then $\mathrm{f}(x) = 0$ has at least one root between $a$ and $b$
2. Say that one is negative and the other positive
3. State that $\mathrm{f}$ is continuous
4. Conclude that a root lies between them
Show that $x^3 - 4x - 1 = 0$ has a root between $2$ and $3$.
Find the root of $x^3 - 4x - 1 = 0$ between $2$ and $3$, correct to $2$ decimal places.
Each step halves the interval containing the root.
The equation $x^3 + x - 4 = 0$ has a root in $(1, 2)$. Use bisection four times to narrow the interval.
A student claims $\mathrm{f}(x) = \dfrac{1}{x-2}$ has a root between $1$ and $3$, because $\mathrm{f}(1) = -1$ and $\mathrm{f}(3) = 1$. Explain the error.
$\mathrm{f}(x) = x^2 - 4x + 3$. Show that $\mathrm{f}(0)$ and $\mathrm{f}(4)$ have the same sign, yet there are two roots in $(0, 4)$.
Explain why a change of sign search will never find the root of $(x-3)^2 = 0$.
The test
Opposite signs at the ends of the interval.
Condition
The function must be continuous.
Write it out
Values, signs, continuity, conclusion.
Decimal search
Step through tenths, then hundredths.
Midpoint check
Needed to settle the rounding.
Bisection
Halves the interval every step.
After $n$ steps
Interval is $\dfrac{1}{2^n}$ of the original.
Fails at asymptotes
Sign change but no root.
Fails for two roots
Same sign at both ends.
Fails for touching roots
Root with no sign change.
Show that $x^3 - 7 = 0$ has a root between $1$ and $2$.
▶ Show solution
$\mathrm{f}(1) = 1 - 7 = -6 < 0$
$\mathrm{f}(2) = 8 - 7 = 1 > 0$
The sign changes and $\mathrm{f}$ is a continuous polynomial, so a root lies in $(1, 2)$.
Show that $x^3 + 3x - 10 = 0$ has a root between $1$ and $2$.
▶ Show solution
$\mathrm{f}(1) = 1 + 3 - 10 = -6 < 0$
$\mathrm{f}(2) = 8 + 6 - 10 = 4 > 0$
Sign change with a continuous function, so a root lies in $(1, 2)$.
For $\mathrm{f}(x) = x^3 - 7$, evaluate $\mathrm{f}(1.9)$ and $\mathrm{f}(1.95)$, and hence give the root to $1$ d.p.
▶ Show solution
$\mathrm{f}(1.9) = 6.859 - 7 = -0.141 < 0$
$\mathrm{f}(1.95) = 7.415 - 7 = 0.415 > 0$
The root is in $(1.9, 1.95)$, so it rounds to $\mathbf{1.9}$ to $1$ d.p.
The root of $x^3 + x - 6 = 0$ lies in $(1, 2)$. Perform one bisection step.
▶ Show solution
$\mathrm{f}(1.5) = 3.375 + 1.5 - 6 = -1.125 < 0$
$\mathrm{f}(2) = 8 + 2 - 6 = 4 > 0$
So the root now lies in $(1.5, 2)$.
Starting from an interval of width $1$, how many bisections are needed to locate a root to within $0.01$?
▶ Show solution
After $n$ steps the width is $\dfrac{1}{2^n}$.
$\dfrac{1}{2^6} = 0.015625$ — not yet small enough.
$\dfrac{1}{2^7} = 0.0078125 < 0.01$ ✓
So $7$ bisections are needed.
Explain why $\mathrm{f}(x) = \dfrac{1}{x+1}$ having a sign change between $x = -2$ and $x = 0$ does not prove there is a root.
▶ Show solution
$\mathrm{f}(-2) = -1$ and $\mathrm{f}(0) = 1$, so the signs do differ.
But $\mathrm{f}$ has an asymptote at $x = -1$, so it is not continuous on $[-2, 0]$.
The change of sign test requires continuity, so no conclusion can be drawn. In fact $\dfrac{1}{x+1}$ is never zero.
Show that $\mathrm{f}(x) = x^2 - 6x + 8$ has the same sign at $x = 1$ and $x = 5$, and explain what this hides.
▶ Show solution
$\mathrm{f}(1) = 1 - 6 + 8 = 3 > 0$
$\mathrm{f}(5) = 25 - 30 + 8 = 3 > 0$
No sign change — yet $\mathrm{f}(x) = (x-2)(x-4)$ has roots at $x = 2$ and $x = 4$, both inside $(1,5)$.
An even number of roots leaves the end signs unchanged, so the test misses them entirely. Testing $\mathrm{f}(3) = -1 < 0$ would reveal both.
Show that $x^3 - 6x^2 + 9x - 2 = 0$ has a root between $0$ and $1$, and find it to $1$ d.p.
▶ Show solution
$\mathrm{f}(0) = -2 < 0$ and $\mathrm{f}(1) = 1 - 6 + 9 - 2 = 2 > 0$, so a root lies in $(0,1)$.
$\mathrm{f}(0.2) = 0.008 - 0.24 + 1.8 - 2 = -0.432 < 0$
$\mathrm{f}(0.3) = 0.027 - 0.54 + 2.7 - 2 = 0.187 > 0$
Root in $(0.2, 0.3)$. Midpoint: $\mathrm{f}(0.25) = 0.016 - 0.375 + 2.25 - 2 = -0.109 < 0$.
So the root is in $(0.25, 0.3)$, giving $\mathbf{0.3}$ to $1$ d.p.
Explain why a change of sign search cannot find the root of $x^2 - 6x + 9 = 0$.
▶ Show solution
$x^2 - 6x + 9 = (x-3)^2$, which is a perfect square.
So the function is never negative, and its only zero is at $x = 3$, where the curve touches the axis rather than crossing it.
With no sign change anywhere, the method has nothing to detect. A repeated root is always invisible to it.
$\mathrm{f}(x) = x^3 - 3x^2 + 1$.
(a) Evaluate $\mathrm{f}(x)$ at $x = -1, 0, 1, 2, 3$. (b) State how many roots the equation $\mathrm{f}(x)=0$ has, and locate each between consecutive integers. (c) Find the largest root to $2$ d.p. (d) Explain how you know you have found all the roots.
▶ Show solution
(a)
| $x$ | $-1$ | $0$ | $1$ | $2$ | $3$ |
|---|---|---|---|---|---|
| $\mathrm{f}(x)$ | $-3$ | $1$ | $-1$ | $-3$ | $1$ |
(b) There are three sign changes: between $-1$ and $0$, between $0$ and $1$, and between $2$ and $3$.
So the equation has three roots, one in each of $(-1, 0)$, $(0, 1)$ and $(2, 3)$.
(c) Searching in $(2,3)$:
$\mathrm{f}(2.8) = 21.952 - 23.52 + 1 = -0.568 < 0$
$\mathrm{f}(2.9) = 24.389 - 25.23 + 1 = 0.159 > 0$
Root in $(2.8, 2.9)$. Refining:
$\mathrm{f}(2.87) = 23.640 - 24.711 + 1 = -0.071 < 0$
$\mathrm{f}(2.88) = 23.888 - 24.883 + 1 = 0.005 > 0$
Root in $(2.87, 2.88)$. Midpoint: $\mathrm{f}(2.875) = 23.764 - 24.797 + 1 = -0.033 < 0$.
So the root lies in $(2.875, 2.88)$ and rounds to $\mathbf{2.88}$ to $2$ d.p.
(d) $\mathrm{f}$ is a cubic, and a cubic has at most three roots. We have located three, so there cannot be any more.
This is the key safeguard: the change of sign test alone can never prove you have found them all — it is the degree of the polynomial that caps the count. Without that argument, extra roots could always be hiding between the points you happened to test.