Most equations that come out of real problems โ $x^3 - 3x - 1 = 0$, $\mathrm{e}^{-x} = x$, $x\sin x = 1$ โ have no neat algebraic solution. We can still pin the answer down as accurately as we like. The first step is always to trap the root inside a small interval.
then $\mathrm{f}(x) = 0$ has at least one root $\alpha$ with $a < \alpha < b$
The reason is visual: a continuous graph has no gaps or jumps, so to get from below the axis to above it the curve must cross the axis somewhere in between.
Exam questions say "Show that the equation โฆ has a root between $a$ and $b$". A full answer has three parts, and the third is the one people forget.
- Rearrange the equation into the form $\mathrm{f}(x) = 0$ and define $\mathrm{f}(x)$.
- Calculate $\mathrm{f}(a)$ and $\mathrm{f}(b)$ and write down their values (a sign alone is not enough).
- Conclude: "There is a change of sign and $\mathrm{f}$ is continuous for $a \leqslant x \leqslant b$, so there is a root between $a$ and $b$."
Show that $x^3 - 3x - 1 = 0$ has a root $\alpha$ with $1 < \alpha < 2$.
Show that the equation $x \sin x = 1$ has a root between $x = 1$ and $x = 1.2$.
By sketching $y = 2^x$ and $y = 3x$, explain why $2^x = 3x$ has exactly two roots, and locate each between consecutive integers.
Once a root is trapped, you can shrink the interval by testing values in between โ a decimal search. Each new sign check tells you which half of the interval the root is in.
Find the root of $x^3 - 3x - 1 = 0$ between $1$ and $2$ correct to $2$ decimal places.
| $x$ | $1.8$ | $1.9$ | $1.87$ | $1.88$ | $1.875$ |
|---|---|---|---|---|---|
| $\mathrm{f}(x)$ | $-0.568$ | $0.159$ | $-0.0708$ | $0.00467$ | $-0.0332$ |
Show that $1.796$ is a root of $x^3 - x - 4 = 0$ correct to $3$ decimal places.
Show that the equation $\mathrm{e}^{-x} = x$ has a root $\alpha = 0.567$ correct to $3$ decimal places.
The principle has two halves: a sign change with continuity guarantees a root, and no sign change guarantees nothing. Both halves can trip you up.
| Situation | What goes wrong | Example |
|---|---|---|
| Vertical asymptote | Sign changes but there is no root โ the function is not continuous. | $\dfrac{1}{x-2}$ for $1 \leqslant x \leqslant 3$ |
| Repeated root | The curve touches the axis without crossing, so there is no sign change. | $(x-1)^2(x+2)$ for $0.5 \leqslant x \leqslant 1.5$ |
| Two (or an even number of) roots close together | The sign changes twice, so the end values have the same sign. | $(x-0.3)(x-0.7)$ for $0 \leqslant x \leqslant 1$ |
| Interval too wide | A sign change may hide three roots, not one. | $x^3 - 3x - 1$ for $-2 \leqslant x \leqslant 2$ |
$\mathrm{f}(x) = \dfrac{1}{x - 2}$. Show that $\mathrm{f}(1)$ and $\mathrm{f}(3)$ have opposite signs, and explain why $\mathrm{f}(x) = 0$ has no root with $1 \leqslant x \leqslant 3$.
$\mathrm{f}(x) = x^3 - 3x + 2$. Find $\mathrm{f}(0.5)$ and $\mathrm{f}(1.5)$, and comment.
The principle
$\mathrm{f}$ continuous for $a \leqslant x \leqslant b$ and $\mathrm{f}(a)\mathrm{f}(b) < 0$ $\Rightarrow$ a root with $a < x < b$.
Set up
Rearrange to $\mathrm{f}(x) = 0$ before testing signs.
Show the values
Write $\mathrm{f}(a)$ and $\mathrm{f}(b)$ as numbers, then conclude.
Conclusion
"Change of sign, $\mathrm{f}$ continuous, so a root lies in the interval."
Accuracy
For $\alpha = 1.796$ (3 d.p.), test $1.7955$ and $1.7965$.
Radians
Always, for trigonometric functions.
Asymptote
Sign change but no root โ discontinuity.
Repeated root
Root but no sign change โ the curve touches.
Sketch first
Tells you how many roots to look for.
Show that $x^3 - 5x + 1 = 0$ has a root between $x = 2$ and $x = 3$.
โถ Show solution
Let $\mathrm{f}(x) = x^3 - 5x + 1$.
$\mathrm{f}(2) = 8 - 10 + 1 = -1 < 0$ and $\mathrm{f}(3) = 27 - 15 + 1 = 13 > 0$.
There is a change of sign and $\mathrm{f}$ is continuous, so there is a root between $2$ and $3$.
Show that the equation $\mathrm{e}^x = 4 - x$ has a root with $1 < x < 1.1$.
โถ Show solution
Let $\mathrm{f}(x) = \mathrm{e}^x + x - 4$.
$\mathrm{f}(1) = \mathrm{e} - 3 = -0.2817\ldots < 0$
$\mathrm{f}(1.1) = \mathrm{e}^{1.1} - 2.9 = 0.1042\ldots > 0$
Change of sign and $\mathrm{f}$ is continuous, so there is a root with $1 < x < 1.1$.
Show that the equation $\ln x = 3 - x$ has a root between $2$ and $3$.
โถ Show solution
Let $\mathrm{f}(x) = \ln x + x - 3$, which is continuous for $x > 0$.
$\mathrm{f}(2) = \ln 2 - 1 = -0.3068\ldots < 0$ and $\mathrm{f}(3) = \ln 3 = 1.0986\ldots > 0$.
Change of sign, so there is a root with $2 < x < 3$.
Show that the equation $x\sin x = 1$ has a root $\alpha = 1.11$ correct to $2$ decimal places.
โถ Show solution
Let $\mathrm{f}(x) = x\sin x - 1$ (radians). Test the bounds $1.105$ and $1.115$.
$\mathrm{f}(1.105) = -0.0127\ldots < 0$ and $\mathrm{f}(1.115) = 0.00117\ldots > 0$.
Change of sign and $\mathrm{f}$ is continuous, so $1.105 < \alpha < 1.115$, giving $\alpha = 1.11$ to $2$ d.p.
Show that $x^3 - 2x - 5 = 0$ has a root $\alpha = 2.095$ correct to $3$ decimal places.
โถ Show solution
Let $\mathrm{f}(x) = x^3 - 2x - 5$. Test $2.0945$ and $2.0955$.
$\mathrm{f}(2.0945) = -0.000575\ldots < 0$
$\mathrm{f}(2.0955) = 0.0106\ldots > 0$
Change of sign, $\mathrm{f}$ continuous, so $2.0945 < \alpha < 2.0955$ and $\alpha = 2.095$ to $3$ d.p.
Note: $\mathrm{f}(2.0945)$ is very small โ the root ($2.09455\ldots$) is only just above the lower bound. That is why you must work to enough significant figures.
A student finds that $\tan 1 > 0$ and $\tan 2 < 0$ and concludes that $\tan x = 0$ has a root between $1$ and $2$. Explain the error.
โถ Show solution
$\tan 1 = 1.557\ldots$ and $\tan 2 = -2.185\ldots$, so the signs are indeed different.
But $\tan x$ has a vertical asymptote at $x = \dfrac{\pi}{2} = 1.5707\ldots$, which lies between $1$ and $2$. The function is not continuous for $1 \leqslant x \leqslant 2$, so the change of sign principle does not apply.
In fact the roots of $\tan x = 0$ are $x = 0, \pm\pi, \ldots$ โ none lies between $1$ and $2$.
$\mathrm{f}(x) = 4x^2 - 4x + 1$. Show that $\mathrm{f}(0)$ and $\mathrm{f}(1)$ have the same sign. Does this mean there is no root with $0 \leqslant x \leqslant 1$?
โถ Show solution
$\mathrm{f}(0) = 1$ and $\mathrm{f}(1) = 4 - 4 + 1 = 1$, both positive.
No: $\mathrm{f}(x) = (2x - 1)^2$, which has a repeated root at $x = 0.5$. The graph touches the axis there without crossing, so there is no sign change even though there is a root.
$\mathrm{f}(x) = x^3 - 3x - 1$. Given $\mathrm{f}(-2) = -3$ and $\mathrm{f}(2) = 1$, a student says "there is one root between $-2$ and $2$". By evaluating $\mathrm{f}(-1)$ and $\mathrm{f}(0)$, show that there are in fact three roots, and give an interval of width $1$ for each.
โถ Show solution
$\mathrm{f}(-1) = -1 + 3 - 1 = 1$ and $\mathrm{f}(0) = -1$.
| $x$ | $-2$ | $-1$ | $0$ | $1$ | $2$ |
|---|---|---|---|---|---|
| $\mathrm{f}(x)$ | $-3$ | $1$ | $-1$ | $-3$ | $1$ |
Sign changes for $-2 < x < -1$, for $-1 < x < 0$ and for $1 < x < 2$: three roots. A cubic has at most three, so these are all of them.
The sign change between $-2$ and $2$ only guarantees at least one root.
The equation $2^x = 3x$ has two roots. Show that the larger root lies between $3.3$ and $3.4$.
โถ Show solution
Let $\mathrm{f}(x) = 2^x - 3x$.
$\mathrm{f}(3.3) = 2^{3.3} - 9.9 = -0.0508\ldots < 0$
$\mathrm{f}(3.4) = 2^{3.4} - 10.2 = 0.356\ldots > 0$
Change of sign and $\mathrm{f}$ continuous, so a root lies in $3.3 < x < 3.4$. (The other root satisfies $0 < x < 1$, as in Worked Example 3.)
The amount of a drug in a patient's bloodstream, in mg, $t$ hours after an injection is modelled by $D = 20t\,\mathrm{e}^{-0.5t}$. The drug is effective while $D \geqslant 10$.
(a) Show that $D = 10$ when $20t\,\mathrm{e}^{-0.5t} - 10 = 0$ has a root between $t = 0.7$ and $t = 0.8$.
(b) Show that there is a second root between $t = 4$ and $t = 5$.
(c) Show that the first root is $0.71$ correct to $2$ decimal places.
(d) Given that the second root is $4.31$ to $2$ d.p., estimate for how long the drug is effective.
โถ Show solution
Let $\mathrm{g}(t) = 20t\,\mathrm{e}^{-0.5t} - 10$, which is continuous.
(a) $\mathrm{g}(0.7) = -0.134\ldots < 0$ and $\mathrm{g}(0.8) = 0.725\ldots > 0$. Change of sign, so a root lies in $0.7 < t < 0.8$.
(b) $\mathrm{g}(4) = 0.827\ldots > 0$ and $\mathrm{g}(5) = -1.79\ldots < 0$. Change of sign, so a root lies in $4 < t < 5$.
(c) $\mathrm{g}(0.705) = -0.0887\ldots < 0$ and $\mathrm{g}(0.715) = 0.00174\ldots > 0$. Change of sign, so the root lies in $0.705 < t < 0.715$ and equals $0.71$ to $2$ d.p.
(d) $D \geqslant 10$ between the two roots, so the drug is effective for about $4.31 - 0.71 = 3.6$ hours.
Check: at $t = 2$ (between the roots) $D = 40\mathrm{e}^{-1} = 14.7 > 10$ โ