B. Derivatives and Their Applications · Curve Sketching: Connecting f, f′ and f″

Concavity and Points of Inflection

B1.3 B1.5About 70 minutes

Learning goals

I can use a sign chart for f″ to find intervals where f is concave up or concave down.
I can find points of inflection algebraically and check that the concavity really changes.
I can explain why f″(c) = 0 is not enough for an inflection point, and why an inflection point can occur where f″ does not exist.

Words to know

test for concavity
If f″ > 0 on an interval, f is concave up there; if f″ < 0, f is concave down there.

candidate for inflection
An x-value in the domain where f″ = 0 or f″ does not exist.

horizontal point of inflection
An inflection point where the tangent is also horizontal (f′ = 0), as for x³ at the origin.

Where does a curve change its bend?

When a road bends left and then right, like an S, there is a spot where the bend switches direction. The place where the bend switches is a point of inflection. In this lesson you will find these points exactly, from the equation, using a sign chart for f″.

The method is the same as for f′, one level higher:

the sign of f′ tells you increasing or decreasing;
the sign of f″ tells you concave up or concave down.

Test for concavity

If f″(x) > 0 for all x in an interval, f is concave up there. If f″(x) < 0 for all x in an interval, f is concave down there.

Finding points of inflection

Find f″(x) and factor it.
Find the candidates: x-values in the domain where f″(x) = 0 or f″(x) does not exist. Also note x-values not in the domain.
Make a sign chart for f″.
A candidate is a point of inflection only if f″ changes sign there (and f is continuous there).
Find the y-coordinate by substituting into the original f, not into f″.

Worked example

Example 1: A cubic

Find the intervals of concavity and the point of inflection of f(x) = x³ − 6x² + 9x + 2.

f′(x) = 3x² − 12x + 9, f″(x) = 6x − 12 = 6(x − 2).
f″ = 0 at x = 2. For x < 2, f″ < 0; for x > 2, f″ > 0.
The sign changes, so x = 2 gives an inflection point. f(2) = 8 − 24 + 18 + 2 = 4.

Concave down on (−∞, 2), concave up on (2, ∞); point of inflection (2, 4). Every cubic has exactly one inflection point, halfway (in x) between its turning points when it has two (here 1 and 3).

Worked example

Example 2: A quartic

Find the intervals of concavity and the points of inflection of f(x) = x⁴ − 6x².

f′(x) = 4x³ − 12x, f″(x) = 12x² − 12 = 12(x − 1)(x + 1).
Candidates: x = −1 and x = 1.
Signs of f″: + for x < −1, − for −1 < x < 1, + for x > 1. Both candidates change sign.
f(±1) = 1 − 6 = −5.

Concave up on (−∞, −1) and (1, ∞); concave down on (−1, 1); inflection points (−1, −5) and (1, −5).

The same sign-chart method you used for f′, now applied to f″. The bottom row shows the concavity: ∪ where f″ > 0, ∩ where f″ < 0. The concavity changes at both x = −1 and x = 1, so both are inflection points.

Concave down on the shaded interval −1 < x < 1, concave up outside it. Notice that the inflection points (±1, −5) are not the turning points (0, 0) and (±√3, −9).

f″ = 0 is not enough

Just as f′ = 0 does not guarantee a turning point, f″ = 0 does not guarantee an inflection point. The sign of f″ must change. A factor with an even power, like x² in f″, signals a candidate that probably is not an inflection point.

Worked example

Example 3: Two warnings

(a) Show that f(x) = x⁴ has no inflection point even though f″(0) = 0. (b) Show that g(x) = x^(1/3) has an inflection point at x = 0 even though g″(0) does not exist.

(a) f″(x) = 12x², which is positive for every x ≠ 0. No sign change, so no inflection point: x⁴ is concave up everywhere.
(b) g′(x) = (1/3)x^(−2/3), g″(x) = −(2/9)x^(−5/3) = −2/(9x^(5/3)).
For x < 0, x^(5/3) < 0, so g″ > 0 (concave up). For x > 0, g″ < 0 (concave down).
g is continuous at 0 and g(0) = 0, and the concavity changes, so (0, 0) is an inflection point. (The tangent there is vertical.)

x⁴: no inflection point; ∛x: inflection point at (0, 0).

Left: f″(0) = 0 for x⁴, but the graph is concave up on both sides, so (0, 0) is not an inflection point. Right: for the cube root, f″(0) does not exist, but the concavity does change (with a vertical tangent), so (0, 0) is an inflection point.

Worked example

Example 4: Inflection points and a horizontal tangent

Find the intervals of concavity and the inflection points of f(x) = x⁴ − 4x³.

f′(x) = 4x³ − 12x², f″(x) = 12x² − 24x = 12x(x − 2).
Candidates: x = 0 and x = 2. Signs: + for x < 0, − for 0 < x < 2, + for x > 2.
Both change sign. f(0) = 0 and f(2) = 16 − 32 = −16.
Notice that f′(0) = 0 too. In the lesson on critical numbers you saw that x = 0 is a flat spot with no turn: now you know it is a horizontal point of inflection.

Concave up on (−∞, 0) and (2, ∞), concave down on (0, 2); inflection points (0, 0) and (2, −16).

Worked example

Example 5: A rational function

Describe the concavity of f(x) = (x + 1)/(x − 1).

Divide: x + 1 = (x − 1) + 2, so f(x) = 1 + 2(x − 1)⁻¹.
f′(x) = −2(x − 1)⁻², f″(x) = 4(x − 1)⁻³ = 4/(x − 1)³.
f″ is never 0. It is undefined at x = 1, which is not in the domain (a vertical asymptote).
For x < 1, (x − 1)³ < 0 so f″ < 0; for x > 1, f″ > 0.

Concave down on (−∞, 1), concave up on (1, ∞). No inflection point, because the concavity changes across an asymptote, not at a point of the graph.

Just for fun

A point of inflection walks into a karaoke bar. The host says, "Great voice, but every time you hit the middle of the song you change your tone." (In English, a change in the pitch of your voice is also called an inflection.)

Hands-on activity

Inflection hunt

You need: a pencil, paper, a flexible ruler or a piece of string, and a few everyday curves: the letter S on a cereal box or sign, a garden hose or extension cord lying on the floor, a playground slide (in a photo or in person).

Lay the string or flexible ruler along each curve, or trace the curve onto paper.
Walk a pencil along the tracing from left to right. Mark where it bends like ∪ and where it bends like ∩.
Circle every place where the bend switches. These are the points of inflection.
Draw a short straight line (a tangent) through each circled point. Notice that the curve crosses its tangent there.

What to look for: a letter S has one inflection point near its middle. A typical slide that curves down from its flat top, gets steep, and flattens out at the bottom has one where it is steepest. At every inflection point the curve passes from one side of the tangent line to the other; at ordinary points it stays on one side.

Your turn

Find the intervals of concavity and the inflection point of f(x) = x³ + 3x² − 9x.

Find the inflection point of f(x) = −2x³ + 6x² + 1.

Find the intervals of concavity of f(x) = x⁴ − 24x².

Find the inflection points of f(x) = x⁴ − 24x².

Does f(x) = (x − 3)⁴ + 2 have an inflection point?

Describe the concavity of f(x) = 1/(x + 2).

More practice: warm-up

f″(x) = x − 5. Where is f concave up?

f″ changes from + to − at x = 2. What is at x = 2?

Find f″ for f(x) = x³.

To find the y-coordinate of an inflection point, substitute into f, f′ or f″?

How many inflection points does every cubic polynomial have?

More practice: on your own

Find the intervals of concavity and inflection points of f(x) = x⁴ − 2x³.

Find the inflection points of f(x) = x⁵ − 10x³.

Find the inflection points of f(x) = 3x⁵ − 5x⁴.

Describe the concavity of f(x) = x/(x − 3).

Find the inflection points of f(x) = 1/(x² + 3).

Challenge

Prove that the inflection point of any cubic f(x) = ax³ + bx² + cx + d is at x = −b/(3a).

For which values of c does f(x) = x⁴ + cx² have two inflection points?

Show that the inflection points of f(x) = x⁴ − 6x² + 5 and the point (0, 5) are NOT collinear, but that the inflection points of g(x) = x³ − 3x lie on the line through its turning points.

Check your understanding

Quiz

What is the inflection point of f(x) = x³ − 3x² + 4?

Where is f(x) = x⁴ − 12x² concave down?

f″(x) = (x − 4)². At x = 4, f has:

Why does f(x) = 1/x have no point of inflection, even though its concavity changes?

To be a point of inflection, a point (c, f(c)) must have:

How many inflection points does f(x) = x⁴ − 4x³ have?

For parents and teachers

Big idea (B1.3, B1.5): to find where a graph bends up or down, make a sign chart for f″. An inflection point is where the bend actually changes, which you must check; f″ = 0 is only a candidate.

How to help: the inflection hunt activity is quick and visual. Ask your teen to show you the tangent line crossing the curve at each inflection point.

Common difficulties:

Declaring an inflection point wherever f″ = 0 (x⁴ is the classic trap).
Substituting into f″ instead of f to get the y-coordinate.
Calling a change of concavity across a vertical asymptote an inflection point.
Confusing inflection points with turning points.