Using change of sign to locate roots of equations, interval bisection.
Answer at least 3 of 3 correctly to complete this section.
Many equations cannot be solved exactly. Try solving x3−3x+1=0 or ex=3x algebraically — you will not find a neat closed-form answer. Yet these equations have real roots that we might need to know for engineering, physics, or further mathematics.
Numerical methods give us systematic ways to find roots to any desired accuracy. The change of sign method is the simplest and most intuitive: if a continuous function is positive at one point and negative at another, it must cross zero somewhere in between.
This topic appears in every A2 exam paper and is worth reliable marks once you learn the method.
If f(x) is continuous on the interval [a,b] and f(a) and f(b) have opposite signs, then f(x)=0 has at least one root in the interval (a,b).
This is a consequence of the Intermediate Value Theorem: a continuous function that goes from negative to positive (or vice versa) must pass through zero.
Show that the equation x3−3x+1=0 has a root in the interval [1,2].
Let f(x)=x3−3x+1.
f(1)=1−3+1=−1
f(2)=8−6+1=3
Since f(1)=−1<0 and f(2)=3>0, there is a change of sign.
f(x) is a polynomial, so it is continuous on [1,2].
Therefore, f(x)=0 has at least one root in the interval (1,2). □
Exam essential: Always write three things: (1) the values of f(a) and f(b), (2) “change of sign”, and (3) ”f is continuous”. Missing any of these can cost marks.
Show that the equation ex=3x has a root between x=0 and x=1.
Rearrange to f(x)=ex−3x=0.
f(0)=e0−0=1>0
f(1)=e1−3=2.718…−3=−0.282…<0
Change of sign, and f(x)=ex−3x is continuous (exponential and polynomial functions are both continuous).
Therefore, ex=3x has a root in (0,1). □
Answer at least 4 of 5 correctly to complete this section.
The change of sign method does not always work. You need to know when and why it fails.
1. Discontinuities
f(x)=x1 has f(−1)=−1 and f(1)=1 — a change of sign. But f is not continuous on [−1,1] (it has a vertical asymptote at x=0), and there is no root.
2. Repeated (double) roots
f(x)=(x−2)2 touches the x-axis at x=2 but does not cross it. Since f(x)≥0 for all x, there is never a change of sign, even though x=2 is a root.
3. Multiple roots in an interval
f(x)=x2−1 has roots at x=−1 and x=1. But f(−2)=3>0 and f(2)=3>0 — no change of sign, because the function crosses zero an even number of times.
| Situation | What happens | Change of sign? |
|---|---|---|
| One root in [a,b], f continuous | Method works | Yes |
| Odd number of roots | Method detects a root exists | Yes |
| Even number of roots | Method misses them | No |
| Repeated root (touches axis) | Method fails | No |
| Discontinuity in [a,b] | Method is invalid | Possibly, but misleading |
Exam tip: If a question asks “explain why the change of sign method fails here”, look for one of these three cases. Draw a sketch to help you see what is happening.
Answer at least 3 of 3 correctly to complete this section.
Interval bisection is a systematic way to narrow down the location of a root. Starting with an interval [a,b] where f(a) and f(b) have opposite signs:
Each step halves the width of the interval, so after n steps the interval has width 2nb−a.
The equation x3−3x+1=0 has a root in [1,2]. Use interval bisection to find this root correct to 1 decimal place.
Let f(x)=x3−3x+1.
We already know: f(1)=−1<0 and f(2)=3>0.
Step 1: Midpoint c=1.5.
f(1.5)=3.375−4.5+1=−0.125<0
Sign change between f(1.5) and f(2), so the root is in [1.5,2].
Step 2: Midpoint c=1.75.
f(1.75)=5.359375−5.25+1=1.109375>0
Sign change between f(1.5) and f(1.75), so the root is in [1.5,1.75].
Step 3: Midpoint c=1.625.
f(1.625)=4.291…−4.875+1=0.416…>0
Sign change between f(1.5) and f(1.625), so the root is in [1.5,1.625].
Step 4: Midpoint c=1.5625.
f(1.5625)=3.815…−4.6875+1=0.127…>0
Sign change between f(1.5) and f(1.5625), so the root is in [1.5,1.5625].
Since both endpoints round to 1.5 to 1 decimal place, the root is 1.5 (to 1 d.p.).
To guarantee d decimal places of accuracy, you need the interval width to be less than 21×10−d.
After n bisections, the interval width is 2nb−a. Solving:
2nb−a<21×10−d
Why bisection over trial and improvement? Bisection is guaranteed to converge and halves the error at every step. Trial and improvement can be faster if you make clever guesses, but bisection is systematic and reliable — examiners love systematic methods.
Answer at least 3 of 3 correctly to complete this section.
Lock in what you've learned with exam-style questions and spaced repetition.
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Using the Newton-Raphson formula to find approximate roots, graphical interpretation.