Applications of Differentiation | FractionRush A-Level

Welcome to Applications of Differentiation!

Now that you can differentiate, it is time to put derivatives to work. This module covers the most powerful and frequently examined applications of differentiation in Cambridge A-Level Pure Mathematics 1: finding equations of tangents and normals, identifying where functions increase or decrease, locating and classifying stationary points, solving optimisation problems, and using the chain rule to connect rates of change.

Tangent: y − y₁ = f'(a)(x − a) | Normal: y − y₁ = −1/f'(a) · (x − a) | dy/dt = (dy/dx)·(dx/dt)

Learning Objectives

Find the gradient of a curve at a given point using f'(a)
Write the equation of a tangent and a normal to a curve at a given point
Determine intervals where a function is increasing or decreasing
Use the second derivative to identify concavity
Find stationary points by solving f'(x) = 0
Classify stationary points using the second derivative test or sign change of f'
Identify points of inflection
Set up and solve optimisation problems in context
Check endpoints on closed intervals for global maxima/minima
Apply the chain rule to connect two related rates of change

Tangents & Normals

Gradient = f'(a); perpendicular gradient = −1/f'(a)

Increasing/Decreasing

f'(x) > 0 increasing; f'(x) < 0 decreasing

Stationary Points

Solve f'(x) = 0; use f'' or sign change to classify

Optimisation

Model, differentiate, solve — check it's really max or min

Connected Rates

dy/dt = (dy/dx) × (dx/dt) via chain rule

Second Derivative

f'' > 0: concave up (min); f'' < 0: concave down (max)

Learn 1 — Tangents & Normals

Gradient of a Tangent

The tangent to a curve y = f(x) at the point where x = a has gradient equal to the derivative f'(a). This is the instantaneous rate of change at that exact point.

Gradient of tangent at x = a = f'(a)

Equation of the Tangent

Once you have the gradient m = f'(a) and the point (a, f(a)) on the curve, use the point-slope form:

y − y₁ = m(x − x₁) where m = f'(a) and (x₁, y₁) = (a, f(a))

Step-by-Step: Tangent at a Given Point

Find the equation of the tangent to y = x³ − 2x + 1 at x = 2.

Step 1 — Find y-coordinate: y = (2)³ − 2(2) + 1 = 8 − 4 + 1 = 5. Point is (2, 5).
Step 2 — Differentiate: dy/dx = 3x² − 2
Step 3 — Find gradient: m = 3(2)² − 2 = 12 − 2 = 10
Step 4 — Write equation: y − 5 = 10(x − 2) → y = 10x − 15

Gradient of the Normal

The normal to a curve at a point is perpendicular to the tangent at that point. Perpendicular lines have gradients whose product is −1:

Gradient of normal = −1 / (gradient of tangent) = −1 / f'(a)

If f'(a) = 0 (tangent is horizontal), the normal is vertical (undefined gradient, write x = a). If f'(a) is undefined (vertical tangent), the normal is horizontal.

Equation of the Normal

y − y₁ = (−1/m)(x − x₁) where m = f'(a)

Step-by-Step: Normal at a Given Point

Find the equation of the normal to y = x² − 3x + 4 at x = 1.

Step 1 — Find y-coordinate: y = 1 − 3 + 4 = 2. Point is (1, 2).
Step 2 — Differentiate: dy/dx = 2x − 3
Step 3 — Tangent gradient: m = 2(1) − 3 = −1
Step 4 — Normal gradient: m_n = −1/(−1) = 1
Step 5 — Write equation: y − 2 = 1(x − 1) → y = x + 1

Special Cases

Tangent parallel to x-axis: f'(a) = 0. The tangent is y = f(a) (a horizontal line).
Normal parallel to x-axis: This occurs when f'(a) is undefined (vertical tangent), giving normal y = f(a).
Finding point given gradient: If the tangent has gradient k, solve f'(x) = k to find x, then find y.
Two tangents from a point: Set up y − b = f'(x₀)(x − a) using the general point (x₀, f(x₀)), then solve.

Full Worked Example: Both Lines

Curve y = √x = x^(1/2). Find tangent and normal at (4, 2).

dy/dx = (1/2)x^(−1/2) = 1/(2√x)
At x = 4: m = 1/(2·2) = 1/4

Tangent: y − 2 = (1/4)(x − 4) → y = x/4 + 1
Normal: y − 2 = −4(x − 4) → y = −4x + 18

Always find the y-coordinate before differentiating — it is easy to forget. Write down the point (a, f(a)) explicitly in your working. In Cambridge mark schemes, the point coordinates earn a method mark even if the gradient is wrong.

Learn 2 — Increasing & Decreasing Functions

Definitions

A function f(x) is increasing on an interval if its graph rises from left to right across that interval — as x increases, y increases. It is decreasing if the graph falls.

f'(x) > 0 for all x in (a, b) ⟹ f increasing on (a, b)
f'(x) < 0 for all x in (a, b) ⟹ f decreasing on (a, b)

Finding Intervals of Increase and Decrease

Method:
1. Differentiate to find f'(x).
2. Solve f'(x) = 0 to find the critical x-values (boundaries between intervals).
3. Test a value in each interval: is f'(x) positive or negative?
4. State the intervals using inequality notation or interval notation.

Example: f(x) = x³ − 3x² − 9x + 5

f'(x) = 3x² − 6x − 9 = 3(x² − 2x − 3) = 3(x − 3)(x + 1)

f'(x) = 0 when x = 3 or x = −1.

Test x = −2 (in x < −1): f'(−2) = 3(9 + 2 − 3) — or just: 3(−2−3)(−2+1) = 3(−5)(−1) = 15 > 0 ✓ increasing
Test x = 0 (in −1 < x < 3): f'(0) = 3(0−3)(0+1) = −9 < 0 ✓ decreasing
Test x = 4 (in x > 3): f'(4) = 3(1)(5) = 15 > 0 ✓ increasing

Increasing: x < −1 and x > 3 | Decreasing: −1 < x < 3

Example: f(x) = 2x³ + 3x² − 12x

f'(x) = 6x² + 6x − 12 = 6(x² + x − 2) = 6(x + 2)(x − 1)
Zeros: x = −2, x = 1
x < −2: f'(−3) = 6(−1)(−4) > 0 → increasing
−2 < x < 1: f'(0) = 6(2)(−1) = −12 < 0 → decreasing
x > 1: f'(2) = 6(4)(1) > 0 → increasing

Concavity and the Second Derivative

The second derivative f''(x) tells you how the gradient is changing — whether the curve is concave up (bowl-shaped) or concave down (arch-shaped):

f''(x) > 0 ⟹ concave up (gradient increasing) f''(x) < 0 ⟹ concave down (gradient decreasing)

For f(x) = x³ − 3x:
f'(x) = 3x² − 3 f''(x) = 6x
f''(x) > 0 when x > 0 → concave up for x > 0
f''(x) < 0 when x < 0 → concave down for x < 0
f''(0) = 0 → point of inflection at x = 0

Strict vs Non-Strict Inequalities

Cambridge 9709 accepts both "increasing for x > 3" and "x ≥ 3". At isolated points where f'(x) = 0 but the function does not change direction, you may include or exclude the boundary. However, at local maxima or minima, the function genuinely stops increasing (or decreasing), so use strict inequality there for accuracy.

Learn 3 — Stationary Points

What Is a Stationary Point?

A stationary point is where the gradient of the curve is zero — where the tangent is horizontal. There are three types: local maximum, local minimum, and point of inflection.

Stationary points occur where f'(x) = 0

Finding Stationary Points: The Procedure

1. Differentiate: find f'(x).
2. Solve f'(x) = 0 to find x-coordinates.
3. Substitute each x back into f(x) to find y-coordinates.
4. Classify each point (see below).

The Second Derivative Test

f''(a) > 0 ⟹ local minimum f''(a) < 0 ⟹ local maximum f''(a) = 0 ⟹ inconclusive

Intuition: At a minimum, the gradient is increasing (negative before, zero at, positive after), so f'' > 0 — concave up. At a maximum, the gradient is decreasing, so f'' < 0 — concave down.

Example: y = x³ − 3x

dy/dx = 3x² − 3 = 3(x²−1) = 3(x−1)(x+1)
Stationary points at x = 1 and x = −1.

At x = 1: y = 1 − 3 = −2. Point (1, −2).
At x = −1: y = −1 + 3 = 2. Point (−1, 2).

d²y/dx² = 6x
At x = 1: d²y/dx² = 6 > 0 → local minimum at (1, −2)
At x = −1: d²y/dx² = −6 < 0 → local maximum at (−1, 2)

The Sign Change Method (Nature Table)

When the second derivative test is inconclusive (f''(a) = 0), examine the sign of f'(x) on either side of the stationary point:

Local maximum: f' changes + → 0 → − (gradient decreases through zero)
Local minimum: f' changes − → 0 → + (gradient increases through zero)
Point of inflection: f' has same sign on both sides, + → 0 → + or − → 0 → −

Example Using Sign Change: y = x⁴

dy/dx = 4x³ → stationary at x = 0.
d²y/dx² = 12x² → at x = 0: d²y/dx² = 0. Inconclusive!

Sign table:
x = −0.1: f'(−0.1) = 4(−0.001) < 0
x = 0: f'(0) = 0
x = 0.1: f'(0.1) = 4(0.001) > 0
Sign changes − → 0 → + ⟹ local minimum at (0, 0)

Points of Inflection

A point of inflection is where the concavity changes. It requires f''(x) = 0 AND a sign change in f''(x). Not all stationary points of inflection have f'(x) = 0 — but stationary points of inflection (also called saddle points) do.

Example: y = x³. f''(x) = 6x. f''(0) = 0, and f'' changes sign at x = 0 → inflection at (0, 0).

Never stop at f''(a) = 0 and write "inconclusive" without following up with a sign table. Cambridge examiners expect you to identify the nature fully. A missing classification costs marks even if the coordinates are correct.

Learn 4 — Optimisation

What Is an Optimisation Problem?

Optimisation means finding the maximum or minimum value of a real-world quantity. This could be the largest volume a box can have, the least material needed to make a container, or the maximum profit from selling goods.

The Five-Step Method

Step 1 — Define the variable: Identify the single variable you will differentiate with respect to (e.g., x = length of one side).
Step 2 — Write an expression: Express the quantity to maximise or minimise (e.g., Volume, Area, Cost) as a function of your variable. Use any given constraints to eliminate extra unknowns.
Step 3 — Differentiate: Find dy/dx (or dV/dx, dA/dx, etc.).
Step 4 — Solve = 0: Set the derivative equal to zero and solve. Reject any solutions outside the feasible domain (e.g., negative lengths).
Step 5 — Classify and conclude: Use the second derivative (or sign change) to confirm whether you have found a maximum or minimum. State the maximum/minimum value with appropriate units.

Example 1: Open Box from a Sheet

A square sheet of cardboard has side 12 cm. Equal squares of side x cm are cut from each corner, and the flaps are folded up to make an open box. Find the value of x that maximises the volume.

Step 1: Variable is x (corner cut size), 0 < x < 6.
Step 2: Base side = 12 − 2x. Height = x.
V = x(12 − 2x)² = x(144 − 48x + 4x²) = 144x − 48x² + 4x³
Step 3: dV/dx = 144 − 96x + 12x²
Step 4: 12x² − 96x + 144 = 0 → x² − 8x + 12 = 0 → (x−2)(x−6) = 0
x = 2 or x = 6. Since 0 < x < 6, reject x = 6. So x = 2.
Step 5: d²V/dx² = 24x − 96. At x = 2: 48 − 96 = −48 < 0 → maximum.
V_max = 2(8)² = 128 cm³

Example 2: Fencing Problem

A farmer has 120 m of fencing to enclose a rectangular pen against a straight wall (wall forms one side). Maximise the enclosed area.

Let x = width (two widths used), y = length along wall (one length used).
Constraint: 2x + y = 120 → y = 120 − 2x
A = xy = x(120 − 2x) = 120x − 2x²
dA/dx = 120 − 4x = 0 → x = 30
d²A/dx² = −4 < 0 → maximum.
y = 120 − 60 = 60. A_max = 30 × 60 = 1800 m²

Example 3: Cylinder Surface Area

A closed cylinder has volume 250π cm³. Find the radius that minimises the total surface area.

V = πr²h = 250π → h = 250/r²
S = 2πr² + 2πrh = 2πr² + 2πr(250/r²) = 2πr² + 500π/r
dS/dr = 4πr − 500π/r²
Set = 0: 4πr = 500π/r² → r³ = 125 → r = 5
d²S/dr² = 4π + 1000π/r³ > 0 → minimum.
h = 250/25 = 10. Note: h = 2r at minimum (cylinder is as tall as it is wide).

Checking Endpoints on Closed Intervals

On a closed interval [a, b], a global maximum or minimum may occur at the boundary (x = a or x = b), not at a stationary point. Always evaluate f at any stationary points inside the interval AND at both endpoints, then compare values.

The most common error in optimisation is forgetting to verify that a stationary point is actually a maximum (or minimum) — especially in context where the question says "show that..." or "prove that...". Always use the second derivative or end-value comparison to justify your conclusion.

Learn 5 — Connected Rates of Change

The Idea

Two quantities that both depend on time may be related to each other. If you know how fast one is changing, you can find how fast the other is changing using the chain rule.

dy/dt = (dy/dx) · (dx/dt)

This says: the rate of change of y with time equals the rate of change of y with x, multiplied by the rate of change of x with time.

Setting Up the Relationship

Step 1: Identify the two changing quantities (say A and r).
Step 2: Write the geometric or physical relationship between them (e.g., A = πr²).
Step 3: Differentiate that relationship with respect to r (or whatever the linking variable is) to get dA/dr.
Step 4: Apply the chain rule: dA/dt = (dA/dr) · (dr/dt).
Step 5: Substitute the known values of r and dr/dt.

Example 1: Expanding Circle

The radius of a circular oil slick is increasing at 0.5 m/s. Find the rate at which the area is increasing when the radius is 4 m.

A = πr² → dA/dr = 2πr
Given: dr/dt = 0.5
dA/dt = (dA/dr)(dr/dt) = 2πr × 0.5 = πr
When r = 4: dA/dt = 4π ≈ 12.57 m²/s

Example 2: Filling a Spherical Tank

Water fills a sphere of radius 10 cm. When the volume is increasing at 5 cm³/s, find the rate at which the radius is increasing when r = 3 cm.

V = (4/3)πr³ → dV/dr = 4πr²
Given: dV/dt = 5
dV/dt = (dV/dr)(dr/dt) → 5 = 4π(9)(dr/dt)
dr/dt = 5/(36π) ≈ 0.0442 cm/s

Example 3: Shadow Length

A 2 m tall person walks away from a 6 m lamppost at 1.5 m/s. Find the rate at which the shadow length is increasing.

Let x = distance from post to person, s = shadow length.
By similar triangles: 6/(x + s) = 2/s → 6s = 2(x + s) → 4s = 2x → s = x/2
ds/dt = (1/2)(dx/dt) = (1/2)(1.5) = 0.75 m/s

Example 4: Conical Tank

Water drains from a conical tank (radius 3 m, height 9 m) at 2 m³/min. Find the rate at which the depth decreases when depth h = 6 m.

By similar triangles: r/h = 3/9 → r = h/3
V = (1/3)πr²h = (1/3)π(h/3)²h = πh³/27
dV/dh = 3πh²/27 = πh²/9
dV/dt = (dV/dh)(dh/dt) → −2 = (π·36/9)(dh/dt)
dh/dt = −2/(4π) = −1/(2π) ≈ −0.159 m/min

Direction of the Chain Rule

Always write the chain rule in the correct order: if you want dy/dt, you need dy/d(link variable) times d(link variable)/dt. Drawing a small diagram of which quantities connect through which variable prevents direction errors. Also: be careful about signs — a decreasing quantity has a negative rate of change.

Worked Examples

Eight fully worked examples covering every application topic.

Example 1 — Tangent at a Point

Find the equation of the tangent to y = x³ − 4x + 1 at the point where x = −1.

M1 Find y: y = (−1)³ − 4(−1) + 1 = −1 + 4 + 1 = 4. Point (−1, 4).

M1 Differentiate: dy/dx = 3x² − 4.

M1 Gradient at x = −1: m = 3(1) − 4 = −1.

A1 Tangent: y − 4 = −1(x + 1) → y = −x + 3

Example 2 — Normal Equation

Find the equation of the normal to y = (2x − 1)³ at the point (1, 1).

M1 Verify: y = (2·1−1)³ = 1³ = 1 ✓

M1 dy/dx = 3(2x−1)² · 2 = 6(2x−1)² (chain rule)

M1 At x = 1: m_tangent = 6(1)² = 6. Normal gradient = −1/6.

A1 Normal: y − 1 = −(1/6)(x − 1) → y = −x/6 + 7/6 or 6y + x = 7

Example 3 — Intervals of Increase/Decrease

Find the ranges of x for which f(x) = 2x³ − 9x² + 12x − 3 is increasing.

M1 f'(x) = 6x² − 18x + 12 = 6(x² − 3x + 2) = 6(x−1)(x−2)

M1 f'(x) = 0 at x = 1 and x = 2. Test intervals: x < 1, 1 < x < 2, x > 2.

M1 f'(0) = 6(−1)(−2) = 12 > 0 ✓ · f'(1.5) = 6(0.5)(−0.5) = −1.5 < 0 · f'(3) = 6(2)(1) = 12 > 0 ✓

A1 f increasing for x < 1 and x > 2

Example 4 — Classify Stationary Points

Find and classify all stationary points of y = x⁴ − 8x² + 3.

M1 dy/dx = 4x³ − 16x = 4x(x² − 4) = 4x(x−2)(x+2) = 0 → x = 0, 2, −2

M1 y-values: y(0) = 3, y(2) = 16 − 32 + 3 = −13, y(−2) = −13

M1 d²y/dx² = 12x² − 16. At x=0: −16 < 0 → max. At x=±2: 48−16 = 32 > 0 → min.

A1 Local max (0, 3); local minima (2, −13) and (−2, −13)

Example 5 — Optimise Open Box

A rectangular piece of card 10 cm × 8 cm has equal squares of side x cut from each corner and is folded into an open box. Show that V = 4x³ − 36x² + 80x and find the maximum volume.

M1 Dimensions: length = 10 − 2x, width = 8 − 2x, height = x (domain: 0 < x < 4)

M1 V = x(10−2x)(8−2x) = x(80 − 20x − 16x + 4x²) = x(80 − 36x + 4x²) = 4x³ − 36x² + 80x ✓

M1 dV/dx = 12x² − 72x + 80 = 0 → 3x² − 18x + 20 = 0 → x = (18 ± √(324−240))/6 = (18 ± √84)/6

A1 x ≈ 1.473 (the other root x ≈ 4.527 is outside domain). V_max ≈ 52.5 cm³

Example 6 — Optimise Perimeter/Area

A window consists of a rectangle surmounted by a semicircle. The perimeter is 12 m. Show that the area is A = 6r − r² − πr²/2 (where r is the radius of the semicircle) and find r for maximum area.

M1 Let r = radius, h = height of rectangle. Perimeter = 2h + 2r + πr = 12 → h = (12 − 2r − πr)/2 = 6 − r − πr/2

M1 A = 2rh + πr²/2 = 2r(6 − r − πr/2) + πr²/2 = 12r − 2r² − πr² + πr²/2 = 12r − 2r² − πr²/2

M1 Hmm — note: A = 6r − r² − πr²/2 requires dividing by 2. Let's recheck using half-perimeter = 6: then h = 6 − r(1 + π/2) and A = 2rh + πr²/2. dA/dr = 6 − 2r − πr/2 = 0

A1 r = 6/(2 + π/2) = 12/(4 + π) ≈ 1.68 m. Verify d²A/dr² = −2 − π/2 < 0 → maximum.

Example 7 — Connected Rate: Expanding Circle

The area of a circle increases at 10 cm²/s. Find the rate of increase of the circumference when the radius is 5 cm.

M1 C = 2πr → dC/dr = 2π. A = πr² → dA/dr = 2πr.

M1 dA/dt = (dA/dr)(dr/dt) → 10 = 2πr · (dr/dt) → dr/dt = 10/(2πr) = 5/(πr)

M1 dC/dt = (dC/dr)(dr/dt) = 2π · 5/(πr) = 10/r

A1 When r = 5: dC/dt = 2 cm/s

Example 8 — Connected Rate: Conical Tank

Water fills a cone (vertex down) of semi-vertical angle 30°. Water enters at 4 cm³/s. Find dh/dt when h = 6 cm.

M1 tan 30° = r/h → r = h tan 30° = h/√3. V = (1/3)πr²h = (1/3)π(h²/3)h = πh³/9

M1 dV/dh = 3πh²/9 = πh²/3

M1 dV/dt = (dV/dh)(dh/dt) → 4 = (πh²/3)(dh/dt)

A1 When h = 6: 4 = (36π/3)(dh/dt) = 12π(dh/dt) → dh/dt = 1/(3π) ≈ 0.106 cm/s

Common Mistakes

These are the errors Cambridge examiners see most often. Review each one carefully.

Mistake 1 — Inverting the Normal Gradient

Wrong: Tangent gradient is 3, so normal gradient is also 3 (or just −3 without reciprocal).

Correct: Normal gradient = −1/3. You must take the NEGATIVE RECIPROCAL. If tangent gradient is m, normal is −1/m.

Mistake 2 — Not Finding the y-Coordinate

Wrong: "Gradient at x = 2 is 5, so tangent is y = 5x + c." (Forgetting to find the actual point on the curve.)

Correct: First substitute x = 2 into y = f(x) to get the y-coordinate. Then use y − y₁ = m(x − x₁) with the actual point.

Mistake 3 — Wrong Sign in Second Derivative Classification

Wrong: "f''(a) > 0 means local maximum because the curve is 'going up'."

Correct: f''(a) > 0 means MINIMUM (concave up, bowl shape). f''(a) < 0 means MAXIMUM (concave down, arch shape). Think of the smiley face (min) vs sad face (max).

Mistake 4 — Stopping at f''(a) = 0

Wrong: "f''(a) = 0, so the nature is inconclusive" — and leaving it there without further analysis.

Correct: When f''(a) = 0, use a sign change table for f'(x) near x = a to determine whether it is a max, min, or point of inflection.

Mistake 5 — Not Checking Feasibility in Optimisation

Wrong: Accepting x = −2 as the optimal corner cut for a box, even though negative lengths are impossible.

Correct: Always state the domain constraint (e.g., 0 < x < 6) before solving, then reject solutions outside the domain.

Mistake 6 — Not Verifying Max vs Min in Context

Wrong: "Setting dA/dx = 0 gives x = 5, which is the maximum area." (No verification.)

Correct: Always compute d²A/dx² or use a sign change. State clearly: "Since d²A/dx² < 0, this is a maximum." Cambridge requires this justification.

Mistake 7 — Chain Rule Direction in Connected Rates

Wrong: dA/dt = (dr/dt) ÷ (dA/dr) — dividing instead of multiplying, or flipping the chain rule.

Correct: dA/dt = (dA/dr) × (dr/dt). The "dr" cancels on paper. Always write the chain rule explicitly before substituting numbers.

Mistake 8 — Ignoring Endpoints on Closed Intervals

Wrong: "The only stationary point is at x = 3, so the maximum on [0, 5] is f(3)." (Not checking boundaries.)

Correct: Evaluate f at every stationary point AND at both endpoints. The global maximum is the largest of all these values. On [0, 5], check f(0), f(3), and f(5).

Mistake 9 — Using Incorrect Similar-Triangle Ratios

Wrong: In a conical tank problem, writing r/h = h/r or mixing up the fixed cone dimensions with the variable water depth.

Correct: Use the fixed cone geometry to relate r and h: if cone has radius R at height H, then r/h = R/H. Substitute this to write V as a function of h only, then differentiate.

Key Formulas

Tangent and Normal Equations

Quantity	Formula
Gradient of curve at x = a	m = f'(a)
Equation of tangent at (a, b)	y − b = m(x − a)
Gradient of normal	m_n = −1/m = −1/f'(a)
Equation of normal at (a, b)	y − b = (−1/m)(x − a)
Tangent parallel to x-axis	f'(a) = 0 → tangent is y = b
Normal is vertical	When f'(a) = 0, normal is x = a

Increasing and Decreasing

f'(x) > 0 → increasing | f'(x) < 0 → decreasing | f''(x) > 0 → concave up | f''(x) < 0 → concave down

Stationary Points and Classification

Condition	Type
f'(a) = 0 and f''(a) > 0	Local minimum
f'(a) = 0 and f''(a) < 0	Local maximum
f'(a) = 0 and f''(a) = 0	Inconclusive — use sign change of f'
f' changes + → 0 → −	Local maximum
f' changes − → 0 → +	Local minimum
f' same sign both sides of a	Point of inflection

Optimisation Checklist

Connected Rates of Change

Relation	Chain Rule Form
Area of circle A = πr²	dA/dt = 2πr · (dr/dt)
Volume of sphere V = (4/3)πr³	dV/dt = 4πr² · (dr/dt)
Volume of cone V = (1/3)πr²h (fixed r/h ratio)	Express V(h) first, then dV/dt = dV/dh · (dh/dt)
General two-variable	dy/dt = (dy/dx)(dx/dt)
Inverse rate	dt/dy = 1 / (dy/dt)

Key Derivatives for Applications

d/dx(xⁿ) = nxⁿ⁻¹ | d/dx(sin x) = cos x | d/dx(cos x) = −sin x | d/dx(eˣ) = eˣ | d/dx(ln x) = 1/x

Proof Bank

Rigorous justifications of the key results used in applications of differentiation.

Proof 1 — Normal Gradient Is the Negative Reciprocal

Claim: If a tangent has gradient m, the perpendicular (normal) has gradient −1/m.

Proof:

Two lines L₁ and L₂ are perpendicular if and only if the product of their gradients equals −1.

Geometrically: if L₁ makes angle θ with the x-axis, then L₂ makes angle θ + 90° (or θ − 90°). Using tan(θ + 90°) = −1/tan θ:

m₂ = tan(θ + 90°) = −cot θ = −1/tan θ = −1/m₁.

Algebraically: if L₁ has gradient m and L₂ has gradient k, rotate the vector (1, m) by 90° to get (−m, 1). The gradient of L₂ is then 1/(−m) ... wait, slope of direction vector (−m, 1) is 1/(−m) = −1/m. Hence m · k = m · (−1/m) = −1. QED.

Note: This fails when m = 0 (horizontal tangent → vertical normal, x = a) or m is undefined (vertical tangent → horizontal normal, y = b).

Proof 2 — Second Derivative Test (from Taylor Expansion)

Claim: If f'(a) = 0 and f''(a) > 0, then f has a local minimum at x = a.

Proof via Taylor Series:

Expand f near x = a to second order:

f(a + h) = f(a) + f'(a)·h + (1/2)f''(a)·h² + O(h³)

Since f'(a) = 0:

f(a + h) − f(a) = (1/2)f''(a)·h² + O(h³)

For small enough |h|, the O(h³) term is negligible compared to h². If f''(a) > 0, then (1/2)f''(a)·h² > 0 for all h ≠ 0. Therefore f(a + h) > f(a) for all small h ≠ 0, which is precisely the definition of a local minimum.

Similarly, if f''(a) < 0, then f(a + h) < f(a) for small h ≠ 0 — a local maximum.

If f''(a) = 0, the leading term is O(h³) or higher, and we cannot conclude from f'' alone — hence the test is inconclusive and the sign change method is required.

Proof 3 — Increasing Function Theorem

Claim: If f'(x) > 0 for all x in (a, b), then f is strictly increasing on (a, b).

Proof (Mean Value Theorem approach):

Take any two points x₁, x₂ ∈ (a, b) with x₁ < x₂.

By the Mean Value Theorem, there exists c ∈ (x₁, x₂) such that:

f(x₂) − f(x₁) = f'(c) · (x₂ − x₁)

Since f'(c) > 0 (given) and x₂ − x₁ > 0 (since x₁ < x₂), we have f(x₂) − f(x₁) > 0, i.e., f(x₂) > f(x₁). Since x₁ < x₂ implies f(x₁) < f(x₂), f is strictly increasing. QED.

Proof 4 — Chain Rule for Connected Rates

Claim: If y = f(x) and x = g(t), then dy/dt = (dy/dx) · (dx/dt).

Informal Proof:

The chain rule states d/dt[f(g(t))] = f'(g(t)) · g'(t).

In Leibniz notation with y = f(x), x = g(t):

dy/dt = (dy/dx) · (dx/dt)

The "dx" in the denominator of dy/dx and the "dx" in the numerator of dx/dt "cancel" symbolically, leaving dy/dt.

This is valid whenever dy/dx and dx/dt both exist. The result extends to chains of any length: if y depends on x, x on u, and u on t, then dy/dt = (dy/dx)(dx/du)(du/dt).

Visualiser — Tangent & Stationary Points on y = x³ − 3x

Drag the slider to move the point along the curve. The tangent line at that point is drawn in pink. Stationary points are marked in purple.

x = 0.00

Gradient: —

y-value: —

Nature: —

Move the slider to see the tangent equation.

Exercise 1 — Tangents & Normals (10 questions)

Enter the gradient value or constant term as requested. Tolerance ±0.01.

Q1. The curve y = x² + 3x − 2 passes through (1, 2). Find the gradient of the tangent at x = 1.

Q2. Find the gradient of the normal to y = 2x² − x at x = 2.

Q3. The tangent to y = x³ at (2, 8) has equation y = 12x + c. Find c.

Q4. Find the x-coordinate of the point on y = x² − 4x + 3 where the tangent is parallel to the x-axis.

Q5. The curve y = √x has a tangent at (4, 2). Find the y-intercept of this tangent.

Q6. Find the gradient of the tangent to y = (2x+1)³ at x = 0.

Q7. The normal to y = x² + 1 at (2, 5) has gradient m. Find m.

Q8. The tangent to y = 1/x at (2, 0.5) has equation y = −x/4 + c. Find c.

Q9. For y = x² − 6x, find the x-value where the tangent has gradient 2.

Q10. The tangent to y = eˣ at x = 0 has equation y = x + c. Find c.

Exercise 2 — Increasing & Decreasing Intervals (10 questions)

Enter the boundary x-value(s) where the function changes behaviour. If two boundaries, enter the smaller one.

Q1. f(x) = x² − 4x + 1. Find the x-value where f changes from decreasing to increasing.

Q2. f(x) = −x² + 6x. Find the x-value where f changes from increasing to decreasing.

Q3. f(x) = x³ − 3x². Find the smaller x-value where f' = 0.

Q4. f(x) = x³ − 3x² (continued). Find the larger x-value where f' = 0.

Q5. f(x) = 2x³ − 3x² − 12x + 1. Find the smaller boundary of the decreasing interval.

Q6. f(x) = 2x³ − 3x² − 12x + 1. Find the larger boundary of the decreasing interval.

Q7. f(x) = x⁴ − 8x². Find the positive x where f' = 0 (not x = 0).

Q8. f(x) = x + 1/x for x > 0. Find the x-value where f' = 0.

Q9. f(x) = 3x − x³. Find the positive x-value where f' = 0.

Q10. f(x) = x² e^(−x). Find the positive x-value where f' = 0 (other than x = 0).

Exercise 3 — Stationary Points (10 questions)

Enter the y-value at the stationary point, or the x-value as specified.

Q1. y = x² − 4x + 7. Find the y-value at the stationary point.

Q2. y = −x² + 8x − 3. Find the maximum y-value.

Q3. y = x³ − 12x. Find the y-value at the local maximum (x = −2).

Q4. y = x³ − 12x. Find the y-value at the local minimum (x = 2).

Q5. y = 2x³ − 3x² − 12x. Find the x-value of the local minimum.

Q6. y = x⁴ − 4x³. Find the y-value at the minimum where x = 3.

Q7. y = x + 4/x. Find the y-value at the local minimum for x > 0.

Q8. y = x²e^x. The stationary point at x = 0 is a min/max/inflection. Enter −1 for min, 1 for max, 0 for inflection.

Q9. y = x³. How many stationary points does it have? Enter the number.

Q10. y = x⁴ − 2x². Find the y-value at x = 0 stationary point.

Exercise 4 — Optimisation (10 questions)

Enter the maximum or minimum value as requested.

Q1. A square has perimeter P = 4x. The area is A = x². If P = 20, find the value of A.

Q2. A rectangle has perimeter 40 cm. Find the maximum area (cm²).

Q3. A box with square base (side x) and no lid has surface area 300 cm². Express height h in terms of x, then find the x that maximises volume. Enter x to 2 d.p.

Q4. The sum of two positive numbers is 10. Find the minimum value of the sum of their squares.

Q5. A 10 cm × 10 cm sheet has equal squares of side x cut from corners to form an open box. Find x (to 2 d.p.) that maximises volume.

Q6. A farmer has 60 m of fence for 3 sides of a rectangle (wall forms the 4th). Find the maximum area (m²).

Q7. Revenue R(x) = 50x − x². Find the maximum revenue.

Q8. A cylinder has volume 100π cm³. Find the radius r (cm) that minimises the total surface area. Give to 2 d.p.

Q9. y = x³ − 6x² + 9x on [0, 4]. Find the maximum y-value on this closed interval.

Q10. A right-angle triangle has legs summing to 10. Find the maximum hypotenuse squared.

Exercise 5 — Connected Rates of Change (10 questions)

Enter the rate of change as requested. Tolerance ±0.01.

Q1. Area of circle A = πr². If dr/dt = 2 cm/s, find dA/dt when r = 3 cm. (Leave in terms of π as a decimal to 2 d.p.)

Q2. Volume of sphere V = (4/3)πr³. If dr/dt = 1 cm/s, find dV/dt when r = 2 cm. (Give as a multiple of π, enter number only.)

Q3. A = πr². If dA/dt = 10π cm²/s, find dr/dt when r = 5 cm.

Q4. A square has side x cm, area A = x². If dx/dt = 3 cm/s, find dA/dt when x = 4 cm.

Q5. Volume V = x³. If dV/dt = 12 cm³/s, find dx/dt when x = 2 cm.

Q6. y = x² + 1. If dx/dt = 4, find dy/dt when x = 3.

Q7. A cone has V = (1/3)πr²h with fixed r/h = 1/2 (so r = h/2). If dV/dt = 5 cm³/s, find dh/dt when h = 4 cm. Give to 3 d.p.

Q8. Perimeter of square P = 4x. If dP/dt = 8 cm/s, find dA/dt when x = 3 cm.

Q9. y = sin x. If dx/dt = 2 rad/s, find dy/dt when x = π/6.

Q10. The radius of a circle decreases at 0.3 cm/s. Find the rate at which circumference C = 2πr decreases. Give in terms of π (enter the coefficient).

Practice — 30 Mixed Questions

Mixed questions across all five topics. Get 100% for confetti!

P1. Gradient of tangent to y = 3x² − 2 at x = 1.

P2. y = x³ − x at x = −1. Find the tangent gradient.

P3. Normal gradient to y = 4x − x² at x = 1.

P4. y = x² − 5x. x-value where f' = 0.

P5. f(x) = x³ − 6x. Number of stationary points.

P6. y = x³ − 6x. y at local maximum.

P7. y = x³ − 6x. y at local minimum.

P8. f(x) = x² − 4x + 7. Minimum y-value.

P9. y = −2x² + 8x. Maximum y-value.

P10. The tangent to y = x² at (3, 9) has gradient m. Find m.

P11. y = 1/x. Tangent at (1,1) has equation y = −x + c. Find c.

P12. dA/dt when A = πr², dr/dt = 3, r = 2. (Express as multiple of π, enter coefficient.)

P13. Rectangle perimeter 80 m. Maximum area (m²).

P14. y = x³ − 3x + 2. y-value at local maximum.

P15. y = x³ − 3x + 2. y-value at local minimum.

P16. f(x) = 2x³ + 3x² − 12x. x-value of the local maximum (the negative one).

P17. f''(x) at a stationary point is −5. Is this a max or min? Enter 1 for max, −1 for min.

P18. V = x(10−2x)², 0 < x < 5. Find dV/dx at x = 2.

P19. y = (x−1)². Find the x-value of the stationary point.

P20. A = πr², r increases at 0.5 m/s. Rate of increase of A when r = 4 m. (Multiple of π, enter coefficient.)

P21. Normal to y = x³ at (2, 8): normal gradient.

P22. f(x) = x⁴ − 8x². How many stationary points are there?

P23. y = x + 9/x for x > 0. Minimum y-value.

P24. The product of two positive numbers is 16. Minimum sum of the two numbers.

P25. V = (4/3)πr³. dV/dt when r = 3 and dr/dt = 2. (Express as multiple of π.)

P26. y = sin x. Gradient at x = π/3. Give to 2 d.p.

P27. y = eˣ. Normal gradient at x = 0.

P28. f(x) = x² − 4. For what x is f' = 0?

P29. y = x³ − 6x² + 12x − 8. How many stationary points does it have?

P30. A cube has side x cm. If dx/dt = 2 cm/s, find dV/dt when x = 3 cm.

Challenge — 15 Harder Questions

These require multi-step reasoning. Tolerance ±0.01 unless otherwise stated.

C1. Find the x-coordinates where the tangent to y = x³ − 3x² + 2 is parallel to the line y = 9x − 1.

C2. The tangents to y = x² at two points have gradients that sum to 6. If the points are (a, a²) and (b, b²), find a + b.

C3. y = x³ − 3x + k. For what value of k does the curve have a stationary point at y = 0? (Give the positive answer.)

C4. A cylinder (no top) has volume 250π. Find the minimum total surface area divided by π. (Enter integer.)

C5. A cone and sphere share the same volume πr³ (sphere radius = r). If V_cone = (1/3)πR²H where H = 4r, find R/r.

C6. y = x⁴ − 4x³ + 6x². Find the x-value of the inflection point where f'' = 0 and f'' changes sign.

C7. Rate of change of the volume of a sphere (dV/dt) is always numerically equal to the surface area. Find dr/dt. (Just enter the number.)

C8. A rectangle is inscribed in a circle of radius 5. Find the maximum area of the rectangle.

C9. f(x) = x³ + ax² + bx has a local max at x = −1 and local min at x = 3. Find a.

C10. The normal to y = x² − 3x + 5 at the point P is parallel to the line x + y = 7. Find the x-coordinate of P.

C11. y = (x−2)²(x+1). Find the x-coordinate of the point of inflection.

C12. Two sides of a triangle are 6 and 8, with included angle θ. Maximise the area A = (1/2)(6)(8)sin θ. What is the maximum area?

C13. Water fills a hemisphere (radius 10 cm) at 20 cm³/s. Find dh/dt when h = 5 cm, where V = π(10h² − h³/3). Give to 3 d.p.

C14. The curve y = ax³ + bx² has a stationary point at (2, −4). Find a.

C15. Prove that y = x³ + 3x is an increasing function for all x. What is the minimum value of dy/dx?

Exam Style Questions (8)

Cambridge-style multi-part questions. Show your working. Reveal mark schemes after attempting each.

Question 1 [8 marks]

A curve has equation y = 2x³ − 9x² + 12x − 4.

(a) Find dy/dx. 2

(b) Find the coordinates of the two stationary points. 3

(c) Determine the nature of each stationary point. 3

(a) dy/dx = 6x² − 18x + 12 [B2]
(b) 6x² − 18x + 12 = 0 → x² − 3x + 2 = 0 → (x−1)(x−2) = 0 [M1]
x = 1: y = 2 − 9 + 12 − 4 = 1. Point (1, 1). [A1]
x = 2: y = 16 − 36 + 24 − 4 = 0. Point (2, 0). [A1]
(c) d²y/dx² = 12x − 18 [M1]
At x=1: d²y/dx² = −6 < 0 → local maximum at (1, 1). [A1]
At x=2: d²y/dx² = 6 > 0 → local minimum at (2, 0). [A1]

Question 2 [7 marks]

The curve C has equation y = x² − 4x + 5.

(a) Find the equation of the tangent to C at the point (3, 2). 3

(b) Find the equation of the normal to C at the point (3, 2). 2

(c) The normal meets the x-axis at point Q. Find the coordinates of Q. 2

(a) dy/dx = 2x − 4. At x=3: m = 2. [M1]
Tangent: y − 2 = 2(x − 3) → y = 2x − 4 [A2]
(b) Normal gradient = −1/2. y − 2 = −(1/2)(x − 3) → y = −x/2 + 7/2 [A2]
(c) Set y = 0: 0 = −x/2 + 7/2 → x = 7. Q = (7, 0). [M1 A1]

Question 3 [9 marks]

A rectangular garden is to be enclosed with 100 m of fencing. One side of the garden is against a straight wall and requires no fencing. Let the width perpendicular to the wall be x m.

(a) Show that the area A = 100x − 2x². 2

(b) Find the value of x that gives the maximum area. 3

(c) Find the maximum area. 1

(d) Justify that this is indeed a maximum. 3

(a) Fencing constraint: 2x + L = 100 → L = 100 − 2x [M1]. A = xL = x(100 − 2x) = 100x − 2x² [A1]
(b) dA/dx = 100 − 4x = 0 → x = 25 [M2 A1]
(c) A = 100(25) − 2(625) = 2500 − 1250 = 1250 m² [A1]
(d) d²A/dx² = −4 < 0 for all x → concave down → x = 25 gives a maximum [B1].
OR: check x = 0: A = 0; x = 50: A = 0; x = 25: A = 1250. Maximum confirmed [B2].

Question 4 [8 marks]

A circle has area A cm² and radius r cm. The radius is increasing at 0.2 cm/s.

(a) Find dA/dr. 1

(b) Find the rate of increase of the area when r = 5 cm. 2

(c) The circumference is C = 2πr. Find the rate of increase of the circumference when r = 5 cm. 2

(d) Show that the rate of increase of A = C × (dr/dt). 3

(a) dA/dr = 2πr [B1]
(b) dA/dt = (dA/dr)(dr/dt) = 2π(5)(0.2) = 2π cm²/s [M1 A1]
(c) dC/dt = (dC/dr)(dr/dt) = 2π(0.2) = 0.4π cm/s [M1 A1]
(d) dA/dt = 2πr · (dr/dt) = C · (dr/dt) since C = 2πr [M1 M1 A1]

Question 5 [10 marks]

A function f is defined by f(x) = x³ − 6x² + 9x + 2.

(a) Find f'(x) and f''(x). 2

(b) Find the intervals where f is increasing. 3

(c) Find and classify all stationary points. 4

(d) State the range of f on the closed interval [0, 4]. 1

(a) f'(x) = 3x² − 12x + 9 = 3(x−1)(x−3) [B1]; f''(x) = 6x − 12 [B1]
(b) f'(x) > 0: 3(x−1)(x−3) > 0 → x < 1 or x > 3 [M2 A1]
(c) Stationary at x=1 and x=3. [M1]
f(1) = 1 − 6 + 9 + 2 = 6. f''(1) = −6 < 0 → local max (1, 6). [A1]
f(3) = 27 − 54 + 27 + 2 = 2. f''(3) = 6 > 0 → local min (3, 2). [A2]
(d) Check endpoints: f(0) = 2, f(4) = 64 − 96 + 36 + 2 = 6. Range is [2, 6]. [A1]

Question 6 [6 marks]

A closed cylinder has volume V = 300π cm³.

(a) Express the total surface area S in terms of the radius r only. 3

(b) Show that dS/dr = 0 gives r = ∛150. 3

(a) V = πr²h = 300π → h = 300/r² [M1]
S = 2πr² + 2πrh = 2πr² + 2πr(300/r²) = 2πr² + 600π/r [A2]
(b) dS/dr = 4πr − 600π/r² [M1]
Set = 0: 4πr = 600π/r² → 4r³ = 600 → r³ = 150 → r = ∛150 [M1 A1]

Question 7 [7 marks]

The gradient of a curve at any point (x, y) is given by dy/dx = 3x² − 4x + 1. The curve passes through (2, 3).

(a) Find the equation of the curve. 3

(b) Find the equation of the tangent to the curve at (2, 3). 2

(c) Determine the x-values of the stationary points and their nature. 2

(a) y = ∫(3x² − 4x + 1) dx = x³ − 2x² + x + C [M1]
At (2, 3): 3 = 8 − 8 + 2 + C → C = 1. y = x³ − 2x² + x + 1 [A2]
(b) At x=2: dy/dx = 12 − 8 + 1 = 5. Tangent: y − 3 = 5(x−2) → y = 5x − 7 [M1 A1]
(c) 3x² − 4x + 1 = 0 → (3x−1)(x−1) = 0 → x = 1/3 or x = 1 [M1]
d²y/dx² = 6x − 4. At x=1/3: −2 < 0 → max. At x=1: 2 > 0 → min. [A1]

Question 8 [8 marks]

Water pours into a conical funnel (vertex at bottom). The cone has height 20 cm and radius at top 10 cm. Water enters at a rate of 5 cm³/s.

(a) Using similar triangles, express r in terms of h. 1

(b) Show that V = πh³/12. 2

(c) Find dh/dt when h = 8 cm. Give your answer to 3 significant figures. 5

(a) r/h = 10/20 = 1/2, so r = h/2. [B1]
(b) V = (1/3)πr²h = (1/3)π(h/2)²h = (1/3)π(h²/4)h = πh³/12. [M2]
(c) dV/dh = 3πh²/12 = πh²/4. [M1]
dV/dt = (dV/dh)(dh/dt) → 5 = (π·64/4)(dh/dt) = 16π(dh/dt). [M2]
dh/dt = 5/(16π) ≈ 0.0995 cm/s [M1 A1]

Past Paper Questions

Five Cambridge 9709 style past paper questions with full solutions on reveal.

PP1 — Tangent and Normal [Cambridge 9709 P1 style]

The curve C has equation y = x³ − 2x² + x + 1.

(i) Find the equation of the tangent to C at the point (2, 3).

(ii) Find the equation of the normal to C at the point (2, 3), giving your answer in the form ax + by + c = 0 where a, b, c are integers.

dy/dx = 3x² − 4x + 1. At x=2: m = 12 − 8 + 1 = 5. [M1]
(i) Tangent: y − 3 = 5(x − 2) → y = 5x − 7. [A2]
(ii) Normal gradient = −1/5. y − 3 = −(1/5)(x − 2) [M1]
5y − 15 = −(x − 2) → 5y − 15 = −x + 2 → x + 5y − 17 = 0. [A2]

PP2 — Stationary Points [Cambridge 9709 P1 style]

A curve has equation y = 2x³ + 3x² − 12x + 1.

(i) Find the coordinates of each stationary point.

(ii) Determine the nature of each stationary point.

(iii) State the set of values of x for which y is decreasing.

dy/dx = 6x² + 6x − 12 = 6(x² + x − 2) = 6(x+2)(x−1). [M1]
Stationary at x = −2 and x = 1. [A1]
y(−2) = −16 + 12 + 24 + 1 = 21. y(1) = 2 + 3 − 12 + 1 = −6. [A2]
(i) Stationary points: (−2, 21) and (1, −6).
d²y/dx² = 12x + 6. [M1]
At x=−2: −24 + 6 = −18 < 0 → local maximum at (−2, 21). [A1]
At x=1: 12 + 6 = 18 > 0 → local minimum at (1, −6). [A1]
(iii) y decreasing when dy/dx < 0: 6(x+2)(x−1) < 0 → −2 < x < 1. [A2]

PP3 — Optimisation [Cambridge 9709 P1 style]

A solid cylinder has radius r cm and height h cm. Its volume is 54π cm³.

(i) Express h in terms of r.

(ii) Show that the total surface area S = 2πr² + 108π/r.

(iii) Find the value of r that minimises S, and find the minimum value of S.

(i) πr²h = 54π → h = 54/r². [B1]
(ii) S = 2πr² + 2πrh = 2πr² + 2πr(54/r²) = 2πr² + 108π/r. [M1 A1] ✓
(iii) dS/dr = 4πr − 108π/r² = 0 → 4r³ = 108 → r³ = 27 → r = 3. [M2 A1]
d²S/dr² = 4π + 216π/r³ > 0 → minimum confirmed. [B1]
S_min = 2π(9) + 108π/3 = 18π + 36π = 54π cm². [A2]

PP4 — Connected Rates [Cambridge 9709 P1 style]

A spherical balloon is being inflated. When the radius is 6 cm, the volume is increasing at 12 cm³/s.

(i) Find the rate of increase of the radius at this instant.

(ii) Find the rate of increase of the surface area at this instant. (S = 4πr²)

V = (4/3)πr³ → dV/dr = 4πr². [B1]
(i) dV/dt = (dV/dr)(dr/dt) → 12 = 4π(36)(dr/dt) → dr/dt = 12/(144π) = 1/(12π) ≈ 0.0265 cm/s. [M2 A1]
(ii) S = 4πr² → dS/dr = 8πr. [M1]
dS/dt = (dS/dr)(dr/dt) = 8π(6) · 1/(12π) = 48π/(12π) = 4 cm²/s. [M1 A1]

PP5 — Complete Applications [Cambridge 9709 P1 style]

The function f is defined for x ∈ ℝ by f(x) = x³ − 6x² + 9x + 1.

(i) Find f'(x) and the x-coordinates of the stationary points of y = f(x).

(ii) Determine the nature of each stationary point, showing your reasoning.

(iii) Find the range of f on the interval [0, 5].

(iv) On which subinterval is f increasing? State in set notation.

(i) f'(x) = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x−1)(x−3). [M1]
Stationary at x = 1 and x = 3. [A1]
(ii) f''(x) = 6x − 12. At x=1: f''(1) = −6 < 0 → local maximum. [A1]
At x=3: f''(3) = 6 > 0 → local minimum. [A1]
f(1) = 1 − 6 + 9 + 1 = 5. f(3) = 27 − 54 + 27 + 1 = 1. [A1]
(iii) Check endpoints: f(0) = 1, f(5) = 125 − 150 + 45 + 1 = 21. [M1]
Values: f(0) = 1, f(1) = 5 (max), f(3) = 1 (min), f(5) = 21 (endpoint max). [M1]
Range is [1, 21]. [A1]
(iv) f'(x) > 0 when x < 1 or x > 3. {x : x < 1} ∪ {x : x > 3}. [A2]

Applications of Differentiation A-Level Pure 1