward) and $f^{''} < 0$ for $x > 0$ (convex upward); inflection at the origin $(0, 0)$ .
x-intercepts x(12−x2)=0, so x=0 and x=±23≈±3.46.
Sketch falls to the minimum (−2,−16), climbs through the origin to the maximum (2,16), then falls through (23,0).

Quadratic functions

For a parabola f(x)=ax2+bx+c f′′(x)=2a is constant, so there is one turning point at x=−b2a and no inflection.
It is a cup if $a > 0$ and a cap if $a < 0$ .

Sketching from given properties

✍️ Worked examples

Example 11 Sketch a continuous $f$ on $[1, 7]$ with $f (1) = - 2$ , $f (5) = 4$ , $f^{'} (5) = 0$ , $f^{'} > 0$ for $x < 5$ , $f^{'} < 0$ for $x > 5$ and $f^{''} < 0$ throughout.
Solution:
Plot $(1, - 2)$ and $(5, 4)$ ; the tangent is horizontal at $x = 5$ .
$f$ rises on $[1, 5]$ and falls on $[5, 7]$ , so $(5, 4)$ is a local (and absolute) maximum.
$f^{''} < 0$ everywhere means the whole curve is convex upward (a cap), with no inflection point.

Some curves, such as $f (x) = \frac{x^{2}}{x^{2} + 1}$ , level off. Since $\lim_{x \to \pm \infty} \frac{x^{2}}{x^{2} + 1} = 1$ , the line $y = 1$ is a horizontal asymptote that the two ends approach.
A graphing program is the best help for such functions, to confirm extrema, convexity and asymptotes.

Convexity, Inflection Points, Curve Sketching and Optimisation · 5 / 14

2-4 (a) Mathematical modelling

The modelling steps

Define the problem and its aim (largest area, least cost, shortest length, greatest profit).
Draw a figure and name the variables (the unknowns).
Write the quantity to optimise as a function (the model).
Use the constraints (given volume, perimeter, similar triangles) to reduce it to one variable, and state its domain.
Differentiate, find the critical points, and decide using the first- or second-derivative test (or the end points on a closed interval).
Interpret the result in the language of the problem and give all the required quantities with units.

If a continuous function has only one critical point in an interval and it is a local maximum (or minimum), then it is the absolute maximum (or minimum) on that interval.
If $f^{''}$ keeps the same sign on the whole interval, the unique critical point is the absolute extremum.

Convexity, Inflection Points, Curve Sketching and Optimisation · 6 / 14

2-4 (b) Geometric optimisation

Example 14 A closed cylinder must hold a fixed volume $V$ . Find the shape of least surface area.
Solution:
$V = π r^{2} h$ , so $h = \frac{V}{π r^{2}}$ .
Surface $S = 2 π r^{2} + 2 π r h = 2 π r^{2} + \frac{2 V}{r}$ .
$S^{'} = 4 π r - \frac{2 V}{r^{2}} = 0$ gives $r^{3} = \frac{V}{2 π}$ .
$S^{''} = 4 π + \frac{4 V}{r^{3}} > 0$ , so this is a minimum.
Since V=2πr3, h=2πr3πr2=2r the height equals the diameter.
Example 15 A sector has area 16 cm $^{2}$ . Find the radius that makes its perimeter least, and the angle then.
Solution:
Let the arc be $ℓ$ . Area $= \frac{1}{2} r ℓ = 16$ , so $ℓ = \frac{32}{r}$ .
Perimeter $P = 2 r + ℓ = 2 r + \frac{32}{r}$ .
$P^{'} = 2 - \frac{32}{r^{2}} = 0$ gives $r = 4$ ; $P^{''} = \frac{64}{r^{3}} > 0$ .
So $r = 4$ cm, $ℓ = 8$ cm, the angle is $θ = \frac{ℓ}{r} = 2$ rad, and the least perimeter is $16$ cm.
Example 16 A sector has perimeter 12 cm. Find its angle when its area is greatest.
Solution:
$2 r + ℓ = 12$ , so $ℓ = 12 - 2 r$ .
$A = \frac{1}{2} r ℓ = 6 r - r^{2}$ , so $A^{'} = 6 - 2 r = 0$ and $r = 3$ ; $A^{''} = - 2 < 0$ .
$ℓ = 6$ and $θ = \frac{6}{3} = 2$ rad; the greatest area is 9 cm $^{2}$ .

Convexity, Inflection Points, Curve Sketching: a cardboard sheet folded into an open box shows convexity, curve sketching and optimisation: cutting 2 cm squares gives the maximum volume 128 cm³.

✍️ Worked examples

Example 12:
A rectangle is inscribed in a triangle of base 16 cm and height 12 cm, with one side on the base and the two upper vertices on the other sides.
Find the dimensions of the rectangle of greatest area.
Solution:
Let the rectangle have height $x$ and length $y$ . Similar triangles give $\frac{y}{16} = \frac{12 - x}{12}$ , so $y = \frac{4}{3} (12 - x)$ .
Area $A = x y = \frac{4}{3} (12 x - x^{2})$ with $0 < x < 12$ .
$A^{'} = \frac{4}{3} (12 - 2 x) = 0$ gives $x = 6$ ; $A^{''} = - \frac{8}{3} < 0$ .
Then $y = \frac{4}{3} (6) = 8$ .
The best rectangle is 6 cm by 8 cm, with area 48 cm $^{2}$ .
Example 13:
A square sheet of side 24 cm has equal squares cut from its corners and the sides are folded up to make an open box.
Find the cut that gives the largest volume.
Solution:
Let the cut square have side $x$ ; the box has base side $24 - 2 x$ and height $x$ , so $V = x (24 - 2 x)^{2}$ , with $0 < x < 12$ .
$V^{'} = (24 - 2 x)^{2} - 4 x (24 - 2 x) = (24 - 2 x) (24 - 6 x)$ .
$V^{'} = 0$ gives $x = 4$ (the value $x = 12$ is rejected).
$V^{'} > 0$ for $x < 4$ and $V^{'} < 0$ for $x > 4$ , so $x = 4$ gives the maximum.
$V = 4 \times 1 6^{2} = 1024$ cm $^{3}$ .

Convexity, Inflection Points, Curve Sketching and Optimisation · 7 / 14

2-4 (c) Cost, shortest length and other applications

✍️ Worked examples

Example 17:
A silo is a vertical cylinder topped by a hemisphere.
Its cylindrical part holds $108 π$ m $^{3}$ of grain.
A unit area of the top costs twice a unit area of the wall.
Find the radius and height that minimise the cost.
Solution:
Let the wall cost $c$ per m $^{2}$ and the roof cost $2 c$ .
Volume of the cylinder πr2h=108π, so h=108r2.
Wall area $2 π r h$ ; hemisphere area $2 π r^{2}$ .
Cost $K = c (2 π r h) + 2 c (2 π r^{2}) = 2 π c r (h + 2 r) = 2 π c (\frac{108}{r} + 2 r^{2})$ .
$K^{'} = 2 π c (- \frac{108}{r^{2}} + 4 r) = 0$ gives $r^{3} = 27$ , so $r = 3$ .
$K^{''} = 2 π c (\frac{216}{r^{3}} + 4) > 0$ , so this is a minimum.
Then h=1089=12 radius 3 m, cylinder height 12 m.
Example 18:
A 2 m high fence stands 2 m from a wall.
Find the shortest ladder that rests on the ground, touches the top of the fence and leans on the wall.
Solution:
Let the foot of the ladder be $x$ m from the fence and the top be $y$ m up the wall. Similar triangles: $\frac{y}{x + 2} = \frac{2}{x}$ , so $y = \frac{2 (x + 2)}{x}$ .
Length squared L2=(x+2)2+y2=(x+2)2(1+4x2).
Differentiate L2 d(L2)dx=2(x+2)(1+4x2)+(x+2)2(−8x3)=2(x+2)(x3−8)x3.
It is zero at $x = 2$ (since $x + 2 > 0$ ) and changes from negative to positive there, a minimum.
$L^{2} = 16 (1 + 1) = 32$ , so $L = 4 \sqrt{2} \approx 5.66$ m.

Example 19 A line through $C (3, 2)$ meets the axes at $A$ and $B$ . Show that the least area of triangle $A O B$ is 12 square units.
Solution:
Let $A = (a, 0)$ and $B = (0, b)$ . The line passes through $C$ : $\frac{3}{a} + \frac{2}{b} = 1$ , so $b = \frac{2 a}{a - 3}$ with $a > 3$ .
Area $S = \frac{1}{2} a b = \frac{a^{2}}{a - 3}$ .
$S^{'} = \frac{2 a (a - 3) - a^{2}}{(a - 3)^{2}} = \frac{a (a - 6)}{(a - 3)^{2}}$ .
$S^{'} = 0$ at $a = 6$ , changing from negative to positive, so $S$ is least there.
Then $b = 4$ and $S = \frac{36}{3} = 12$ square units.
Example 20 The sum of two positive numbers is 20. Find them if the sum of their squares is least.
Solution:
Let the numbers be $x$ and $20 - x$ ; $S = x^{2} + (20 - x)^{2}$ .
$S^{'} = 2 x - 2 (20 - x) = 4 x - 40 = 0$ gives $x = 10$ ; $S^{''} = 4 > 0$ .
Both numbers are 10 and the least sum of squares is 200.

Convexity, Inflection Points, Curve Sketching and Optimisation · 8 / 14

Checking the answer

Test the end points or the second derivative to be sure you found the required type of extremum.
Check that the value lies in the physical domain (lengths positive, angles in range).
Answer every part of the question the optimum value and the dimensions that produce it.

Convexity, Inflection Points, Curve Sketching and Optimisation · 9 / 14

📌 Comprehensive Scientific Terminology & Definitions:

Comparison aspect	Main idea	Keywords
Concavity / Convexity (التحدب لأعلى ولأسفل)	A curve is concave upward (convex downward) on an interval if all its tangents lie below the curve (f”(x) > 0) concave downward (convex upward) if all its tangents lie above the curve (f”(x) < 0)	curveconcaveupwarddownward
Inflection Point (نقطة الانقلاب)	A point (c f(c)) on a continuous curve where the concavity changes from upward to downward or vice versa requiring f”(c) = 0 or f”(c) is undefined AND f”(x) must change algebraic sign across c	continuouscurveconcavitychanges
Vertical and Horizontal Asymptotes (خطوط التقارب)	Horizontal asymptote y = L if lim(x→±∞) f(x) = L Vertical asymptote x = a if lim(x→a) f(x) = ±∞

Comparison aspect	Main idea	Keywords
Applied Optimization (تطبيقات القيم العظمى والصغرى)	Mathematical modeling to maximize profit area or efficiency or minimize cost material or travel time subject to geometric/physical constraints	Mathematicalmodelingmaximizeprofit
Steps for Applied Optimization	1. Define variables and draw a diagram 2. Formulate the primary objective function Q to optimize 3. Formulate the secondary constraint equation to eliminate extra variables 4. Find dQ/dx = 0 5. Verify maximum/minimum using second derivative	Definevariablesdrawdiagram

Convexity, Inflection Points, Curve Sketching and Optimisation · 10 / 14

⚖️ Governing Scientific Laws, Principles & Mathematical Formulas:

Comparison aspect	Main idea	Keywords
(“Inflection Condition	f”(c) = 0 (or undefined) AND f”(c – ε) · f”(c + ε) < 0 (must strictly change sign)."	undefinedmuststrictlychange
(“Cost-Distance Optimization (Snell’s Law Analogy)	Total Time T = (d₁ / v₁) + (d₂ / v₂) → dT/dx = 0 leads to sin θ₁ / v₁ = sin θ₂ / v₂.”	TotalTimeleads
(‘Box Volume Maximization	For a square sheet of side L with cut corner squares of side x: V(x) = x(L – 2x)² → dV/dx = (L – 2x)(L – 6x) = 0 → x = L/6 yields maximum volume Vmax = 2L³/27.’	squaresheetsidecorner

Convexity, Inflection Points, Curve Sketching and Optimisation · 11 / 14

🔍 In-Depth Explanations, Deductions & Core Concepts:

Why f”(c) = 0 does not automatically guarantee an inflection point: For f(x) = x⁴, f”(x) = 12x² so f”(0) = 0. However, 12x² ≥ 0 for all x ≠ 0 (no sign change, remaining concave up everywhere), so (0, 0) is a local minimum, not an inflection point.
Optimum dimensions of a cylinder of fixed volume V to minimize total surface area: Area A = 2π r² + 2π r h. Since V = π r² h → h = V / (π r²), substituting gives A(r) = 2π r² + 2V/r. Setting dA/dr = 4π r – 2V/r² = 0 yields r³ = V / (2π), leading to h = 2r (height equals diameter).
Systematic 7-step guide to curve sketching 1. Domain; 2. Intercepts (x and y); 3. Symmetry (even/odd/periodic); 4. Asymptotes; 5. Intervals of increase/decrease (f’); 6. Local extrema; 7. Concavity and inflection points (f”).

Convexity, Inflection Points, Curve Sketching and Optimisation · 12 / 14

📊 Classic Applied Optimization Problems and Their Golden Ratios:

Optimization Objective	Constraint Condition	Optimal Geometric Ratio / Dimension	Proven Mathematical Result
Maximum Rectangular Area with Fixed Perimeter P	2x + 2y = P	Square: x = y = P / 4	Area_max = P² / 16
Minimum Surface Area of Closed Cylinder with Fixed Volume V	V = π r² h	Height equals diameter: h = 2r	A_min = 6π r²
Maximum Open Box Volume from Square Sheet of Side L	V = x(L – 2x)²	Cut square corners: x = L / 6	V_max = 2L³ / 27
Maximum Inscribed Cylinder in Cone of Radius R and Height H	Similar Triangles r/R + h/H = 1	r = (2/3)R, h = (1/3)H	V_max = (4/27) V_cone

Convexity, Inflection Points, Curve Sketching and Optimisation · 13 / 14

💡 High-Yield Exam Traps & Key Takeaways:

Comparison aspect	Main idea	Keywords
Exam Secret	In rectangular fencing problems against an existing wall (3-sided fence P = 2x + y) the maximum area occurs when the side parallel to the wall is twice the perpendicular side (y = 2x)	rectangularfencingproblemsagainst
Inflection Tangent Tip	The tangent line drawn at an inflection point crosses/cuts through the curve (inflectional tangent)	linedrawnpointcrosses

Convexity, Inflection Points, Curve Sketching and Optimisation · 14 / 14

Curriculum

Mathematics

Unit 1

Unit 2

Unit 3

Unit 4

Unit 5

Unit 6

Unit 7

Convexity, Inflection Points, Curve Sketching and Optimisation

🌳 Concept tree

├── 2-3 (a) Convexity of a curve

│ ├── Direction of convexity

│ └── The second-derivative test for convexity

├── 2-3 (b) Inflection points

│ ├── Definition

│ └── Finding inflection points

│ └── Rule

├── 2-3 (d) Sketching a polynomial curve

│ ├── Plan

│ ├── Quadratic functions

│ ├── Sketching from given properties

│ └── Asymptotes (supplementary)

├── 2-4 (a) Mathematical modelling

│ ├── The modelling steps

├── 2-4 (b) Geometric optimisation

└── Checking the answer

Direction of convexity

✍️ Worked examples

Definition

✍️ Worked examples

f(x)={x2−4x<−2x3−3x+2x≥−2

Rule

✍️ Worked examples

Plan

✍️ Worked examples

✍️ Worked examples

Quadratic functions

Sketching from given properties

✍️ Worked examples

The modelling steps

✍️ Worked examples

✍️ Worked examples

Well done! You finished this lesson