Stationary points: 7 exam questions solved step by step

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What this lesson covers

  • A stationary point is where the derivative is zero; concavity is the sign of the second derivative.
  • Set the full derivative equal to zero before clearing any denominator.
Key idea

7 questions, one method

  • 1. Write dy/dx = 0.
  • 2. Solve: find every root.
  • 3. Each y from the original function.
  • 4. Classify with d²y/dx².
Your turn

Question 1 · Achieved

  • y = 2x³ + 3x² − 12x + 5
  • Find the coordinates of
  • the stationary points.
Worked example

Q1: y = 2x³ + 3x² − 12x + 5

  • Gradient zero.
  • Both coordinates.
  1. dy/dx = 6x² + 6x − 12differentiate
  2. 6x² + 6x − 12 = 0stationary: dy/dx = 0
  3. 6(x² + x − 2) = 0
  4. 6(x + 2)(x − 1) = 0factorise
  5. x = −2 or x = 1
  6. y(−2) = −16 + 12 + 24 + 5 = 25
  7. y(1) = 2 + 3 − 12 + 5 = −2
  8. (−2, 25) and (1, −2)
Your turn

Question 2 · Achieved

  • The same curve. Classify each
  • stationary point, and state where
  • the curve is decreasing.
Worked example

Q2: classify, then decreasing

  • d²y/dx² < 0: a maximum.
  • Decreasing: dy/dx < 0.
  1. d²y/dx² = 12x + 6differentiate again
  2. x = −2: −24 + 6 = −18 < 0concave down
  3. (−2, 25) is a maximum
  4. x = 1: 12 + 6 = 18 > 0concave up
  5. (1, −2) is a minimum
  6. decreasing: dy/dx < 0the condition
  7. 6(x + 2)(x − 1) < 0
  8. x = 0: 6(2)(−1) = −12 < 0test between
  9. decreasing for −2 < x < 1
Your turn

Question 3 · Merit

  • y = (x² + 5)/(x − 2)
  • Find the coordinates of
  • the stationary points.
Worked example

Q3: the derivative = 0

  • Quotient rule.
  • Whole fraction = 0 first.
  1. u = x² + 5, u′ = 2x
  2. v = x − 2, v′ = 1
  3. top: 2x(x − 2) − (x² + 5)u′v − uv′
  4. = 2x² − 4x − x² − 5expand
  5. = x² − 4x − 5collect
  6. dy/dx = (x² − 4x − 5)/(x − 2)²
  7. (x² − 4x − 5)/(x − 2)² = 0write this first
Worked example

Q3: then clear the fraction

  • Fraction = 0: top = 0.
  • x = 2 is not allowed.
  1. (x² − 4x − 5)/(x − 2)² = 0
  2. x² − 4x − 5 = 0top = 0
  3. (x − 5)(x + 1) = 0factorise
  4. x = 5 or x = −1neither is 2
  5. y(5) = 30/3 = 10
  6. y(−1) = 6/(−3) = −2
  7. (5, 10) and (−1, −2)
Your turn

Question 4 · Merit

  • y = (x² − 3)eˣ
  • Find the exact coordinates of
  • the stationary points.
Worked example

Q4: y = (x² − 3)eˣ

  • Product rule.
  • eˣ is never zero.
  1. u = x² − 3, u′ = 2xproduct rule
  2. v = eˣ, v′ = eˣ
  3. dy/dx = 2xeˣ + (x² − 3)eˣ
  4. = (x² + 2x − 3)eˣtake out eˣ
  5. (x + 3)(x − 1)eˣ = 0stationary
  6. eˣ > 0, so x = −3 or x = 1
  7. y(−3) = 6e⁻³ ≈ 0.299
  8. y(1) = −2e ≈ −5.44
  9. (−3, 6e⁻³) and (1, −2e)
Your turn

Question 5 · Merit

  • y = 2sin x + cos 2x, 0 ≤ x ≤ π
  • Find and classify
  • the stationary points.
  • There are three.
Worked example

Q5: dy/dx = 0

  • Double angle, factorise.
  • Never divide by cos x.
  1. dy/dx = 2cos x − 2sin 2xchain: × 2
  2. = 2cos x − 4sin x cos xsin 2x = 2sin x cos x
  3. = 2cos x(1 − 2sin x)factorise
  4. 2cos x(1 − 2sin x) = 0stationary
  5. ÷ cos xloses cos x = 0
  6. cos x = 0: x = π/2
  7. sin x = 1/2: x = π/6, 5π/6
Worked example

Q5: y-values and nature

  • y from the original.
  • Sign of d²y/dx².
  1. x = π/6, π/2, 5π/6
  2. y(π/6) = 2(1/2) + 1/2 = 3/2
  3. y(π/2) = 2(1) + (−1) = 1
  4. y(5π/6) = 2(1/2) + 1/2 = 3/2
  5. d²y/dx² = −2sin x − 4cos 2x
  6. π/6, 5π/6: −1 − 2 = −3 < 0
  7. π/2: −2 + 4 = 2 > 0concave up
  8. max (π/6, 3/2), (5π/6, 3/2)
  9. min (π/2, 1)
Your turn

Question 6 · Excellence

  • y = x³ − 3kx² + 4, k > 0
  • The minimum point lies
  • on the x-axis. Find k.
Worked example

Q6: which point is the minimum?

  • k > 0 fixes the signs.
  • d²y/dx² > 0: minimum.
  1. dy/dx = 3x² − 6kx
  2. 3x(x − 2k) = 0stationary
  3. x = 0 or x = 2k
  4. d²y/dx² = 6x − 6k
  5. x = 0: −6k < 0maximum, as k > 0
  6. x = 2k: 12k − 6k = 6k > 0the minimum
Worked example

Q6: the minimum on the x-axis

  • On the x-axis: y = 0.
  • One real cube root.
  1. minimum at x = 2k
  2. y(2k) = 8k³ − 12k³ + 4
  3. = 4 − 4k³
  4. 4 − 4k³ = 0on the x-axis
  5. k³ = 1, so k = 1one real cube root
  6. k = 1: y(2) = 8 − 12 + 4 = 0
Your turn

Question 7 · Excellence, in context

  • Medicine in the blood, t hours after
  • a dose: C = 20te^−0.5t mg/L
  • Find the greatest concentration.
  • When is C falling fastest?
Worked example

Q7: the greatest concentration

  • Product rule.
  • Justify: + then −.
  1. u = 20t, v = e^−0.5tproduct rule
  2. dC/dt = 20e^−0.5t − 10te^−0.5t
  3. = 10e^−0.5t(2 − t)factorise
  4. 2 − t = 0, so t = 2e^−0.5t is never 0
  5. C(2) = 40e⁻¹ = 14.715
  6. sign of 2 − t: + then −a maximum
  7. 14.7 mg/L, 2 hours after dose
Worked example

Q7: falling fastest

  • Fastest: an inflection.
  • d²C/dt² changes sign.
  1. dC/dt = 10e^−0.5t(2 − t)
  2. d²C/dt² = −5e^−0.5t(2 − t)product rule
  3. − 10e^−0.5t
  4. = 5e^−0.5t(t − 4)collect
  5. t − 4 = 0, so t = 4d²C/dt² = 0
  6. sign of t − 4: − then +an inflection
  7. dC/dt = −20e⁻² ≈ −2.71at t = 4
  8. fastest at 4 h: 2.71 mg/L per h
Summary

The method that solved all seven

  • dy/dx = 0, every root, y from the curve, then classify.
StepHow
Stationarywrite dy/dx = 0, then solve
Fractionswhole derivative = 0 first
Factorisetake out cos x, never ÷ by it
Both valuesy from the curve, not dy/dx
Classifyd²y/dx² < 0 max, > 0 min
Inflectiond²y/dx² = 0, sign changes

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