First derivative test
Walking over a hill, you climb up (positive slope) and then descend (negative slope). The top is where the slope changes from plus to minus.
At a critical point c (f′(c) = 0): if f′ changes from positive to negative, c is a local maximum; if f′ changes from negative to positive, c is a local minimum; if f′ does not change sign, c is neither (a point of inflexion).
Formula: f′: + to − ⇒ local maximum; f′: − to + ⇒ local minimum
In the figure: The graph of f(x) = x³ − 3x has a local maximum 2 at x = −1 and a local minimum −2 at x = 1.
Rule 1: Apply the test — Find critical points, then check the sign change of f′.
- Easy: f(x) = x² − 4x + 5: f′ = 2x − 4 changes from − to + at x = 2, so the local minimum value is 1.
- Medium: f(x) = x³ − 3x: f′ = 3(x − 1)(x + 1): local maximum f(−1) = 2 and local minimum f(1) = −2.
- Exam-type: f(x) = x³: f′ = 3x² ≥ 0 does not change sign at 0, so there is no extreme value.
Rule 2: Compute the extreme values — Substitute the critical point in f.
- Easy: f(x) = −x² + 4x + 1: f′ = −2x + 4; the sign changes from + to − at 2, so the local maximum value is 5.
- Medium: f(x) = x³ − 6x² + 9x: maximum f(1) = 4 and minimum f(3) = 0.
- Exam-type: f(x) = x + 1/x (x > 0): minimum at x = 1, with value 2.
A local maximum or minimum is a height in the neighbourhood, not necessarily the biggest or smallest overall.
Treating every point where f′ = 0 as a maximum or minimum.
Plus-then-minus is a hill top; minus-then-plus is a valley bottom.
Practice questions
- 1Find the critical point of f(x) = x² − 10x.
- 2Find the local minimum value of f(x) = x² − 10x.
- 3Find the local maximum and minimum values of f(x) = x³ − 12x.
- 4Does f(x) = x³ + 1 have a local maximum or minimum?
- 5Find the local extreme value of f(x) = −x² + 4x + 1.
- 6Find the local extreme values of f(x) = x³ − 6x² + 9x.
- 7Find the minimum value of f(x) = x + 1/x for x > 0.
- 8Find all local extreme values of f(x) = x⁴ − 8x².
Show answers (try them yourself first)
- 1f′ = 2x − 10 = 0, so x = 5
- 2f(5) = 25 − 50 = −25
- 3f′ = 3(x − 2)(x + 2); local maximum f(−2) = 16 and local minimum f(2) = −16
- 4No: f′ = 3x² does not change sign at 0
- 5Local maximum 5 at x = 2
- 6Local maximum f(1) = 4 and local minimum f(3) = 0
- 7f′ = 1 − 1/x² = 0 at x = 1; the minimum value is 2
- 8f′ = 4x(x − 2)(x + 2): local maximum 0 at x = 0; local minimum −16 at x = 2 and x = −2
Second derivative test and simple optimisation
Shaped like a cup, the graph holds water at its lowest point; shaped like a cap, it spills. The sign of the second derivative tells whether you are in a cup or a cap.
If f′(c) = 0 and f″(c) < 0, then c is a local maximum; if f″(c) > 0, then c is a local minimum; if f″(c) = 0 the test fails and the first derivative test is used. In a word problem, write the quantity as a function of one variable and find its maximum or minimum.
Formula: f′(c) = 0, f″(c) < 0 ⇒ maximum; f″(c) > 0 ⇒ minimum
Rule 1: Second derivative test — Find f′, f″, then check f″ at each critical point.
- Easy: f(x) = x² − 6x: f″ = 2 > 0, so the minimum is f(3) = −9.
- Medium: f(x) = 2x³ − 9x² + 12x: f″ = 12x − 18, so the maximum is f(1) = 5 and the minimum is f(2) = 4.
- Exam-type: f(x) = x⁴: f″(0) = 0, so the test fails; f′ = 4x³ changes from − to +, so 0 is a local minimum.
Rule 2: Optimisation — Express as one variable, then find the maximum or minimum.
- Easy: Split 12 into two parts with maximum product: x(12 − x) is largest at x = 6, product 36.
- Medium: A rectangle with perimeter 40 m has maximum area 100 m² when it is a 10 m square.
- Exam-type: Two positive numbers add to 20: the sum of their squares is least at 10 and 10, equal to 200.
If f″(c) = 0 do not conclude anything; use the first derivative test.
Calling f″(c) = 0 a point of inflexion without checking the sign change.
Positive f″ is a smile (minimum); negative f″ is a frown (maximum).
Practice questions
- 1Find the extreme value of f(x) = x² − 6x using f″.
- 2Find the extreme value of f(x) = 3 − x².
- 3Find the local extreme values of f(x) = x³ − 3x².
- 4Find the local extreme values of f(x) = 2x³ − 3x² − 12x.
- 5Split 20 into two parts whose product is largest. State the product.
- 6A rectangle has perimeter 36 m. Find its maximum area.
- 7Two positive numbers have sum 16. Find the least value of the sum of their squares.
- 8A rectangle has area 36 m². Find its least perimeter.
Show answers (try them yourself first)
- 1f″ = 2 > 0: minimum value f(3) = −9
- 2f′ = −2x = 0, f″ = −2 < 0: maximum value f(0) = 3
- 3f′ = 3x(x − 2), f″ = 6x − 6: maximum f(0) = 0, minimum f(2) = −4
- 4f′ = 6(x − 2)(x + 1), f″ = 12x − 6: minimum f(2) = −20, maximum f(−1) = 7
- 5x(20 − x) is largest at x = 10: parts 10 and 10, product 100
- 6Sides x and 18 − x: maximum area at x = 9, area 81 m²
- 7x² + (16 − x)² is least at x = 8: 64 + 64 = 128
- 8Sides x and 36/x: the perimeter is least at x = 6: 4 × 6 = 24 m