Papers 2 and 3. Equations such as \(\ln x = 3 - x\) cannot be solved with algebra. A sketch tells you how many roots there are and roughly where; a change of sign in a continuous function pins each root between two numbers, as tightly as you need.
Write the equation as \(f(x) = 0\). If \(f\) is continuous on \([a, b]\) and \(f(a)\) and \(f(b)\) have opposite signs, then there is at least one root between \(a\) and \(b\). To show a root is \(\alpha\) correct to 2 d.p., show the sign changes between \(\alpha - 0.005\) and \(\alpha + 0.005\).
Fig. 1: y = ln x and y = 3 − x cross exactly once, so ln x = 3 − x has one root α, between 2 and 3 (α ≈ 2.21)
Section 01
Sketch first: how many roots?
In the lab, choose a curve and move the line across it. Watch the number of crossing points change: each crossing is one root of the equation curve = line. That is exactly how you count the roots of an equation like \(\ln x = 3 - x\): sketch both sides on one diagram and count the intersections.
Split the equation into two familiar graphs: \(e^{-x} = x\) becomes \(y = e^{-x}\) and \(y = x\).
Count intersections: the number of crossing points is the number of real roots.
Read rough positions: the \(x\)-coordinate of each crossing shows roughly where the root lies, for example between 2 and 3.
Worked example Graphs and a sign change
By sketching suitable graphs, show that the equation \(\ln x = 3 - x\) has exactly one real root. Verify by calculation that this root lies between 2 and 3.
Sketch \(y = \ln x\) (increasing, through \((1, 0)\)) and \(y = 3 - x\) (decreasing straight line). An increasing curve and a decreasing line cross exactly once (Fig. 1).
Let \(f(x) = \ln x + x - 3\), continuous for \(x > 0\).
The sign changes and \(f\) is continuous, so the root lies between 2 and 3.
Answer: One root, \(2 < \alpha < 3\)
Examiner trap
In a sketch-and-count question the marks are for two correct graph shapes in the right places AND a stated conclusion ("one point of intersection, so one root"). A sketch with no statement does not earn the final mark.
Section 02
The change of sign test.
Rearrange so that one side is zero. A continuous graph cannot get from below the \(x\)-axis to above it without crossing it, so a sign change between \(a\) and \(b\) traps a root. Always give the values (to 2 or 3 s.f.) and write a conclusion.
Worked example Between consecutive integers
Show that the equation \(x^3 - 4x + 1 = 0\) has a root between 1 and 2.
Let \(f(x) = x^3 - 4x + 1\), a polynomial, so continuous.
\(f(1) = 1 - 4 + 1 = -2 < 0\).
\(f(2) = 8 - 8 + 1 = 1 > 0\).
Change of sign, so there is a root between 1 and 2.
Answer: \(f(1) = -2\), \(f(2) = 1\): sign change, so a root lies in \((1, 2)\)
Worked example Radians
Show that the equation \(x = 2\cos x\) has a root between 1 and 1.1.
Let \(f(x) = x - 2\cos x\), continuous. Use radians.
Show that the equation \(x^3 + x = 5\) has a root between 1.5 and 1.6.
\(f(x) = x^3 + x - 5\). \(f(1.5) = 3.375 + 1.5 - 5 = -0.125\) and \(f(1.6) = 4.096 + 1.6 - 5 = 0.696\). Sign change and \(f\) continuous, so a root lies between 1.5 and 1.6.
Examiner trap
Compare with zero, not with each other: you must rearrange to \(f(x) = 0\) first. Checking that \(\ln 2 < 3 - 2\) and \(\ln 3 > 3 - 3\) is fine, but only if you say clearly that the sign of \(\ln x - (3 - x)\) has changed.
Stage 1 · Learn
Check your understanding
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A continuous function has \(f(1) = -3\) and \(f(2) = 4\). What can you conclude?
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Why does \(f(x) = \tan x\) changing sign between 1.5 and 1.7 not prove a root there?
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How many real roots does \(e^{-x} = x\) have?
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To show that a root is 2.34 correct to 2 decimal places, evaluate \(f\) at
Section 03
Accuracy, and when the test fails.
To confirm a root to a given accuracy, test the bounds of the rounding interval. And remember the test can mislead: a sign change needs a continuous function, and some roots produce no sign change at all.
Worked example Correct to 2 decimal places
The equation \(e^{x} = 3x\) has a root \(\alpha\) between 1 and 2. Show that \(\alpha = 1.51\) correct to 2 decimal places.
Let \(f(x) = e^{x} - 3x\). If \(\alpha = 1.51\) to 2 d.p., then \(1.505 \le \alpha < 1.515\).
\(f(1.505) = e^{1.505} - 4.515 = -0.0108\).
\(f(1.515) = e^{1.515} - 4.545 = 0.00442\).
Sign change, so \(1.505 < \alpha < 1.515\), and \(\alpha\) rounds to 1.51.
Answer: \(\alpha = 1.51\) (2 d.p.)
Worked example Sign change, but no root
For \(f(x) = \dfrac{1}{x-2}\), \(f(1) = -1\) and \(f(3) = 1\). Does \(f(x) = 0\) have a root between 1 and 3?
\(\dfrac{1}{x-2}\) is never zero: a fraction is zero only when its numerator is zero.
The sign changes because the graph jumps across a vertical asymptote at \(x = 2\), where \(f\) is not continuous.
Likewise \(\tan x\) changes sign across \(x = \dfrac{\pi}{2}\) without a root there.
Answer: No: the function is not continuous on \([1, 3]\), so the test does not apply
Examiner trap
No sign change does NOT mean no root. \(f(x) = (x-1)^2\) has a root at \(x = 1\) but is positive on both sides; and two roots between \(a\) and \(b\) cancel out the sign change. A sketch protects you from both traps.
Stage 2 · Exam
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The equation \(x^3 - 2x - 5 = 0\) has one real root. In which interval does it lie?
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How many real roots does the equation \(x^2 = 2^{x}\) have?
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Which statement about \(f(x) = (x - 3)^2\) on the interval \([2, 4]\) is correct?
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The equation \(x^3 + 2x - 7 = 0\) has a root \(\alpha\). Which calculation shows that \(\alpha = 1.57\) correct to 2 decimal places?
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