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

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 ${}{\frac {7}{4}}={\frac {14}{8}}$.