9.1 Locating roots by change of sign — Practise

Work through these on your own. For each bracket, write down the two values and their signs before you decide.

The questions and the answers below come from a set that is checked by computer every time the book is built.

Show that $x^3 + x - 1 = 0$ has a root between $0$ and $1$ .
Show that $x^3 + 3x - 2 = 0$ has a root between $0$ and $1$ .
Show that $e^x = 3x$ has a root between $1$ and $2$ .
Show that $x + \ln x = 0$ has a root between $0.5$ and $0.6$ .
Show that the root of $x^3 + x - 1 = 0$ lies between $0.6$ and $0.7$ .
Show that the root of $x^3 + x - 1 = 0$ lies between $0.68$ and $0.69$ .
For $f(x) = (x - 1)^2$ , work out $f(0)$ and $f(2)$ . Explain why the change of sign does not find the root at $x = 1$ .
For $f(x) = \dfrac{1}{x}$ on the interval $[-1, 1]$ , work out $f(-1)$ and $f(1)$ . Explain why the sign change does not prove a root.
A function has $f(1) = -3$ , $f(2) = 1$ and $f(3) = 4$ . Between which two consecutive values does a root lie?
Show that $x^3 = 20$ has a root between $2$ and $3$ .
Show that $2^x = x + 3$ has a root between $2$ and $3$ .
For $f(x) = x^2 - 9$ on the interval $[3, 4]$ , work out $f(3)$ . What does this tell you about the root?

Answers and concise solutions

1. The two values

Answer: At $x = 0$ the value is $-1 < 0$ , and at $x = 1$ it is $1 > 0$ . The signs differ, so the graph crosses the axis between $0$ and $1$ .

2. The two values

Answer: At $x = 0$ the value is $-2 < 0$ , and at $x = 1$ it is $2 > 0$ . The signs differ, so a root lies in $(0, 1)$ .

3. The two values

Answer: Rearrange to $e^x - 3x = 0$ . At $x = 1$ the value is $e - 3 = -0.2817\ldots < 0$ , and at $x = 2$ it is $e^2 - 6 = 1.3891\ldots > 0$ . The signs differ, so a root lies in $(1, 2)$ .

4. The two values

Answer: At $x = 0.5$ the value is $0.5 + \ln 0.5 = -0.1931\ldots < 0$ , and at $x = 0.6$ it is $0.6 + \ln 0.6 = 0.0892\ldots > 0$ . The signs differ, so a root lies in $(0.5, 0.6)$ .

5. The two values

Answer: At $x = 0.6$ the value is $0.216 + 0.6 - 1 = -0.184 < 0$ , and at $x = 0.7$ it is $0.343 + 0.7 - 1 = 0.043 > 0$ . The sign changes, so the root lies in $(0.6, 0.7)$ .

6. The two values

Answer: At $x = 0.68$ the value is $0.314432 + 0.68 - 1 = -0.005568 < 0$ , and at $x = 0.69$ it is $0.328509 + 0.69 - 1 = 0.018509 > 0$ . The root lies in $(0.68, 0.69)$ .

7. The two values

Answer: $f(0) = 1$ and $f(2) = 1$ , so the sign does not change. The root at $x = 1$ is repeated: the graph touches the axis and turns back, so no interval around it has a sign change.

8. The two values

Answer: $f(-1) = -1$ and $f(1) = 1$ , so the signs differ. But $1/x$ is undefined at $x = 0$ and is never zero, so there is no root. The function is not continuous on $[-1, 1]$ , so the test does not apply.

9. The bracket

Answer: The value at $x = 1$ is negative and the value at $x = 2$ is positive, so the sign changes between them. The root lies in $(1, 2)$ .

10. The two values

Answer: At $x = 2$ , $x^3 = 8 < 20$ . At $x = 3$ , $x^3 = 27 > 20$ . The sign of $x^3 - 20$ changes between the two values, so the root lies in $(2, 3)$ .

11. The two values

Answer: Rearrange to $2^x - x - 3 = 0$ . At $x = 2$ the value is $4 - 5 = -1 < 0$ , and at $x = 3$ it is $8 - 6 = 2 > 0$ . The signs differ, so a root lies in $(2, 3)$ .

12. The value

Answer: $f(3) = 9 - 9 = 0$ , so $x = 3$ is itself a root of $f(x) = 0$ . When an endpoint gives zero, the root is that endpoint and no sign change is needed.