# The second derivative: rates of change, concavity and inflection · Grade 11 · 5 units

> Explanation, worked examples and free practice following the Israeli curriculum.

In The second derivative: rates of change, concavity and inflection (Grade 11 · 5 units) you practise 7 skills: Concavity: where a function is concave up or down · Working out the second derivative · Points of inflection · The type of turning point from the second derivative · A function and its two derivatives on graphs · Velocity and acceleration · The rate of change of a rate in problems.

The second derivative is the derivative of the derivative: it tells you how fast the rate of change is itself changing. For a moving body the derivative of the position is the velocity, and the second derivative is the acceleration. The speed increases when the velocity and the acceleration have the same sign.

Where the second derivative is positive the derivative is rising, and the graph is concave up: the tangent lies below the graph and the chord above it. Where it is negative the graph is concave down. A point of inflection is where the concavity changes, so the second derivative changes sign there. A polynomial of degree two never has a point of inflection, one of degree three always has exactly one, and one of degree four may or may not have any. The square root of x has a second derivative too; it is negative, so its graph is concave down.

There are four ways to decide whether a point where the derivative is 0 is a maximum or a minimum: the sign chart of the derivative, values of the function on both sides, the graph, and the second derivative. If the second derivative is positive at the point it is a minimum, and if it is negative it is a maximum. If it is 0 the test is inconclusive, so go back to the sign chart of the derivative.

In problems the second derivative answers the question of when something changes fastest. The length of the day increases fastest at the point of inflection of its graph, between the shortest day and the longest. After waste is poured into a lake the oxygen level falls and then rises again, and it rises fastest where the second derivative goes from positive to negative. Points of inflection come in four kinds: with a horizontal tangent or a sloped one, on a rising graph or a falling one.

**Tip:** For a point of inflection it is not enough that the second derivative is 0: check on a sign chart that it also changes sign.

## What you learn

### Concavity: where a function is concave up or down

- Example: Here is the function f(x) = −3x² + 6x − 7 . What is true of its concavity?
- Options: Concave up for every value of x · Concave up only for x > 0 · Concave down for every value of x
- Answer: Concave down for every value of x
- Solution: Differentiate: f′(x) = −6x + 6 , so f″(x) = −6 , negative for every value of x . So the function is concave down for every value of x .

### Working out the second derivative

- Example: Here is the function f(x) = −3x³ + 5x² − 7x . Find its second derivative.
- Answer: −18x + 10
- Solution: Differentiate: f′(x) = −9x² + 10x − 7 , so f″(x) = −18x + 10 .

### Points of inflection

- Example: Here is the function f(x) = x³ − 12x² − 5x . At which value of x does it have a point of inflection?
- Answer: 4
- Solution: Differentiate: f′(x) = 3x² − 24x − 5 and f″(x) = 6x − 24 . The second derivative f″ is 0 at x = 4 : before it, it is negative, and after it, positive. So there is a point of inflection at x = 4 .

### The type of turning point from the second derivative

- Example: A function f has f′(−4) = 0 and f″(−4) = 6 . What kind of point does it have at x = −4 ?
- Options: A maximum point · There is no way to tell · A minimum point
- Answer: A minimum point
- Solution: The derivative is 0 and the second derivative is positive, so it is a minimum point.

### A function and its two derivatives on graphs

- Example: The first picture shows the graph of the function y = f(x) , a polynomial of degree three. Which of the graphs after it is the graph of the second derivative f″ ? (The graph of the function, a polynomial of degree three) Graph A: (Graph A, a rising line) Graph B: (Graph B, a falling line) Graph C: (Graph C, a parabola opening upwards)
- Options: Graph A · Graph B · Graph C
- Answer: Graph A
- Solution: The graph is concave down and then up, and the point of inflection is at x = 1 . So f″ is a rising line that crosses the horizontal axis at x = 1 : that is Graph A. The parabola is the graph of f′ .

### Velocity and acceleration

- Example: A body moves along a straight line, and its position relative to the start after t seconds is s(t) = −2t³ + 5t² − 7t + 4 metres. What is its velocity when t = 5 ?
- Answer: −107
- Solution: The velocity: v(t) = −6t² + 10t − 7 . Put in t = 5 : −6 × 5² + 10 × 5 − 7 = −107 .

### The rate of change of a rate in problems

- Example: The graph of the length of the day against the date rises from the shortest day of the year to the longest, then falls again. Where on the graph does the day length increase fastest?
- Options: At the maximum point of the graph · At the minimum point of the graph · At a point of inflection of the graph
- Answer: At a point of inflection of the graph
- Solution: At the turning points of the graph the length of the day hardly changes: the derivative is 0 there. On the way from the minimum to the maximum the change is fastest where the graph changes its concavity: at a point of inflection.

The exercises are made by rules and the answers are worked out in code. The topic follows the Israeli Ministry of Education’s high school maths curriculum at 5 units (grades 10 to 12).

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