Stationary points, concavity and points of inflection

y = x³ − 6x² + 9x + 1

• A peak, then a dip.
• Flat tangents at both.
• An inflection between them.

Finding the stationary points

• Write dy/dx = 0.
• Factorise, then solve.
• Give both coordinates.

1. y = x³ − 6x² + 9x + 1
2. dy/dx = 3x² − 12x + 9differentiate
3. 3x² − 12x + 9 = 0stationary: dy/dx = 0
4. 3(x² − 4x + 3) = 0
5. 3(x − 1)(x − 3) = 0factorise
6. x = 1 or x = 3
7. x = 1: y = 1 − 6 + 9 + 1 = 5
8. x = 3: y = 27 − 54 + 27 + 1 = 1
9. (1, 5) and (3, 1)

Classifying with d²y/dx²

• Negative: concave down.
• Positive: concave up.

1. d²y/dx² = 6x − 12differentiate again
2. x = 1: 6 − 12 = −6 < 0concave down
3. (1, 5) is a maximum
4. x = 3: 18 − 12 = 6 > 0concave up
5. (3, 1) is a minimum

Increasing and decreasing

• Increasing: dy/dx > 0.
• Decreasing: dy/dx < 0.
• Test each interval.

1. dy/dx = 3(x − 1)(x − 3)
2. x = 0: 3(−1)(−3) = 9 > 0x < 1
3. x = 2: 3(1)(−1) = −3 < 01 < x < 3
4. x = 4: 3(3)(1) = 9 > 0x > 3
5. decreasing for 1 < x < 3
6. increasing for x < 1, x > 3

Concavity and inflection

• Zero, then a sign change.
• A zero alone is not enough.

1. 6x − 12 = 0, so x = 2d²y/dx² = 0
2. x < 2: 6x − 12 < 0concave down
3. x > 2: 6x − 12 > 0concave up
4. sign changes at x = 2
5. y = 8 − 24 + 18 + 1 = 3
6. inflection at (2, 3)

y = x/(x² + 4)

• Quotient rule first.
• Whole fraction = 0 first.

1. y = x/(x² + 4)
2. u = x, v = x² + 4quotient rule
3. u′ = 1, v′ = 2x
4. dy/dx = (vu′ − uv′)/v²
5. = (x² + 4 − 2x²)/(x² + 4)²
6. = (4 − x²)/(x² + 4)²simplify the top
7. (4 − x²)/(x² + 4)² = 0write this first

Then clear the denominator

• Fraction = 0: top = 0.
• Give both coordinates.

1. (4 − x²)/(x² + 4)² = 0
2. 4 − x² = 0top = 0
3. x² = 4, so x = ±2
4. x = 2: y = 2/(4 + 4) = 1/4
5. x = −2: y = −2/8 = −1/4
6. (2, 1/4) and (−2, −1/4)

y = xe^−x

• Product rule.
• e^−x is never zero.

1. u = x, v = e^−xproduct rule
2. u′ = 1, v′ = −e^−xchain rule
3. dy/dx = e^−x − xe^−xu′v + uv′
4. = (1 − x)e^−xfactorise
5. (1 − x)e^−x = 0stationary
6. e^−x > 0, so x = 1e^−x is never 0
7. y = 1 × e^−1 = 1/e
8. stationary at (1, 1/e)

The second derivative

• Product rule again.
• Check the sign change.

1. dy/dx = (1 − x)e^−x
2. d²y/dx² = −e^−x − (1 − x)e^−x
3. = (x − 2)e^−xcollect
4. x = 1: −e^−1 < 0a maximum
5. (x − 2)e^−x = 0 at x = 2
6. x − 2 goes from − to +e^−x > 0
7. y = 2e^−2 = 2/e²
8. inflection at (2, 2/e²)

y = x + 2cos x, 0 ≤ x ≤ 2π

• sin x = 1/2 in the range.
• Find both y-values.

1. dy/dx = 1 − 2sin x
2. 1 − 2sin x = 0stationary
3. sin x = 1/2
4. x = π/6 or x = 5π/60 ≤ x ≤ 2π
5. y(π/6) = π/6 + 2(√3/2)
6. = π/6 + √3 ≈ 2.26
7. y(5π/6) = 5π/6 + 2(−√3/2)
8. = 5π/6 − √3 ≈ 0.886

Classifying with −2cos x

• Negative: a maximum.
• Positive: a minimum.

1. x = π/6 or x = 5π/6
2. d²y/dx² = −2cos x
3. π/6: −2(√3/2) = −√3 < 0
4. maximum at (π/6, π/6 + √3)
5. 5π/6: −2(−√3/2) = √3 > 0
6. minimum at (5π/6, 5π/6 − √3)

The answers: questions 1 and 2

1. 1. dy/dx = 3x² − 6x − 9 = 0
2. 3(x − 3)(x + 1) = 0factorise
3. x = −1: y = −1 − 3 + 9 + 5 = 10
4. x = 3: y = 27 − 27 − 27 + 5 = −22
5. (−1, 10) and (3, −22)
6. 2. decreasing for −1 < x < 3dy/dx < 0 between roots
7. d²y/dx² = 6x − 6 = 0, x = 1changes sign
8. y = 1 − 3 − 9 + 5 = −6
9. inflection at (1, −6)

The answers: question 3

1. 3. top: (x² + 3)(1) − (x + 1)(2x)quotient rule
2. dy/dx = (3 − 2x − x²)/(x² + 3)²simplify
3. (3 − 2x − x²)/(x² + 3)² = 0write this first
4. x² + 2x − 3 = 0top = 0
5. (x − 1)(x + 3) = 0factorise
6. x = 1: y = 2/4 = 1/2
7. x = −3: y = −2/12 = −1/6
8. (1, 1/2) and (−3, −1/6)

The answers: question 4

1. 4. dy/dx = ln x + x(1/x)product rule
2. = ln x + 1
3. ln x + 1 = 0, ln x = −1stationary
4. x = e^−1 = 1/e
5. y = (1/e) ln(1/e) = −1/eln(1/e) = −1
6. d²y/dx² = 1/x = e > 0concave up
7. minimum at (1/e, −1/e)