Locating Roots by Change of Sign

SPEC 9.1 - NUMERICAL METHODS

Learn how to find approximate solutions to equations that cannot be solved algebraically

Why Numerical Methods?

Many equations cannot be solved algebraically using standard techniques. For example:

Equations We Cannot Solve

x = cos x

e^{x} = 3 x

x^{5} - 3 x + 1 = 0

For these equations, we need numerical methods to find approximate solutions to a required degree of accuracy.

Key Concept

The roots of an equation

f (x) = 0

are the x-coordinates where the curve

y = f (x)

crosses the x-axis.

The Sign Change Principle

Sign Change Theorem

If

f (x)

is continuous on the interval

[a, b]

and

f (a)

and

f (b)

have opposite signs, then there is at least one root in the interval

(a, b)

.

f(a) > 0, f(b) < 0 → Root exists

f(a) < 0, f(b) > 0 → Root exists

Method for Showing a Root Exists

Follow these steps to show that a root exists in an interval:

StepAction

Step 1Define

f (x)

clearly

Step 2Calculate

f (a)

and

f (b)

for the given interval

Step 3Show that

f (a)

and

f (b)

have opposite signs

Step 4State: "Since

f (x)

is continuous and there is a sign change, a root exists in

(a, b)

"

Important

Always show the actual numerical values of

f (a)

and

f (b)

, not just their signs!

Worked Example01

Show that the equation $x^{3} - 3 x + 1 = 0$ has a root between $x = 1$ and $x = 2$ .

1

Let

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

2

f (1) = (1)^{3} - 3 (1) + 1 = 1 - 3 + 1 = - 1

3

f (2) = (2)^{3} - 3 (2) + 1 = 8 - 6 + 1 = 3

4

Since

f (1) = - 1 < 0

and

f (2) = 3 > 0

, there is a sign change.

Conclusion: Since

f (x)

is continuous (it's a polynomial) and there is a sign change, there is a root in the interval

(1, 2)

.

Worked Example02

Show that $e^{x} - 4 x = 0$ has a root in the interval $[0, 1]$ .

1

Let

f (x) = e^{x} - 4 x

2

f (0) = e^{0} - 4 (0) = 1 - 0 = 1

3

f (1) = e^{1} - 4 (1) = e - 4 \approx 2.718 - 4 = - 1.282

4

Since

f (0) = 1 > 0

and

f (1) = e - 4 < 0

, there is a sign change.

Conclusion: Since

f (x) = e^{x} - 4 x

is continuous and there is a sign change, there is a root in

(0, 1)

.

Worked Example03

Show that the equation $x = cos x$ has a root between $x = 0.7$ and $x = 0.8$ .

1

Rearrange to

x - cos x = 0

. Let

f (x) = x - cos x

2

f (0.7) = 0.7 - cos (0.7) = 0.7 - 0.7648... = - 0.0648...

3

f (0.8) = 0.8 - cos (0.8) = 0.8 - 0.6967... = 0.1033...

4

Since

f (0.7) < 0

and

f (0.8) > 0

, there is a sign change.

Conclusion: Since

f (x) = x - cos x

is continuous and there is a sign change, there is a root in

(0.7, 0.8)

.

Limitations of Sign Change

The sign change method does not always work. Be aware of these situations:

Discontinuous
No root despite sign change — Discontinuous No root despite sign change

Repeated root
No sign change — Repeated root No sign change

Even roots
No sign change — Even roots No sign change

Odd roots (3) Sign change present

Remember

A sign change guarantees at least one root exists (could be 1, 3, 5, ... roots). No sign change does not mean no roots exist - there could be 0, 2, 4, ... roots!

Worked Example04

Explain why the sign change method fails to find a root of $f (x) = x^{2} - 4 x + 4$ in the interval $[1, 3]$ , even though $f (1) > 0$ and $f (3) > 0$ .

1

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

2

f (1) = (1 - 2)^{2} = 1 > 0

3

f (3) = (3 - 2)^{2} = 1 > 0

4

Both values are positive, so there is no sign change.

Explanation: The function

f (x) = (x - 2)^{2}

has a repeated root at

x = 2

. The curve touches the x-axis at

x = 2

but does not cross it. Since the function is always

\geq 0

, there is no sign change even though a root exists.

Verifying a Root to n Decimal Places

To verify that

α

is correct to

n

decimal places, you must show that both bounds round to the same value.

Method

To verify

α

is correct to

n

d.p.:

Find the bounds: lower bound =

α - 0.5 \times 1 0^{- n}

, upper bound =

α + 0.5 \times 1 0^{- n}

Show there is a sign change:

f (lower)

and

f (upper)

have opposite signs

Conclude: both bounds round to

α

to

n

d.p., so the root is correct to

n

d.p.

Finding the Bounds

For 1 d.p.: bounds are

α \pm 0.05

(e.g., for 1.3, test 1.25 and 1.35)

For 2 d.p.: bounds are

α \pm 0.005

(e.g., for 1.57, test 1.565 and 1.575)

For 3 d.p.: bounds are

α \pm 0.0005

(e.g., for 1.571, test 1.5705 and 1.5715)

Worked Example05

The equation $x^{3} + 2 x - 7 = 0$ has a root $α$ . Verify that $α = 1.57$ correct to 2 decimal places.

1

Let

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

2

For 2 d.p., we need bounds

1.57 \pm 0.005

, i.e., test 1.565 and 1.575

3

f (1.565) = (1.565)^{3} + 2 (1.565) - 7 = 3.836... + 3.13 - 7 = - 0.034...

4

f (1.575) = (1.575)^{3} + 2 (1.575) - 7 = 3.908... + 3.15 - 7 = 0.058...

5

Since

f (1.565) < 0

and

f (1.575) > 0

, there is a sign change.

Conclusion: There is a sign change in the interval

(1.565, 1.575)

. Both 1.565 and 1.575 round to 1.57 (to 2 d.p.), therefore

α = 1.57

correct to 2 decimal places.

Worked Example06

The curve $y = x^{3} - 3 x^{2} + 1$ has a stationary point where $x > 0$ . Show that the x-coordinate of this stationary point lies between 1.9 and 2.1, and find this value correct to 1 decimal place.

1

At stationary points,

\frac{d y}{d x} = 0

2

\frac{d y}{d x} = 3 x^{2} - 6 x = 3 x (x - 2)

3

Let

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

. We need to find where

f (x) = 0

for

x > 0

.

4

f (1.9) = 3 (1.9)^{2} - 6 (1.9) = 10.83 - 11.4 = - 0.57

5

f (2.1) = 3 (2.1)^{2} - 6 (2.1) = 13.23 - 12.6 = 0.63

6

Sign change confirms root in

(1.9, 2.1)

. Testing

f (2.0) = 12 - 12 = 0

.

AnswerThe x-coordinate of the stationary point is exactly

x = 2.0

(to 1 d.p.)

Practice

Question 1: Show that

x^{3} - x - 5 = 0

has a root between

x = 1

and

x = 2

.

Let

f (x) = x^{3} - x - 5

f (1) = 1 - 1 - 5 = - 5

(negative)

f (2) = 8 - 2 - 5 = 1

(positive)

Since

f (1) < 0

and

f (2) > 0

, there is a sign change.

Conclusion: Since

f (x)

is continuous and there is a sign change, there is a root in

(1, 2)

.

Practice

Question 2: Show that

e^{x} = x + 2

has a root in the interval

[- 2, - 1]

, and verify the root is

- 1.8

correct to 1 decimal place.

Let

f (x) = e^{x} - x - 2

f (- 2) = e^{- 2} + 2 - 2 = 0.135...

(positive)

f (- 1) = e^{- 1} + 1 - 2 = - 0.632...

(negative)

Sign change confirms root in

(- 2, - 1)

.

To verify

α = - 1.8

to 1 d.p., test bounds

- 1.85

and

- 1.75

:

f (- 1.85) = e^{- 1.85} + 1.85 - 2 = 0.0074...

(positive)

f (- 1.75) = e^{- 1.75} + 1.75 - 2 = - 0.076...

(negative)

AnswerBoth

- 1.85

and

- 1.75

round to

- 1.8

, so

α = - 1.8

to 1 d.p.

Practice

Question 3: The curve

y = x^{4} - 4 x^{3} + 2

has a stationary point where

x > 0

. Show that the x-coordinate lies between 2 and 4.

At stationary points,

\frac{d y}{d x} = 0

\frac{d y}{d x} = 4 x^{3} - 12 x^{2} = 4 x^{2} (x - 3)

Let

f (x) = 4 x^{3} - 12 x^{2}

f (2) = 4 (8) - 12 (4) = 32 - 48 = - 16

(negative)

f (4) = 4 (64) - 12 (16) = 256 - 192 = 64

(positive)

Conclusion: Since

f (2) < 0

and

f (4) > 0

with

f

continuous, there is a stationary point with x-coordinate in

(2, 4)

.

Summary

Key Points

ConceptDetails

Sign Change PrincipleIf

f (a)

and

f (b)

have opposite signs and

f

is continuous, there is at least one root in

(a, b)

Verifying to n d.p.Show both bounds of a sign-change interval round to the same value to n d.p.

Stationary PointsFind where

f^{'} (x) = 0

using sign change on the derivative

LimitationsFails for discontinuous functions, repeated roots, or even numbers of roots

Exam Tips

Always state "f(x) is continuous" in your conclusion

Show actual numerical values, not just signs

To verify to n d.p., test at the (n+1)th decimal place bounds

For stationary points, apply sign change to the derivative