Using the Second Derivative Test to Determine Extrema: Study Guide (Calculus AB 5.7)

Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.

Notation: f′ is the first derivative and f″ is the second derivative. A critical point of f is a value x = c in the domain where f′(c) = 0 or f′(c) does not exist.

The idea: a flat point on a cup or a cap

In Topic 5.4 you classified a critical point by checking the sign of f′ on each side of it. That needs a sign chart. Topic 5.6 gave you a second tool: the sign of f″ tells you the concavity of the graph.

Put the two ideas together. Suppose f′(c) = 0, so the tangent line at x = c is horizontal.

• If the graph is concave up at c, it bends upwards like a cup. Near c the curve lies above its horizontal tangent, so f(c) is lower than the nearby values. That is a relative minimum.
• If the graph is concave down at c, it bends downwards like a cap. Near c the curve lies below its horizontal tangent, so f(c) is higher than the nearby values. That is a relative maximum.

The second derivative test

Let c be a critical point of f with f′(c) = 0, and suppose f″(c) exists.

f′(c)	f″(c)	Conclusion
0	positive	relative minimum at x = c
0	negative	relative maximum at x = c
0	0	no conclusion: use the first derivative test

A memory aid: positive f″ means “holds water”, a cup, so a minimum. Many students instinctively link “positive” with “top”. Resist that.

Why the test works

Suppose f′(c) = 0 and f″(c) > 0. Since f″ is the derivative of f′, a positive f″(c) means f′ is increasing as x passes through c. An increasing function that equals 0 at c must be negative just to the left of c and positive just to the right. So f′ changes from negative to positive at c. By the first derivative test (Topic 5.4), f has a relative minimum at c. The case f″(c) < 0 works the same way: f′ is decreasing through 0, so it changes from positive to negative, giving a relative maximum.

So the second derivative test is a shortcut to the first derivative test. It reads the sign change of f′ from one number, f″(c), instead of a sign chart.

When the test says nothing

There are two situations where you must go back to the first derivative test.

1. f″(c) = 0. The test is inconclusive. Anything can happen. Look at three functions, each with f′(0) = 0 and f″(0) = 0:

• y = x⁴ has a relative minimum at x = 0;
• y = −x⁴ has a relative maximum at x = 0;
• y = x³ has no extremum at x = 0.

2. f′(c) does not exist. The test needs f′(c) = 0, so it does not apply at corners, cusps or vertical tangents. For example, y = |x| and y = x^(2/3) both have a relative minimum at x = 0, but neither has a derivative there, so there is no f″(0) to test. Use the first derivative test.

Writing the justification

A conclusion from this test needs both facts, and it should name the function and the point:

f has a relative maximum at x = c because f′(c) = 0 and f″(c) < 0.

Weak versions lose credit. “f″ is negative” (where? and is f′(c) zero?) is incomplete. “The second derivative changes sign” describes a point of inflection, not an extremum. “The derivative is negative” is ambiguous: say f′ or f″.

Worked example 1: three critical points on an open interval

Question. Let f(x) = 2 sin x + cos 2x for 0 < x < π. Find the critical points of f and use the second derivative test to classify each one.

1. Differentiate. f′(x) = 2 cos x − 2 sin 2x. Use sin 2x = 2 sin x cos x: f′(x) = 2 cos x − 4 sin x cos x = 2 cos x (1 − 2 sin x)
2. Solve f′(x) = 0 on 0 < x < π.
   • cos x = 0 gives x = π/2.
   • 1 − 2 sin x = 0 gives sin x = 1/2, so x = π/6 or x = 5π/6. f′ exists for every x, so these three are the only critical points.
3. Second derivative. f″(x) = −2 sin x − 4 cos 2x.
4. Test each point.
   • x = π/6: f″ = −2(1/2) − 4 cos(π/3) = −1 − 4(1/2) = −3 < 0.
   • x = π/2: f″ = −2(1) − 4 cos π = −2 + 4 = 2 > 0.
   • x = 5π/6: f″ = −2(1/2) − 4 cos(5π/3) = −1 − 2 = −3 < 0.
5. Values. f(π/6) = 2(1/2) + cos(π/3) = 3/2. f(π/2) = 2 + cos π = 1. f(5π/6) = 1 + cos(5π/3) = 3/2.

Answer. f has relative maxima at x = π/6 and x = 5π/6 (value 3/2 each) because f′ = 0 and f″ < 0 there. f has a relative minimum at x = π/2 (value 1) because f′(π/2) = 0 and f″(π/2) > 0.

Check. A maximum, then a minimum, then a maximum: the extrema alternate, as they must for a smooth curve.

Worked example 2: when f″(c) = 0

Question. Let g(x) = x⁴ − 8x³ + 18x² − 5. Classify each critical point of g.

1. Differentiate and factor. g′(x) = 4x³ − 24x² + 36x = 4x(x² − 6x + 9) = 4x(x − 3)². Critical points: x = 0 and x = 3.
2. Second derivative. g″(x) = 12x² − 48x + 36 = 12(x − 1)(x − 3).
3. x = 0. g″(0) = 36 > 0 and g′(0) = 0, so g has a relative minimum at x = 0, with g(0) = −5.
4. x = 3. g″(3) = 0. The test is inconclusive. Switch to the first derivative test.
5. First derivative test at x = 3. In g′(x) = 4x(x − 3)², the factor (x − 3)² is positive on both sides of 3, and 4x is positive near 3. So g′ > 0 just left of 3 and just right of 3. g is increasing on both sides, so there is no extremum at x = 3.

Answer. Relative minimum at x = 0 (value −5); no relative extremum at x = 3.

Interpretation. At x = 3 the graph is momentarily flat, at height g(3) = 22, but keeps rising. Because g″ = 12(x − 1)(x − 3) changes sign at x = 3, that point is a point of inflection with a horizontal tangent, like the centre of y = x³ in Figure 2.

From relative to absolute: the one-critical-point rule

The second derivative test finds relative extrema. Sometimes you need the absolute (global) maximum or minimum on an interval that has no endpoints, or only one, so the candidates test from Topic 5.5 cannot be used. This rule helps:

One-critical-point rule. Suppose f is continuous on an interval and has only one critical point c in that interval. If f has a relative minimum (maximum) at c, then f(c) is the absolute minimum (maximum) of f on the whole interval.

Why it is true: say f has a relative minimum at c but some other point d in the interval has f(d) < f(c). Travelling from c to d, the graph first rises (c is a relative minimum) and later must fall to below f(c). So it turns around somewhere in between, which creates a relative maximum, and that is another critical point. That contradicts “only one critical point”.

Check the conditions every time: continuous, one interval, exactly one critical point inside it.

Worked example 3: lowest energy use on an open interval

Question. A small delivery robot (a fictional model) uses energy at a rate of E(v) = v²/20 + 100/v watt-hours per kilometre when it travels at v kilometres per hour, for v > 0. Find the speed that minimises E, and justify that it gives the absolute minimum.

1. Differentiate. E′(v) = v/10 − 100/v².
2. Critical points. E′(v) = 0 gives v/10 = 100/v², so v³ = 1000 and v = 10. E′ exists for all v > 0, so this is the only critical point.
3. Second derivative test. E″(v) = 1/10 + 200/v³. E″(10) = 0.1 + 0.2 = 0.3 > 0. With E′(10) = 0, E has a relative minimum at v = 10.
4. Absolute. E is continuous on the interval v > 0, and v = 10 is its only critical point there. By the one-critical-point rule, the relative minimum is the absolute minimum on v > 0.
5. Value. E(10) = 100/20 + 100/10 = 5 + 10 = 15 watt-hours per kilometre.

Answer. The robot uses least energy per kilometre at 10 km/h, where E = 15 Wh/km.

Check. E(5) = 21.25 and E(20) = 25. Both are larger than 15.

Why not the candidates test? The interval v > 0 has no endpoints, so there is no finite list of candidates to compare.

Common misconceptions

• “f″(c) > 0 means a maximum.” It is the other way round. Positive f″ means concave up, a cup, so a minimum.
• Using the test where f′(c) ≠ 0. A negative f″ at a point where the graph is still rising tells you about concavity only. The test starts with f′(c) = 0.
• “f″(c) = 0, so there is no extremum.” Wrong: y = x⁴ has f″(0) = 0 and a minimum. The test simply gives no answer.
• “f″(c) = 0, so there is a point of inflection.” Also wrong: y = x⁴ is concave up on both sides of 0. A point of inflection needs a change in concavity.
• Applying the test at a cusp or corner. If f′(c) does not exist, f″(c) does not exist either. Use the first derivative test.
• Calling a relative extremum absolute without checking. The one-critical-point rule needs exactly one critical point on the interval. With two or more, compare values or use the candidates test.
• Mixing up location and value. “The maximum is at x = π/6” gives the location; “the maximum value is 3/2” gives the height. Answer what the question asks.

Where this leads

The second derivative test is a fast way to classify critical points. It returns in Topic 5.11, where optimisation problems often take place on intervals with no endpoints and the one-critical-point rule justifies an absolute answer. In Topic 5.8, Sketching Graphs of Functions and Their Derivatives, you combine the first and second derivatives to draw the whole graph. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.