Applications of Differentiation: Turning Points, Optimisation & Curve Sketching

Why Study Derivative Applications?

Once you know how to differentiate, the next step is knowing what to do with it. Derivatives unlock the ability to find maximum and minimum values, determine where a function is increasing or decreasing, and sketch accurate curves — all critical skills in A-Level, AP and IB examinations.

Learning Goals: By the end of this guide, you should be able to:

Find and classify stationary points using the first and second derivative tests.

Determine intervals where a function is increasing, decreasing or concave.

Solve real-world optimisation problems.

Sketch a curve using derivatives systematically.

Stationary Points

A stationary point occurs where $f'(x) = 0$ . There are three types:

Type	First Derivative	Second Derivative
Local maximum	$f'(x) = 0$	$f''(x) < 0$
Local minimum	$f'(x) = 0$	$f''(x) > 0$
Point of inflection	$f'(x) = 0$	$f''(x) = 0$ (check sign change)

The Second Derivative Test

Find $f'(x)$ and solve $f'(x) = 0$ .
Compute $f''(x)$ at each stationary point.
If $f''(x) > 0$ : minimum. If $f''(x) < 0$ : maximum. If $f''(x) = 0$ : test is inconclusive — use a sign table instead.

Worked Example: Finding Turning Points

Question: Find the stationary points of $f(x) = x^3 - 6x^2 + 9x + 1$ and classify them.

Step 1: $f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3)$

Setting $f'(x) = 0$ : $x = 1$ or $x = 3$ .

Step 2: $f''(x) = 6x - 12$

At $x = 1$ : $f''(1) = -6 < 0$ → local maximum at $(1, 5)$
At $x = 3$ : $f''(3) = 6 > 0$ → local minimum at $(3, 1)$

Optimisation Problems

Optimisation uses derivatives to find the maximum or minimum value of a quantity subject to constraints.

Strategy

Define variables and write the quantity to optimise as a function of one variable.
Use the constraint to eliminate one variable if needed.
Differentiate, set $f'(x) = 0$ , and solve.
Verify it's a max or min using the second derivative test.
Answer in context with correct units.

Example: Fencing Problem

Question: A farmer has 100 m of fencing to enclose a rectangular field against a straight wall. Find the dimensions that maximise the enclosed area.

Let $x$ = width perpendicular to the wall. Then length $= 100 - 2x$ .

A(x) = x(100 - 2x) = 100x - 2x^2

A'(x) = 100 - 4x = 0 \implies x = 25

A''(x) = -4 < 0 \implies \text{maximum}

Dimensions: $25 \text{ m} \times 50 \text{ m}$ , maximum area $= 1250 \text{ m}^2$ .

Curve Sketching Checklist

To sketch $y = f(x)$ systematically:

Intercepts: Set $x = 0$ for y-intercept; set $y = 0$ for x-intercepts.
Stationary points: Solve $f'(x) = 0$ , classify with $f''$ .
Increasing/decreasing: Use sign of $f'(x)$ .
Concavity: Use sign of $f''(x)$ .
End behaviour: What happens as $x \to \pm\infty$ ?
Asymptotes: Vertical (denominator = 0), horizontal (limits at infinity).

Common Mistakes

Finding stationary points but not classifying them — You must determine whether each is a max, min or inflection point.
Forgetting the context in optimisation — Always check that your answer makes physical sense (e.g., dimensions must be positive).
Assuming $f''(x) = 0$ means inflection — The second derivative being zero is necessary but not sufficient. You need a sign change in $f''$ .

Exam Tips (A-Level / AP / IB)

In optimisation questions, clearly define your variable and write the function before differentiating. Examiners award marks for the setup.
If the second derivative test is inconclusive, construct a sign table for $f'(x)$ either side of the stationary point.
For curve sketching, a labelled sketch with all key features (intercepts, turning points, asymptotes) earns full marks even if it's not perfectly to scale.

Frequently Asked Questions

What is the difference between a local and global maximum?

A local maximum is the highest point in its immediate neighbourhood. A global maximum is the absolute highest point on the entire domain. To find global extrema on a closed interval $[a, b]$ , evaluate $f$ at all stationary points AND at the endpoints $a$ and $b$ .

Can a function have stationary points but no turning points?

Yes — the function $f(x) = x^3$ has a stationary point at $x = 0$ where $f'(0) = 0$ , but it's a point of inflection, not a turning point. The function is still strictly increasing.

Derivatives from First Principles — The foundation: where the derivative comes from.
Conic Sections — Optimisation of areas and distances involving conics.
Linear Regression — The least squares method is fundamentally an optimisation problem.

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