First derivative test
At a critical point c, if f′ changes from + to −, c is a local maximum; from − to +, a local minimum; no change, neither.
The idea from the graph
Walk along a graph from left to right. Before a hilltop you climb, so the slope f′(x) is positive. After it you go down, so f′(x) is negative. Before a valley you go down (f′ < 0) and after it you climb (f′ > 0). So the change of sign of f′ tells you the kind of point. If the slope keeps the same sign on both sides, as for y = x³ at x = 0, the curve only flattens for a moment and goes on. That point is neither a maximum nor a minimum.
The test
| Sign of f′ just left of c | Sign of f′ just right of c | Conclusion |
|---|---|---|
| + | − | Local maximum at c; f(c) is the local maximum value |
| − | + | Local minimum at c; f(c) is the local minimum value |
| + | + | Neither (point of inflection) |
| − | − | Neither (point of inflection) |
How to apply it
- 1Find f′(x) and factorise it.
- 2Solve f′(x) = 0, and note where f′ does not exist, to get the critical points.
- 3★For each critical point c, find the sign of f′ a little to its left and a little to its right.
- 4Use the table: + to − is a local maximum, − to + is a local minimum, no change is neither.
- 5Find the local maximum or minimum value by putting c into f(x), not into f′(x).
★ key step
Example 1: a cubic with a peak and a valley
Find all points of local maxima and local minima of f(x) = x³ − 12x + 5, and the local maximum and minimum values.
- f′(x) = 3x² − 12 = 3(x − 2)(x + 2)
- f′(x) = 0 gives x = −2 and x = 2.
- Just left of −2, f′ = 3(−)(−) > 0; just right, f′ = 3(−)(+) < 0. The sign changes from + to −.
- So x = −2 is a point of local maxima, and f(−2) = −8 + 24 + 5 = 21.
- Just left of 2, f′ = 3(−)(+) < 0; just right, f′ = 3(+)(+) > 0. The sign changes from − to +.
- So x = 2 is a point of local minima, and f(2) = 8 − 24 + 5 = −11.
Local maximum value 21 at x = −2; local minimum value −11 at x = 2.
Example 2: no change of sign
Show that f(x) = x³ − 3x² + 3x + 7 has no local maximum or local minimum.
- f′(x) = 3x² − 6x + 3 = 3(x − 1)²
- f′(x) = 0 gives only x = 1.
- (x − 1)² > 0 on both sides of 1, so f′(x) > 0 just left and just right of 1.
- The sign of f′ does not change at x = 1.
x = 1 is neither a point of local maxima nor of local minima; it is a point of inflection. So f has no local extreme values.
When to use it, and how to draw the sign chart
- Use it when f is not differentiable at c. For f(x) = |x|, f′ = −1 for x < 0 and +1 for x > 0, so x = 0 is a local minimum.
- Use it when the second derivative test fails (f″(c) = 0), as for f(x) = x⁴ at x = 0.
- To draw the sign chart: draw a number line and mark the critical points in order.
- Write + or − above each gap, from a test point in that gap.
- Under each gap draw an arrow going up for + and down for −. An up arrow then a down arrow shows a peak.
During a 100 m race at a school sports day, a runner's heart rate climbs, peaks and then falls as she walks back. The peak is where its rate of change switches from + to −.
1 marks: “f′(x) changes from positive to negative as x increases through 4. What kind of point is x = 4?” A point of local maxima, by the first derivative test.
2 marks: “Using the first derivative test, find the local minimum value of f(x) = x² − 6x + 10.” f′(x) = 2x − 6 = 0 gives x = 3. f′ < 0 for x < 3 and f′ > 0 for x > 3, so x = 3 gives a local minimum. Value f(3) = 9 − 18 + 10 = 1.
3 marks: “Find the points of local maxima, local minima and inflection of f(x) = 3x⁴ − 4x³.” f′(x) = 12x³ − 12x² = 12x²(x − 1). Critical points: 0 and 1. At 0, f′ < 0 on both sides: a point of inflection. At 1, f′ changes from − to +: local minimum, f(1) = −1. There is no local maximum.
Up then down, a crown; down then up, a cup.