Practice Maths

Curve Sketching — Maxima, Minima and End Behaviour

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Key Terms

Full curve sketching method (5 steps)
:
  1. y-intercept: set x = 0, find f(0).
  2. x-intercepts: solve f(x) = 0 (factorise or use known factors).
  3. Stationary points: solve f′(x) = 0, find y-values, classify using first derivative test.
  4. End behaviour: determine behaviour as x → +∞ and x → −∞ using the leading term.
  5. Sketch: plot all key points and draw a smooth curve consistent with all features.
End behaviour rules (leading term a·xn)
:
  — Even n, a > 0: y → +∞ both ends (U-shape).
  — Even n, a < 0: y → −∞ both ends (∩-shape).
  — Odd n, a > 0: y → −∞ as x → −∞, y → +∞ as x → +∞.
  — Odd n, a < 0: y → +∞ as x → −∞, y → −∞ as x → +∞.
Optimisation steps
: (1) define the variable; (2) write the objective function; (3) differentiate; (4) solve f′ = 0; (5) verify max/min using first derivative test; (6) state the answer with units.
📚 Hot Tip — Closed-Interval Optimisation
On a closed interval [a, b], the global maximum/minimum could be at a stationary point or at an endpoint. Always check all candidate points: stationary points in (a, b) and both endpoints.
x y 0 1 2 3 2 0 −4 y=x³−3x² (0,0) (3,0) (2,−4) min local max y→−∞ y→+∞
Hot Tip End behaviour is determined entirely by the leading term. For y = x³ − 3x², the leading term is x³ (odd degree, positive coefficient): y → −∞ as x → −∞, and y → +∞ as x → +∞. All other terms become insignificant for very large |x|.

Worked Example — Full Curve Sketch of y = x³ − 3x²

Step 1 — y-intercept: f(0) = 0. Point: (0, 0).

Step 2 — x-intercepts: x³ − 3x² = x²(x − 3) = 0 → x = 0 (double root) or x = 3. Intercepts at x = 0 and x = 3.

Step 3 — Stationary points: f′(x) = 3x² − 6x = 3x(x − 2) = 0 → x = 0 or x = 2.
f(0) = 0; f(2) = 8 − 12 = −4.
Sign test: f′(−1) = 9 > 0; f′(1) = −3 < 0; f′(3) = 9 > 0.
x = 0: +→− → local max at (0, 0).   x = 2: −→+ → local min at (2, −4).

Step 4 — End behaviour: Leading term x³: y → −∞ as x → −∞; y → +∞ as x → +∞.

Step 5 — Sketch: See graph above.

Introduction: Putting It All Together

Curve sketching is the synthesis of all differentiation skills: you use intercepts, stationary points, classification, and end behaviour to produce an accurate, labelled sketch without plotting dozens of individual points. A good sketch communicates the essential shape and all key features of the curve.

The 5-Step Curve Sketching Method

  1. y-intercept: substitute x = 0 into f(x).
  2. x-intercepts: solve f(x) = 0 (factorise completely if possible). Note the multiplicity of roots — a double root means the curve touches the x-axis, not crosses it.
  3. Stationary points: differentiate, set f′(x) = 0, find y-values, classify by the first derivative test.
  4. End behaviour: use the leading term only to determine what happens as x → ±∞.
  5. Sketch: connect all key points with a smooth curve, consistent with end behaviour, stationary points, and intercepts.

Optimisation Problems

Many real-world problems ask us to maximise or minimise a quantity. The calculus approach is systematic:

  1. Define a variable: choose a single variable to express the unknown.
  2. Write the objective function: express the quantity to optimise (e.g. area, volume, profit) as a function of that variable.
  3. State the domain: what are the valid values of the variable?
  4. Differentiate and solve: set f′ = 0.
  5. Verify: use the first derivative test (or check endpoints on a closed domain).
  6. Answer: state with units.
Worked Example — Optimisation: Fencing a Paddock

A farmer has 120 m of fencing to enclose a rectangular paddock against a straight river (no fence needed on the river side). Find the dimensions that maximise the area.

Step 1: Let x = width (m) of paddock (two sides parallel to river). Then length = 120 − 2x.

Step 2: A(x) = x(120 − 2x) = 120x − 2x². Domain: 0 < x < 60.

Step 3: A′(x) = 120 − 4x. Set to zero: 120 − 4x = 0 → x = 30.

Step 4: A′(20) = 40 > 0; A′(40) = −40 < 0. Sign changes +→− → local maximum.

Answer: Width = 30 m, length = 60 m. Maximum area = 30 × 60 = 1800 m².

Worked Example — Optimisation: Open Box

A box is made by cutting squares of side x from the corners of a 20 × 20 cm sheet and folding up the sides. Find x to maximise the volume.

Step 1: After cutting: base = (20−2x) × (20−2x), height = x. Domain: 0 < x < 10.

Step 2: V(x) = x(20−2x)² = x(400 − 80x + 4x²) = 4x³ − 80x² + 400x.

Step 3: V′(x) = 12x² − 160x + 400 = 4(3x² − 40x + 100) = 4(3x − 10)(x − 10).

Step 4: V′ = 0 at x = 10⁄3 ≈ 3.33 or x = 10 (boundary). Since 0 < x < 10, use x = 10⁄3.

Verify: V′(1) = 4(3−10)(1−10) = 4(−7)(−9) = 252 > 0; V′(5) = 4(15−10)(5−10) = 4(5)(−5) = −100 < 0. Maximum at x = 10⁄3.

Answer: Cut squares of side 10⁄3 cm (approx. 3.33 cm) for maximum volume.

💡 Key Reminder: In optimisation, always state the domain clearly. On a closed interval [a, b], check the value of f at the stationary point and at both endpoints a and b, then pick the largest (or smallest).

Summary

Curve sketching uses intercepts, stationary points, classification, and end behaviour to produce a complete picture of a function. Optimisation applies the same calculus tools to real-world problems: define a variable, write the objective function, differentiate, solve, verify, and interpret.

Mastery Practice

See Answers ➔
  1. Fluency

    Use the 5-step method to sketch y = x³ − 3x². Label all intercepts and stationary points with coordinates.

  2. Fluency

    Use the 5-step method to sketch y = −x³ + 4x. Label all intercepts and stationary points with coordinates.

  3. Fluency

    Determine the end behaviour (state what y approaches as x → +∞ and as x → −∞) for each function.

    1. (a) y = 2x4
    2. (b) y = −3x³
    3. (c) y = x5 − 2x
  4. Understanding

    Complete a full sketch of f(x) = x4 − 8x² + 7, labelling all x-intercepts, the y-intercept, and all stationary points.

    Hint: For x-intercepts, try factorising as a quadratic in x².
  5. Understanding

    A rectangular paddock is to be fenced on 3 sides using 120 m of fencing (the fourth side is a river and needs no fence). Find the dimensions that maximise the enclosed area, and state the maximum area.

  6. Understanding

    A box with no lid is made by cutting equal squares of side x cm from the corners of a 20 × 20 cm square sheet and folding up the sides.

    1. (a) Show that the volume is V(x) = 4x³ − 80x² + 400x.
    2. (b) Find the value of x that maximises V.
    3. (c) Find the maximum volume.
  7. Understanding

    The function f(x) = x³ + ax² − 9x + b has a stationary point at (1, −5).

    1. (a) Find the values of a and b.
    2. (b) Find the other stationary point.
    3. (c) Classify both stationary points.
  8. Problem Solving

    A company’s weekly profit (in thousands of dollars) from producing x hundred items is P(x) = −x³ + 15x² − 48x − 10, for 0 ≤ x ≤ 10.

    1. (a) Find all stationary points of P in the domain.
    2. (b) Find the maximum profit and the production level that achieves it.
    3. (c) Check the endpoints x = 0 and x = 10 to confirm your answer is the global maximum.
  9. Problem Solving

    A closed cylinder has radius r cm and height h cm with volume 100π cm³.

    1. (a) Express h in terms of r using the volume constraint.
    2. (b) Show that the total surface area is S = 2πr² + 200π⁄r.
    3. (c) Find the value of r that minimises S and find the minimum surface area.
  10. Problem Solving

    Two positive numbers sum to 12. Find the values of the two numbers that minimise the sum of their squares.