Zero of x³−3x+1 with bisection

Text

Show that $x^3-3x+1=0$ has at least one solution in $[0,1]$ and approximate it with 4 bisection steps.

Solution

Let $f(x)=x^3-3x+1$, a continuous function (a polynomial). At the endpoints: $f(0)=1>0, \qquad f(1)=1-3+1=-1<0.$ Since $f(0)\cdot f(1)<0$, by the zero theorem there exists at least one $c\in(0,1)$ with $f(c)=0$.

Bisection starting from $[0,1]$ (one keeps the interval at whose endpoints $f$ changes sign):

Step	$a$	$b$	$m$	$f(m)$	zero in
1	$0$	$1$	$0{,}5$	$-0{,}375<0$	$[0;\,0{,}5]$
2	$0$	$0{,}5$	$0{,}25$	$+0{,}266>0$	$[0{,}25;\,0{,}5]$
3	$0{,}25$	$0{,}5$	$0{,}375$	$-0{,}072<0$	$[0{,}25;\,0{,}375]$
4	$0{,}25$	$0{,}375$	$0{,}3125$	$+0{,}093>0$	$[0{,}3125;\,0{,}375]$

After 4 steps the zero is in the interval $[0{,}3125;\ 0{,}375]$, of width $0{,}0625$: approximate value $\boxed{x\approx 0{,}34}$, with error $<0{,}0313$. (The exact value is $\approx 0{,}347$.)

Links

Topics: Continuity
Concepts: Bisection algorithm · Zero theorem
Skills: Proving · Estimating
Exercise types: Computing expressions