Lesson · Rates of Change and the Derivative · The derivative as a function

Where the Rate Is Zero

A zero of the derivative is a place where the tangent is level, and that is the only place a polynomial's graph can turn. It is necessary for a turn and not enough: the graph turns only where p′ changes sign. Solve p′ = 0, test the sign on each side of every zero, and each one is a maximum, a minimum or a flat step that keeps going the same way.

Open the interactive lesson Practice

a turn needs p′(t) = 0 p′ = 0 is necessary, not sufficient rising then falling: a maximum falling then rising: a minimum no sign change: a flat step, no turn t³: p′ = 3t² is 0 at 0, yet no turn

Where p′ is zero, and what that does to the graph

Core ideas

What this lesson establishes

Three ideas: why a turn needs a zero, why a zero does not need a turn, and what a maximum or minimum is a maximum or minimum of.

A graph can turn only where the tangent is level

Just before a turn the graph is rising, so p′ is positive; just after, it is falling, so p′ is negative. A polynomial's derivative is a polynomial, and a polynomial does not jump from positive to negative without passing through zero. So every turn sits at a zero of p′.

A level tangent is not always a turn

The converse is false. For p(t) = t³ the derivative 3t² is zero at t = 0, and positive on both sides. The graph flattens for an instant and keeps rising. Whether a zero is a turn is decided by the sign of p′ on each side of it, and by nothing at the zero itself.

Maximum and minimum are local

A maximum at t = c is the highest point of the graph near c: higher than its neighbours on both sides. It need not be the highest point anywhere. A cubic climbs for ever, so any maximum it has is overtaken further along.

The material

Zeros of the derivative, and what they do and do not say

The idea in a sentence, one cubic carried all the way through, the zero that is not a turn, and the case with no zero at all.

“The Derivative as a Function” ended on a warning: a zero of p′ is a place where the tangent is level, and it does not say whether the graph rises or falls on either side. This lesson turns the warning into a method.

Definition · Critical point, maximum and minimum

A critical point of a polynomial p is a place where p′(t) = 0.

It is a local maximum if p′ changes from positive to negative there, and a local minimum if p′ changes from negative to positive. If p′ has the same sign on both sides, it is neither.

Example · A cubic with two turns

Let p(t) = 2t³ − 3t² − 12t + 1. Then p′(t) = 6t² − 6t − 12, and taking out the common factor 6 gives 6(t² − t − 2) = 6(t − 2)(t + 1).

The critical points are t = −1 and t = 2. Test one place in each region they cut the line into, and the whole shape of the graph follows.

p′(t) = 6(t − 2)(t + 1) t < −1 p′ > 0 rising −1 < t < 2 p′ < 0 falling t > 2 p′ > 0 rising

At t = −1 the graph stops rising and starts falling: a local maximum. At t = 2 it stops falling and starts rising: a local minimum. The height comes from p, not from p′, because p′ is zero at both: p(−1) = −2 − 3 + 12 + 1 = 8 and p(2) = 16 − 12 − 24 + 1 = −19.

The word local matters. The maximum height is 8, but further right the graph rises without limit: p(4) = 128 − 48 − 48 + 1 = 33. The height 8 is the highest point near t = −1 and nothing more.

A zero where nothing turns

Now p(t) = t³. Its derivative is 3t², which is zero at t = 0 and positive for every other t, because a square is never negative. So p′ is positive on both sides of its zero, and the graph rises before t = 0 and rises after it, level for an instant and not turning. The lab marks such a zero as having no sign change, and its interval tile reads rising on both sides.

This is the case the necessary-but-not-sufficient wording is for. Solving p′ = 0 finds every place a turn could be, and the sign test sorts the real turns from the flat steps.

A derivative that is never zero

For p(t) = t³ + t the derivative is 3t² + 1. A square is at least 0, so 3t² + 1 is at least 1 and never reaches zero. The graph has no level tangent and no turn: it rises everywhere. The zero tile of the lab says none rational, which is how it reports a polynomial with no zero it can write as a fraction, and the interval tile gives the fuller answer, rising for every t.

Interactive

The derivative

p′ and p″ term by term, and their signs

Find the zeros of p′ and sort them by the sign

Type a polynomial in t. The lab differentiates it, lists the rational zeros of the derivative, and tests the sign of p′ between them. A zero marked no sign change is a flat step and not a turn. The green dots on the plot sit on the graph of p at each zero of p′.

Polynomial

p(t)

Derivatives shown

p′(t)—

p″(t)—

Zeros of p′—

Where p rises and falls—

Inflection points—

A polynomial in t of degree 10 at most: t^3 - 6t^2 + 9t.

Method

Locating and sorting the critical points

Five moves. The fourth is the one that stops a flat step being called a turn.

Differentiate

Apply the power rule to each term and collect the result in descending powers.

Solve p′ = 0

Factor, or use the quadratic formula. The zeros are the only places a turn can be. A zero that appears twice in the factoring, such as 3t², is a warning that the sign may not change.

Test the sign in every region

Between and beyond the zeros, put in one value of t and note whether p′ is positive or negative there. Use the factored form so the sign is easy to read.

Name each zero from its two neighbours

Positive then negative is a maximum, negative then positive is a minimum, and the same sign on both sides is neither.

Find the height from p

Put the zero into p, not p′. The height of the turn is a value of the original polynomial.

Level tangents and turns

Score 0 / 0

Choose the answer you can justify.

A cubic with a maximum and a minimum

The polynomial is p(t) = 2t³ − 3t² − 12t + 1. The derivative, the signs and the heights are each exact.

p′(t) = 6t² − 6t − 12 = 6(t − 2)(t + 1) zeros: t = −1, t = 2 t = −2: p′ = 6·(−4)·(−1) = 24 rising t = 0: p′ = 6·(−2)·1 = −12 falling t = 3: p′ = 6·1·4 = 24 rising t = −1: rising then falling maximum p = 8 t = 2: falling then rising minimum p = −19

The lab's interval tile for this polynomial gives the same three regions as the table, and its zero tile gives the same two places. Neither height was found by plotting a point.

For a rehearsal, take p(t) = t³ − 3t. The derivative is 3t² − 3 = 3(t − 1)(t + 1); work out the signs, then which of t = −1 and t = 1 is the maximum.

Common mistakes

Calling every zero of p′ a maximum or a minimumFor p(t) = t³, p′ = 3t² is zero at t = 0, and the graph is rising on both sides of it. If 0 were a maximum the graph would fall just after it; it does not. The zero says the tangent is level, and only the signs of p′ either side say whether the graph turned.

Reading the height of a turn from p′At t = −1 the derivative 6(t − 2)(t + 1) is zero, and the height of the maximum is 8. The zero of p′ tells you where to look, and the value of p there tells you how high. Writing 0 as the height of a turn is reading the wrong function.

Taking a local maximum for the largest valueA maximum is the highest point near it. The cubic above has a local maximum of 8 at t = −1 and a value of 33 at t = 4. Say local, and where a question asks for the greatest value over an interval, compare the turns with the ends.

Check your understanding

✓

Finish when you can solve p′ = 0 exactly for a polynomial and say, from the signs on either side, whether each zero is a maximum, a minimum or neither.

You should be able to differentiate, factor the result, test the sign of p′ in every region, name each zero from its neighbours, and find the height of each turn from p.

Note: The lab finds the zeros of p′ that are fractions, and for a quadratic derivative it places the others with square roots in the interval tile. The next two lessons build the two rules that let you differentiate something other than a sum of powers, starting with “The Product Rule”.