Test for a Point of Inflection

Mathematics · FE Reference Handbook section

Mathematics

6 formulas

10 exam-style examples

~57 min

All Mathematics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

In a function of two independent variables x and y, a derivative with respect to one of the variables may be found if the other
variable is assumed to remain constant. If y is kept fixed, the function
becomes a function of the single variable x, and its derivative (if it exists) can be found. This derivative is called the partial
derivative of z with respect to x. The partial derivative with respect to x is denoted as follows:
The Curvature of Any Curve
The curvature K of a curve at P is the limit of its average curvature for the arc PQ as Q approaches P. This is also expressed as:
the curvature of a curve at a given point is the rate-of-change of its inclination with respect to its arc length.

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Locating a point of inflection from the second derivative — Test for a Point of Inflection

A deflected shape is modelled by f(x) = 4x³ − 5x² + 1x. Find the x-location of the point of inflection, the ordinate there, and the slope at that point.

Given

$f(x) = 4x^{3} - 5x^{2} + 1x$

Find

x at the point of inflection, f(x), and f′(x) there

Start with the thinking

A point of inflection requires f″(x) = 0 with a sign change in curvature.
The first derivative gives the slope, which need not be zero at an inflection point.

Step-by-step solution

Formula
$f'(x) = 12x^{2} - 10x + 1$
Formula
$f″(x) = 24x - 10$
Set f″ = 0
$24x = 10$
Substituting
$x = 10/24 = 0.4167$
Ordinate
$f(0.417) = 4(0.0723) - 5(0.1736) + 1(0.4167) = -0.1620$
Slope
$f'(0.417) = -1.0833$

Answer:

Point of inflection at x = 0.417, f = -0.162, slope = -1.083

Why the other options are there

x = 0.417 (used f′ = 0 root)
x = 2.400 (ratio inverted)

Reference: FE Reference Handbook — Mathematics → Test for a Point of Inflection

Example 2

Locating a point of inflection from the second derivative — Test for a Point of Inflection (2)

A deflected shape is modelled by f(x) = 1x³ − 5x² + 8x. Find the x-location of the point of inflection, the ordinate there, and the slope at that point.

Given

$f(x) = 1x^{3} - 5x^{2} + 8x$

Find

x at the point of inflection, f(x), and f′(x) there

Start with the thinking

A point of inflection requires f″(x) = 0 with a sign change in curvature.
The first derivative gives the slope, which need not be zero at an inflection point.

Step-by-step solution

Formula
$f'(x) = 3x^{2} - 10x + 8$
Formula
$f″(x) = 6x - 10$
Set f″ = 0
$6x = 10$
Substituting
$x = 10/6 = 1.6667$
Ordinate
$f(1.667) = 1(4.6296) - 5(2.7778) + 8(1.6667) = 4.0741$
Slope
$f'(1.667) = -0.3333$

Answer:

Point of inflection at x = 1.667, f = 4.074, slope = -0.333

Why the other options are there

x = 1.667 (used f′ = 0 root)
x = 0.600 (ratio inverted)

Reference: FE Reference Handbook — Mathematics → Test for a Point of Inflection

Example 3

Locating a point of inflection from the second derivative — Test for a Point of Inflection (3)

A deflected shape is modelled by f(x) = 4x³ − 9x² + 5x. Find the x-location of the point of inflection, the ordinate there, and the slope at that point.

Given

$f(x) = 4x^{3} - 9x^{2} + 5x$

Find

x at the point of inflection, f(x), and f′(x) there

Start with the thinking

A point of inflection requires f″(x) = 0 with a sign change in curvature.
The first derivative gives the slope, which need not be zero at an inflection point.

Step-by-step solution

Formula
$f'(x) = 12x^{2} - 18x + 5$
Formula
$f″(x) = 24x - 18$
Set f″ = 0
$24x = 18$
Substituting
$x = 18/24 = 0.7500$
Ordinate
$f(0.750) = 4(0.4219) - 9(0.5625) + 5(0.7500) = 0.3750$
Slope
$f'(0.750) = -1.7500$

Answer:

Point of inflection at x = 0.750, f = 0.375, slope = -1.750

Why the other options are there

x = 0.750 (used f′ = 0 root)
x = 1.333 (ratio inverted)

Reference: FE Reference Handbook — Mathematics → Test for a Point of Inflection

Example 4

Locating a point of inflection from the second derivative — Test for a Point of Inflection (4)

A deflected shape is modelled by f(x) = 3x³ − 8x² + 7x. Find the x-location of the point of inflection, the ordinate there, and the slope at that point.

Given

$f(x) = 3x^{3} - 8x^{2} + 7x$

Find

x at the point of inflection, f(x), and f′(x) there

Start with the thinking

A point of inflection requires f″(x) = 0 with a sign change in curvature.
The first derivative gives the slope, which need not be zero at an inflection point.

Step-by-step solution

Formula
$f'(x) = 9x^{2} - 16x + 7$
Formula
$f″(x) = 18x - 16$
Set f″ = 0
$18x = 16$
Substituting
$x = 16/18 = 0.8889$
Ordinate
$f(0.889) = 3(0.7023) - 8(0.7901) + 7(0.8889) = 2.0082$
Slope
$f'(0.889) = -0.1111$

Answer:

Point of inflection at x = 0.889, f = 2.008, slope = -0.111

Why the other options are there

x = 0.889 (used f′ = 0 root)
x = 1.125 (ratio inverted)

Reference: FE Reference Handbook — Mathematics → Test for a Point of Inflection

Example 5

Locating a point of inflection from the second derivative — Test for a Point of Inflection (5)

A deflected shape is modelled by f(x) = 1x³ − 5x² + 1x. Find the x-location of the point of inflection, the ordinate there, and the slope at that point.

Given

$f(x) = 1x^{3} - 5x^{2} + 1x$

Find

x at the point of inflection, f(x), and f′(x) there

Start with the thinking

A point of inflection requires f″(x) = 0 with a sign change in curvature.
The first derivative gives the slope, which need not be zero at an inflection point.

Step-by-step solution

Formula
$f'(x) = 3x^{2} - 10x + 1$
Formula
$f″(x) = 6x - 10$
Set f″ = 0
$6x = 10$
Substituting
$x = 10/6 = 1.6667$
Ordinate
$f(1.667) = 1(4.6296) - 5(2.7778) + 1(1.6667) = -7.5926$
Slope
$f'(1.667) = -7.3333$

Answer:

Point of inflection at x = 1.667, f = -7.593, slope = -7.333

Why the other options are there

x = 1.667 (used f′ = 0 root)
x = 0.600 (ratio inverted)

Reference: FE Reference Handbook — Mathematics → Test for a Point of Inflection

Example 6

Locating a point of inflection from the second derivative — Test for a Point of Inflection (6)

A deflected shape is modelled by f(x) = 2x³ − 6x² + 1x. Find the x-location of the point of inflection, the ordinate there, and the slope at that point.

Given

$f(x) = 2x^{3} - 6x^{2} + 1x$

Find

x at the point of inflection, f(x), and f′(x) there

Start with the thinking

A point of inflection requires f″(x) = 0 with a sign change in curvature.
The first derivative gives the slope, which need not be zero at an inflection point.

Step-by-step solution

Formula
$f'(x) = 6x^{2} - 12x + 1$
Formula
$f″(x) = 12x - 12$
Set f″ = 0
$12x = 12$
Substituting
$x = 12/12 = 1.0000$
Ordinate
$f(1.000) = 2(1.0000) - 6(1.0000) + 1(1.0000) = -3.0000$
Slope
$f'(1.000) = -5.0000$

Answer:

Point of inflection at x = 1.000, f = -3.000, slope = -5.000

Why the other options are there

x = 1.000 (used f′ = 0 root)
x = 1.000 (ratio inverted)

Reference: FE Reference Handbook — Mathematics → Test for a Point of Inflection

Example 7

Locating a point of inflection from the second derivative — Test for a Point of Inflection (7)

A deflected shape is modelled by f(x) = 2x³ − 5x² + 6x. Find the x-location of the point of inflection, the ordinate there, and the slope at that point.

Given

$f(x) = 2x^{3} - 5x^{2} + 6x$

Find

x at the point of inflection, f(x), and f′(x) there

Start with the thinking

A point of inflection requires f″(x) = 0 with a sign change in curvature.
The first derivative gives the slope, which need not be zero at an inflection point.

Step-by-step solution

Formula
$f'(x) = 6x^{2} - 10x + 6$
Formula
$f″(x) = 12x - 10$
Set f″ = 0
$12x = 10$
Substituting
$x = 10/12 = 0.8333$
Ordinate
$f(0.833) = 2(0.5787) - 5(0.6944) + 6(0.8333) = 2.6852$
Slope
$f'(0.833) = 1.8333$

Answer:

Point of inflection at x = 0.833, f = 2.685, slope = 1.833

Why the other options are there

x = 0.833 (used f′ = 0 root)
x = 1.200 (ratio inverted)

Reference: FE Reference Handbook — Mathematics → Test for a Point of Inflection

Example 8

Locating a point of inflection from the second derivative — Test for a Point of Inflection (8)

A deflected shape is modelled by f(x) = 4x³ − 2x² + 6x. Find the x-location of the point of inflection, the ordinate there, and the slope at that point.

Given

$f(x) = 4x^{3} - 2x^{2} + 6x$

Find

x at the point of inflection, f(x), and f′(x) there

Start with the thinking

A point of inflection requires f″(x) = 0 with a sign change in curvature.
The first derivative gives the slope, which need not be zero at an inflection point.

Step-by-step solution

Formula
$f'(x) = 12x^{2} - 4x + 6$
Formula
$f″(x) = 24x - 4$
Set f″ = 0
$24x = 4$
Substituting
$x = 4/24 = 0.1667$
Ordinate
$f(0.167) = 4(0.0046) - 2(0.0278) + 6(0.1667) = 0.9630$
Slope
$f'(0.167) = 5.6667$

Answer:

Point of inflection at x = 0.167, f = 0.963, slope = 5.667

Why the other options are there

x = 0.167 (used f′ = 0 root)
x = 6.000 (ratio inverted)

Reference: FE Reference Handbook — Mathematics → Test for a Point of Inflection

Example 9

Locating a point of inflection from the second derivative — Test for a Point of Inflection (9)

A deflected shape is modelled by f(x) = 4x³ − 6x² + 6x. Find the x-location of the point of inflection, the ordinate there, and the slope at that point.

Given

$f(x) = 4x^{3} - 6x^{2} + 6x$

Find

x at the point of inflection, f(x), and f′(x) there

Start with the thinking

A point of inflection requires f″(x) = 0 with a sign change in curvature.
The first derivative gives the slope, which need not be zero at an inflection point.

Step-by-step solution

Formula
$f'(x) = 12x^{2} - 12x + 6$
Formula
$f″(x) = 24x - 12$
Set f″ = 0
$24x = 12$
Substituting
$x = 12/24 = 0.5000$
Ordinate
$f(0.500) = 4(0.1250) - 6(0.2500) + 6(0.5000) = 2.0000$
Slope
$f'(0.500) = 3.0000$

Answer:

Point of inflection at x = 0.500, f = 2.000, slope = 3.000

Why the other options are there

x = 0.500 (used f′ = 0 root)
x = 2.000 (ratio inverted)

Reference: FE Reference Handbook — Mathematics → Test for a Point of Inflection

Example 10

Locating a point of inflection from the second derivative — Test for a Point of Inflection (10)

A deflected shape is modelled by f(x) = 2x³ − 9x² + 6x. Find the x-location of the point of inflection, the ordinate there, and the slope at that point.

Given

$f(x) = 2x^{3} - 9x^{2} + 6x$

Find

x at the point of inflection, f(x), and f′(x) there

Start with the thinking

A point of inflection requires f″(x) = 0 with a sign change in curvature.
The first derivative gives the slope, which need not be zero at an inflection point.

Step-by-step solution

Formula
$f'(x) = 6x^{2} - 18x + 6$
Formula
$f″(x) = 12x - 18$
Set f″ = 0
$12x = 18$
Substituting
$x = 18/12 = 1.5000$
Ordinate
$f(1.500) = 2(3.3750) - 9(2.2500) + 6(1.5000) = -4.5000$
Slope
$f'(1.500) = -7.5000$

Answer:

Point of inflection at x = 1.500, f = -4.500, slope = -7.500

Why the other options are there

x = 1.500 (used f′ = 0 root)
x = 0.667 (ratio inverted)

Reference: FE Reference Handbook — Mathematics → Test for a Point of Inflection