If f(a) and f(b) have different signs, a continuous graph must cross the axis between them. Narrow the interval to find a root.
What you will learn
- Show a root lies between two values
- Narrow the interval with decimal search
- Give a root to 1 decimal place
- Know when a sign change can mislead
The sign changes, so it crosses
f(x) = x³ + 2x − 7. f(1) = −4 (below the axis) and f(2) = 5 (above).
The graph is unbroken, so it crosses the x-axis between 1 and 2: f(x) = 0 has a root there.
To find where to look, make a table of f(0), f(1), f(2), f(3), ... and find where the sign changes. Here f(1) is negative and f(2) is positive, so the root is between 1 and 2, and the whole number just below it is 1.
To narrow it down, use a calculator to work out f at steps of 0.1 (f(1.1), f(1.2), f(1.3), ...) and find the two values where the sign changes.
- f(1.6) = −0.70 and f(1.7) = 0.31
- so the root is between 1.6 and 1.7
- f(1.65) = −0.21, negative
- so the root is between 1.65 and 1.7: it is 1.7 to 1 d.p.
On a graph of y = f(x), the root of f(x) = 0 is where the curve crosses the x-axis. Read off its x value as an estimate, for example 1.6 to 1 decimal place.
The rest of this lesson and its 7 practice questions are locked.
In this lesson
- The sign changes, so it crosses
- Careful
- Try it: Zoom in on a root
7 practice questions, marked as you go; about 25 minutes in all.
In this unit: Proof and iteration
Algebraic proof and counterexamples, then finding roots that do not come out exactly: changes of sign and iteration. Higher tier.
- Algebraic proof
- Counterexamples
- Roots by a change of sign
- Iteration