1
The rate of the rate
If p is a position, p′ is a velocity and p″ is an acceleration: how fast the velocity itself is changing. In general p″ is the derivative of p′, found by applying the power rule a second time: t³ gives 3t² and then 6t.
Lesson · Rates of Change and the Derivative · Second derivatives and the exponential
The Second Derivative
The derivative is itself a polynomial, so it has a derivative: the second derivative p″, the rate at which the slope is changing. Where p″ is positive the slope is rising and the graph bends up; where it is negative the graph bends down. A place where p″ changes sign is an inflection point, and it is a different place from a turn.
p″ = (p′)′, the rate of the rate
p″ > 0: the slope rises, graph bends up
p″ < 0: the slope falls, graph bends down
inflection: p″ changes sign
t³ − 3t²: p″ = 6t − 6, inflection at 1
not where p′ = 0
The second derivative and the bend of a graph
Core ideas
What this lesson establishes
Three ideas: what a second derivative measures, how its sign reads as a bend, and what an inflection point is.
2
Its sign is the bend
Where p″ is positive, p′ is increasing: the tangent slopes get larger as t grows, and the graph bends upward, like a cup. Where p″ is negative the slopes get smaller and the graph bends downward, like a cap. A tangent line lies below a graph that bends up, which is why the estimates in “The Derivative at a Point and the Tangent Line” fell short for t².
3
An inflection is where the bend changes
An inflection point is a place where p″ changes sign: the graph stops bending one way and starts bending the other. It is found from p″ = 0, not from p′ = 0, and it is not a turn.
The material
Differentiating twice, and reading the bend
The definition, one cubic carried through, the slopes that show the bend, and the two other presets of the lab.
Definition · The second derivative
The second derivative of a polynomial p is the derivative of its derivative, written p″.
A graph is concave up where p″ > 0 and concave down where p″ < 0.
Example · A cubic with an S in it
Let p(t) = t³ − 3t². Then p′(t) = 3t² − 6t and p″ = 6t − 6 = 6(t − 1).
The second derivative is negative for t < 1, zero at 1 and positive for t > 1, so the graph bends down and then bends up, and t = 1 is an inflection point.
The slopes make the bend visible. For this cubic the slope p′ = 3t² − 6t is 0 at t = 0, then −3 at t = 1, then 0 at t = 2, then 9 at t = 3. Up to t = 1 the slopes are shrinking from 0 to −3, which is a falling slope, and after it they grow from −3 to 0 to 9. The slope bottoms out at 1, exactly where p″ is zero.
p″ = 6(t − 1)
t < 1 p″ < 0 slope falling bends down
t > 1 p″ > 0 slope rising bends up
An inflection is not a turn
The turns of this graph are at the zeros of p′ = 3t·(t − 2), which are t = 0, a maximum, and t = 2, a minimum. The inflection at t = 1 is between them, and there p′ = −3, not zero: the graph is steeply falling as it passes through. Of the two questions, where does the graph turn and where does it change its bend, the first reads p′ and the second reads p″.
The second derivative also corrects one reading of the sign. Concave up does not mean rising. The parabola t² has p″ = 2, positive for every t, and it falls for every negative t. The bend and the direction are separate facts, one from p″ and the other from p′.
Example · A quartic with two inflections
Let p(t) = t⁴ − 6t². Then p′ = 4t³ − 12t and p″ = 12t² − 12 = 12(t − 1)(t + 1).
The second derivative is positive for t < −1, negative for −1 < t < 1 and positive for t > 1, so there are two inflection points, at t = −1 and t = 1. The graph is a W with its middle hump bending down.
This lab is set to show the second derivative as well, so it prints p″ and the inflection points beside the zeros and intervals of p′. Its zero tile lists only zeros of p′ that are fractions, so for this quartic it names t = 0 and leaves the turns at ±√3 to the interval tile, which places them with square roots.
A parabola is the plainest case: p = t² has p″ = 2, which never changes sign, so the graph bends up everywhere and has no inflection.
Interactive
The derivative
p′ and p″ term by term, and their signs
Method
Reading the bend of a polynomial
Five moves. Differentiate twice, then treat p″ exactly as p′ was treated for turns.
1
Differentiate twice
Apply the power rule to p, and apply it again to the result. Write p′ and p″ in descending powers.
2
Solve p″ = 0
These are the only places the bend can change. Factor the result if it factors.
3
Test the sign of p″ in every region
Put one value of t from each region into p″. Positive is concave up, negative is concave down.
4
Mark an inflection only where the sign changes
A zero of p″ with the same sign on both sides is not an inflection. The bend has to change.
5
Keep it apart from the turns
The turns are zeros of p′ and the inflections are zeros of p″. Check by putting the inflection into p′: it is usually not zero.
Slopes of slopes
Score 0 / 0Choose the answer you can justify.
t³ − 3t², turns and bends
The polynomial is p(t) = t³ − 3t². Both derivatives, the signs and the two kinds of special place are exact.
p′(t) = 3t² − 6t = 3t·(t − 2)
p″ = 6t − 6 = 6(t − 1)
t = 0: p″ = −6 concave down
t = 2: p″ = 6 concave up
p″ changes sign at t = 1: inflection
at t = 1: p = −2, p′ = −3 (not zero: no turn)
turns, from p′ = 0: t = 0 (maximum), t = 2 (minimum)
The lab's second-derivative tile prints 6t − 6 and its inflection tile prints the single place t = 1. The three special places, 0, 1 and 2, are all different.
For a rehearsal, find p″ for t⁴, which is 12t². It is zero at 0 and positive on both sides; decide whether 0 is an inflection.
Common mistakes
1
Taking an inflection point to be where p′ is zeroFor t³ − 3t² the zeros of p′ = 3t·(t − 2) are 0 and 2, and those are the turns. The inflection is at 1, where p′ = −3 is far from zero and the graph is falling steeply. Ask where the graph turns and read p′; ask where the bend changes and read p″.
2
Reading concave up as risingConcave up means the slope is increasing, not that it is positive. The parabola t² has p″ = 2 everywhere and falls for all t < 0. A cup can be descending into its bottom as well as climbing out of it.
3
Calling every zero of p″ an inflection pointThe second derivative of t⁴ is 12t², zero at 0 and positive on both sides. The graph bends up on both sides of 0 and does not change its bend, so there is no inflection. As with the turns, the zero is a place to look, and the sign on each side decides.
Check your understanding
✓
Finish when you can differentiate a polynomial twice, read where its graph bends up and down from the sign of p″, and locate the inflection points.
You should be able to apply the power rule twice, solve p″ = 0, test the sign on each side, keep the inflections apart from the turns, and say in a sentence what a second derivative measures.
Note: With the rules and the second derivative the course has what it needs for polynomials. The last lesson meets a function that is not a polynomial and whose rate is a multiple of itself, in “The Exponential and Its Rate”.