NCEA Level 2Turning pointsAS 91262

Turning points: finding and proving max or min

What this lesson covers

  • At a turning point the gradient is zero.
  • Whether it is a maximum or a minimum is proved by the second derivative or the gradients either side.
Key idea

Where a curve turns

  • At a turning point the curve is flat.
  • Flat means the gradient is zero.
  • So solve f′(x) = 0.
Worked example

f(x) = x³ − 6x² + 9x + 1

  • It rises, turns, falls, then turns again.
  • A maximum at x = 1.
  • A minimum at x = 3.
f(x) = x³ − 6x² + 9x + 1
Worked example

Finding them

  • Set f′(x) = 0.
  • Heights from f, not f′.
  1. f′(x) = 3x² − 12x + 9differentiate
  2. 3x² − 12x + 9 = 0gradient zero
  3. 3(x² − 4x + 3) = 0take out 3
  4. 3(x − 1)(x − 3) = 0factorise
  5. x = 1 or x = 3
  6. f(1) = 1 − 6 + 9 + 1 = 5heights from f
  7. f(3) = 27 − 54 + 27 + 1 = 1
  8. (1, 5) and (3, 1)
Worked example

Method 1: the second derivative

  • f″ < 0: a maximum.
  • f″ > 0: a minimum.
  1. f″(x) = 6x − 12differentiate again
  2. f″(1) = 6 − 12 = −6 < 0a maximum
  3. f″(3) = 18 − 12 = 6 > 0a minimum
  4. max (1, 5), min (3, 1)
Worked example

Method 2: the gradient either side

  • Before x = 1 the gradient is positive; after, negative.
  • Up then down: a maximum.
  • Down then up at x = 3: a minimum.
Method 2: the gradient either side
Worked example

The gradient test, in numbers

  • Test either side.
  • The signs give the nature.
  1. f′(x) = 3x² − 12x + 9
  2. f′(0) = 0 − 0 + 9 = 9 > 0rising
  3. f′(2) = 12 − 24 + 9 = −3 < 0falling
  4. f′(4) = 48 − 48 + 9 = 9 > 0rising
  5. x = 1: + then −, a maximum
  6. x = 3: − then +, a minimum
Worked example

A quadratic: y = −2x² + 8x − 3

  • One turning point.
  • y″ < 0: a maximum.
  1. dy/dx = −4x + 8differentiate
  2. −4x + 8 = 0, so x = 2
  3. y = −2(2)² + 8(2) − 3the height
  4. = −8 + 16 − 3 = 5
  5. y″ = −4 < 0a maximum
  6. maximum at (2, 5)
Key idea

Always justify the nature

  • "It is a maximum" needs a reason.
  • A second derivative, a gradient test or a sketch.
  • Excellence answers do this unasked.
Worked example

Another cubic: y = x³ − 3x²

  • Factorise y′.
  • Classify each with y″.
  1. y′ = 3x² − 6x = 3x(x − 2)factorise
  2. 3x(x − 2) = 0: x = 0 or 2
  3. x = 0: y = 0the height
  4. x = 2: y = 8 − 12 = −4
  5. y″ = 6x − 6
  6. x = 0: y″ = −6 < 0max (0, 0)
  7. x = 2: y″ = 6 > 0min (2, −4)
Key idea

When there are no turning points

  • y = x³ + 3x has y′ = 3x² + 3.
  • 3x² + 3 is never zero: it is always at least 3.
  • So the curve never turns: it always rises.
Key idea

Where the marks are

  • Achieved: find where the gradient is zero.
  • Merit: the coordinates of each point.
  • Excellence: the nature, justified.
Summary

What to take away

  • Find, then prove, every turning point.
Findset f′(x) = 0 and solve
Heightsubstitute into f, not f′
f″ < 0maximum
f″ > 0minimum
Either side+ then − is max, − then + is min
Alwaysstate the nature, with a reason
Your turn

Your turn: four questions

  • Pause here and answer each one.
  • Give coordinates and the nature.
  • Justify with a reason every time.
Question
1Find the turning point of y = x² − 10x + 4.
2Find both turning points of y = x³ − 12x.
3Use the second derivative to classify the one at x = 2.
4Find and classify the turning point of y = −x² + 6x.
Show answer

1. 2x − 10 = 0, x = 5, y = 25 − 50 + 4 = −21: a minimum at (5, −21), since y″ = 2 > 0. 2. 3x² − 12 = 0, x = ±2: (−2, 16) and (2, −16). 3. y″ = 6x = 12 at x = 2, positive, so (2, −16) is a minimum (and (−2, 16) a maximum). 4. −2x + 6 = 0, x = 3, y = 9: a maximum, since y″ = −2 < 0.

Worked example

The answers: questions 1 and 2

  1. 1. dy/dx = 2x − 10 = 0, x = 5
  2. y = 25 − 50 + 4 = −21
  3. y″ = 2 > 0: min (5, −21)
  4. 2. dy/dx = 3x² − 12 = 0
  5. 3x² = 12, x² = 4, x = ±2
  6. x = −2: y = −8 + 24 = 16
  7. x = 2: y = 8 − 24 = −16
  8. (−2, 16) and (2, −16)
Worked example

The answers: questions 3 and 4

  1. 3. y″ = 6xdifferentiate 3x² − 12
  2. x = 2: y″ = 12 > 0min (2, −16)
  3. x = −2: y″ = −12 < 0max (−2, 16)
  4. 4. dy/dx = −2x + 6 = 0, x = 3
  5. y = −9 + 18 = 9
  6. y″ = −2 < 0: max (3, 9)

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