9.1 Locating roots by change of sign — Learn

Prerequisites: substituting a value into a function and reading function notation from Chapter 2; the graphs of $e^x$ and $\ln x$ from Chapter 6; the idea of a continuous curve from Chapter 1.

By the end of this section you can: explain what a root of $f(x) = 0$ is; use the change of sign on an interval to show that a root lies inside it; narrow an interval to locate a root to a given accuracy; and explain three ways a change-of-sign search can fail.

1. What a root is

A root of an equation is a value of the variable that makes the two sides equal. When the equation is written as $f(x) = 0$, a root is a value where the graph of $y = f(x)$ meets the $x$-axis.

Some equations solve exactly. The equation $x^2 - 4 = 0$ has the roots $x = 2$ and $x = -2$, and the graph crosses the axis at both. Other equations have no exact solution at all. The equation

cannot be rearranged to give $x$ by itself. A numerical method finds the root by arithmetic instead of algebra.

2. The change of sign

Here is the key idea. Suppose $f$ is continuous on the interval from $a$ to $b$. That means the graph has no break and no jump between those two values. If $f(a)$ and $f(b)$ have opposite signs, then the graph must cross the axis somewhere between them. There is at least one root in the interval.

Written with the values, the test is

A negative product means one value is positive and the other is negative. The interval $[a, b]$ is called a bracket for the root.

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

Write $f(x) = x^3 + x - 1$ and work out the value at each end.

One value is negative and the other is positive. The function is continuous, so the graph crosses the axis between $0$ and $1$. The root lies in the interval $(0, 1)$.

3. Narrowing the interval

Once a root is bracketed, test a value in the middle. Keep the half of the interval where the sign changes. Each step halves the interval, so the root is trapped more and more tightly.

Example 2. Locate the root of $x^3 + x - 1 = 0$ to one decimal place.

Start from the bracket $(0, 1)$ and test $x = 0.5$:

This is negative, and $f(1)$ is positive, so the root lies in $(0.5, 1)$. Now test $x = 0.7$:

This is positive, and $f(0.5)$ is negative, so the root lies in $(0.5, 0.7)$. Test $x = 0.6$:

Now $f(0.6) < 0$ and $f(0.7) > 0$, so the root lies in $(0.6, 0.7)$. Every number in that interval rounds to $0.7$ or to $0.6$, so the answer to one decimal place is uncertain. One more step settles it. Test $x = 0.68$:

This is just below zero, while $f(0.69) = 0.018509$ is above zero. The root lies in $(0.68, 0.69)$, so to two decimal places it is $0.68$. Section 9.3 finds it to four decimal places with a faster method.

4. Roots of equations with exponentials and logarithms

The same test works for any continuous function, including the exponential and logarithmic functions from Chapter 6. Bring everything to one side first.

Example 3. Show that the equation $e^x = 3x$ has a root between $1$ and $2$.

Rearrange to $e^x - 3x = 0$ and write $f(x) = e^x - 3x$. Then

The values have opposite signs, so a root lies in $(1, 2)$.

Example 4. A function has the values in the table. Show that it has a root between $2$ and $3$.

The value at $x = 2$ is negative and the value at $x = 3$ is positive. The sign changes, so there is a root in $(2, 3)$.

5. When change of sign fails

The test proves a root only when its conditions hold. There are three ways it lets you down.

The graph is not continuous. The function $f(x) = \dfrac{1}{x}$ has $f(-1) = -1$ and $f(1) = 1$, so the signs differ. But $1/x$ is never zero, and the graph jumps from very negative to very positive across the break at $x = 0$. There is no root in $(-1, 1)$.

The root is a touch, not a crossing. The function $f(x) = (x - 1)^2$ has a root at $x = 1$, but it meets the axis and turns back. It is positive on both sides, so no interval has a change of sign. A root like this is a repeated root.

An even number of roots hides the sign change. For this function the values at $-1$ and $2$ are both positive, yet the graph crosses the axis twice in between. Two crossings return the sign to where it started, so the test finds nothing.

There is a fourth case to watch. If $f(a) = 0$ at an end of the interval, then $a$ itself is a root and no sign change is needed.

Watch out. Three mistakes are common.

Reading a sign change as a promise of exactly one root. A sign change proves at least one root. A cubic can cross the axis three times inside one interval. Re-check: after finding a bracket, sketch or plot the curve to see how many crossings it holds.

Testing only the two ends of a wide interval. If the sign is the same at both ends, do not conclude that there is no root. There may be two, four or any even number of crossings. Narrow the interval and test again.

Using a function with a break. The test needs continuity. Check that the function is defined and unbroken across the whole interval before you rely on the sign.

6. Try it: bracketing a root

Predict first. The figure shows the curve $y = x^3 - 3x + 1$. Before you move the sliders, predict whether the sign at $x = -1$ and the sign at $x = 2$ are the same or different. Write your prediction down.

Test. Move the sliders for the two ends of the interval. Watch the values $f(a)$ and $f(b)$ in the readout, and watch the message about the signs.

Explain. Write two sentences. The first explains why two opposite signs guarantee a root between them. The second names an interval where the signs are the same but the graph still crosses the axis twice.

Progress check

Return to the chapter problem. The drone's height is $h(t) = 40e^{-t/4} - 2t$.

1. Work out $h(5)$ and $h(6)$. Give each value to two decimal places.
2. Use the change of sign to show that the drone lands between $5$ and $6$ seconds.
3. Work out $h(5.3)$ and $h(5.31)$. What does the sign tell you about the landing time?

You should now be able to

• explain what a root of $f(x) = 0$ is and where it appears on a graph;
• substitute two values into a function and compare their signs;
• use a change of sign to bracket a root;
• narrow a bracket by testing the midpoint;
• explain why a repeated root, a discontinuity or an even number of roots can defeat the test;
• state that a change of sign proves at least one root, not exactly one.