11.4 Differential Calculus: Gradients and Turning Points

Quick Answer: Differential calculus turns a changing function into its gradient function. On MAT, differentiate a simple polynomial, use f′(a) for the tangent gradient at x = a, solve f′(x) = 0 for stationary points, and read the sign of f′ to decide where a graph increases or decreases. The 2025 National Report explicitly includes these principles in MAT Functions and Graphs.

Why calculus is current MAT content

The official 2025 National Report groups MAT into five current subdomains. MAT M3 — Functions and graphs includes understanding and analysing functions, interpreting graph problems, and applying principles of differential calculus. The official 2015 MAT educator booklet supplies the practical detail: critical points, increasing and decreasing functions, tangents, and interpreting a function from its derivative and vice versa.

The shorter live Test Content page lists function families and graph properties without spelling out calculus. That omission does not cancel the newer national report. This independent OpenExamPrep section teaches the shared official description with original examples; it does not reproduce confidential NBT items.

From average gradient to instantaneous gradient

Between two points (x₁, f(x₁)) and (x₂, f(x₂)), the average gradient is

[f(x₂) − f(x₁)] / (x₂ − x₁).

A derivative asks for the gradient at one instant. Move a second point only h units away and let h approach zero:

f′(x) = lim as h → 0 of [f(x + h) − f(x)] / h.

For f(x) = x²:

• f(x + h) − f(x) = (x + h)² − x² = 2xh + h²
• divide by h to get 2x + h
• let h approach 0, giving f′(x) = 2x

So the parabola's gradient is not one fixed number. At x = −2 it is −4, at x = 0 it is 0, and at x = 3 it is 6. Negative derivative means the graph is falling as x increases; positive derivative means it is rising.

Rules for the simple functions MAT names

For polynomial terms, the power rule is enough for most calculator-free items:

Example: if f(x) = 2x³ − 5x² + 4x − 7, then

f′(x) = 6x² − 10x + 4.

Keep coefficients attached and reduce the exponent by one. The derivative of −7 is 0, not −7. The derivative of 4x is 4, not 4x. A multiple-choice distractor often performs only one of those two steps correctly.

Tangents and normals

At x = a, first find the point (a, f(a)), then the tangent slope m = f′(a). Use point-gradient form:

y − f(a) = f′(a)(x − a).

Worked example: f(x) = x² − 4x + 7 at x = 3.

• Point: f(3) = 9 − 12 + 7 = 4, so (3, 4)
• Derivative: f′(x) = 2x − 4
• Tangent gradient: f′(3) = 2
• Tangent: y − 4 = 2(x − 3), hence y = 2x − 2

If the tangent gradient is nonzero, the normal is perpendicular and has gradient −1/m. Here the normal gradient is −1/2, so its equation is y − 4 = −(1/2)(x − 3). Do not use the reciprocal without changing its sign. If the tangent is horizontal, the normal is vertical and is written x = a.

Stationary points and graph behaviour

A stationary point occurs where f′(x) = 0. It can be a local maximum, a local minimum, or a stationary point of inflection. Solve first; classify second.

Take f(x) = x³ − 3x² − 9x + 5.

f′(x) = 3x² − 6x − 9 = 3(x + 1)(x − 3).

Thus the stationary x-values are −1 and 3. Their coordinates are:

• f(−1) = 10, so (−1, 10)
• f(3) = −22, so (3, −22)

A derivative sign chart classifies them:

The graph changes from increasing to decreasing at x = −1, so (−1, 10) is a local maximum. It changes from decreasing to increasing at x = 3, so (3, −22) is a local minimum.

The second derivative gives a shorter check. Here f″(x) = 6x − 6. A negative f″ at a stationary point means locally concave down and a local maximum; a positive f″ means concave up and a local minimum. Therefore f″(−1) = −12 and f″(3) = 12, confirming the classifications.

Do not confuse a stationary point with an intercept. This cubic's y-intercept is f(0) = 5; neither stationary y-value is 5. A graph question may offer the correct x-values with the wrong y-values, or give the y-intercept as if it were the minimum.

Reading a derivative graph in reverse

Sometimes the stem describes f′ rather than f. Use this translation:

• f′ above the x-axis → f is increasing
• f′ below the x-axis → f is decreasing
• f′ crosses from positive to negative → f has a local maximum
• f′ crosses from negative to positive → f has a local minimum
• f′ touches zero but keeps the same sign → f has a stationary point without a maximum or minimum

For f(x) = x³, f′(x) = 3x². The derivative is zero at x = 0 but positive on both sides. The original graph keeps increasing, so the origin is a stationary point of inflection, not a maximum or minimum. Merely seeing f′(a) = 0 is not enough to classify the point.

Optimization: build the function before differentiating

A wall forms one side of a rectangular enclosure. There are 40 m of fencing for the two equal widths x and the opposite length y. Then 2x + y = 40, so y = 40 − 2x. Area is

A(x) = x(40 − 2x) = 40x − 2x², with 0 < x < 20.

Differentiate: A′(x) = 40 − 4x. Set A′(x) = 0, giving x = 10. Then y = 20 and the maximum area is 200 m². The second derivative A″(x) = −4 is negative, so this stationary point is a maximum. The domain check matters: a negative width might solve badly formed algebra but cannot solve the enclosure problem.

Calculator-free MAT method

1. Write the function and its allowed domain.
2. Differentiate term by term.
3. For a tangent, substitute the x-value into both f and f′.
4. For turning points, solve f′(x) = 0 and classify with a sign change or f″.
5. For optimization, return to the context and compute the requested length, area, or value — not merely the stationary x-coordinate.

Exact factoring and small integer substitutions are deliberate. A calculator cannot decide whether the question asks for a gradient, a coordinate, a maximum value, or an interval of increase.