Locating roots by change of signAQA A-Level Maths: Revision notes

A root of $f(x)=0$ is a value of $x$ at which $f(x)=0$. If $f$ is continuous on $[a,b]$ and $f(a)$ and $f(b)$ have opposite signs, then $f(x)=0$ has at least one root in the interval $a<x<b$. Example: for $f(x)=x^3-5x+3$, $f(1)=-1$ and $f(2)=1$, so there is a root between 1 and 2.

A complete answer has four parts: (1) define $f(x)$ with the right-hand side $0$; (2) work out $f(a)$ and $f(b)$ accurately; (3) state that $f$ is continuous on the interval; (4) conclude that there is a change of sign, so a root lies in the interval. Example: $f(x)=e^x-3x$. $f(0.6)=0.0221>0$ and $f(0.7)=-0.0862<0$. $f$ is continuous and changes sign, so a root lies in $0.6<x<0.7$. To give the root to a given accuracy, use smaller and smaller intervals.

Interval bisection repeatedly halves an interval that contains a root. Evaluate $f$ at the midpoint and keep the half where the sign changes. For $f(x)=e^x-3x$ on $[0.6,0.7]$: the midpoint $0.65$ has $f(0.65)=-0.0345<0$, so the root is in $[0.6,0.65]$; the next midpoint $0.625$ gives $f(0.625)=-0.00675<0$, so the root is in $[0.6,0.625]$. Each step halves the interval width.

If $f$ is not continuous on the interval, a change of sign may come from a break rather than a root. $f(x)=\frac{1}{x-1}$ has $f(0)=-1$ and $f(2)=1$, but $f$ is undefined at $x=1$ and $\frac{1}{x-1}$ is never zero. There is a sign change but no root. Functions with asymptotes, such as $\tan x$, cause this.

If $f(a)$ and $f(b)$ have the same sign, there may still be roots. A repeated root touches the axis without crossing: $f(x)=(x-2)^2(x+2)$ has $f(1)=3$ and $f(3)=5$, yet $f(2)=0$. Two roots in the same interval also cancel the sign change, as does a root exactly at an end-point. A sign change shows an odd number of roots; no sign change shows an even number, which could be zero. Close roots are missed if the interval is too wide, so choose a narrower interval when in doubt.