# Intermediate value theorem/x^3-4x+2/Bisection method/Example

We want to determine approximately a zero for the polynomial

${}f(x)=x^{3}-4x+2\,$ with the help of the method given by the intermediate value theorem. We have ${}f(1)=-1$ and ${}f(2)=2$ , hence by fact there must be a zero inside the interval ${}[1,2]$ . We compute the value of the function at the arithmetic mean of the interval, which is ${}{\frac {3}{2}}$ , and get

${}f{\left({\frac {3}{2}}\right)}={\frac {27}{8}}-4\cdot {\frac {3}{2}}+2={\frac {27-48+16}{8}}={\frac {-5}{8}}<0\,.$ Hence we have to continue with the right half of the interval ${}[{\frac {3}{2}},2]$ . The arithmetic mean thereof is ${}{\frac {7}{4}}$ . The value of the function at this point is

${}f{\left({\frac {7}{4}}\right)}={\left({\frac {7}{4}}\right)}^{3}-4\cdot {\frac {7}{4}}+2={\frac {343}{64}}-5={\frac {343-320}{64}}={\frac {23}{64}}>0\,.$ So now we have to continue with the left part of the interval, namely ${}[{\frac {3}{2}},{\frac {7}{4}}]$ . Its arithmetic mean is ${}{\frac {13}{8}}$ . The value of the function at this point is

${}f{\left({\frac {13}{8}}\right)}={\left({\frac {13}{8}}\right)}^{3}-4\cdot {\frac {13}{8}}+2={\frac {2197}{512}}-{\frac {13}{2}}+2={\frac {2197-3328+1024}{512}}={\frac {-107}{512}}<0\,.$ Therefore we know that there is a zero between ${}{\frac {13}{8}}$ and ${}{\frac {7}{4}}={\frac {14}{8}}$ .