We want to determine approximately a zero for the polynomial
-
![{\displaystyle {}f(x)=x^{3}-4x+2\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/88088312a0ccee7195363023c453f44516a64238)
with the help of the method given by the Intermediate value theorem. We have
and
,
hence by
fact,
there must be a zero inside the interval
. We compute the value of the function at the arithmetic mean of the interval, which is
, and get
-
![{\displaystyle {}f{\left({\frac {3}{2}}\right)}={\frac {27}{8}}-4\cdot {\frac {3}{2}}+2={\frac {27-48+16}{8}}={\frac {-5}{8}}<0\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7b7a99b4c954af28444bce679195f6fb05ef688c)
Hence, we have to continue with the right half of the interval
. The arithmetic mean thereof is
. The value of the function at this point is
-
![{\displaystyle {}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\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/40ae599d036c8e781d91d1e765627bcdcdfdf79b)
So now we have to continue with the left part of the interval, namely
. Its arithmetic mean is
. The value of the function at this point is
-
![{\displaystyle {}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\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cf8da42cdfb04671b3729db6164a95ffb299af16)
Therefore, we know that there is a zero between
and
.