Given $$(x_k,f_k)$$ from a function $$f(x)$$, supose we want to find an approximation of the value of $$x$$ such that $$f(x)=c$$, where $$c$$ is a given value.

We will solve the equation $$x=g(c)$$ where $$g$$ is the inverse function of $$f$$. Then we will interpolate this function $$g(y)$$ and will evaluate it in $$y=c$$, or, in other words, if we use Newton's method we will put in the first column the values $$f_j$$ and in the second one the values $$x_j$$ and proceed the same way.

For example, let's suppose that we want to calculate a zero of the function $$f(x)=x^3-15x+4$$ knowing that this is close to $$x=0.3$$. Then we will do quadratic interpolation, for example, of the inverse of $$f(x)$$. We then first evaluate the function in three points close to $$x=0.3$$:

$$x$$ | $$0.2$$ | $$0.3$$ | $$0.4$$ |

$$f(x)$$ | $$1.008$$ | $$-0.473$$ | $$-1.936$$ |

Now we fill in the table to calculate the divided differences of Newton, but exchanging the columns, obtaining the coefficients of the interpolating polynomial:

$$1.008$$ | $$0.2$$ | ||

$$-0.0675$$ | |||

$$-0.473$$ | $$0.3$$ | $$0.00028963$$ | |

$$-0.0684$$ | |||

$$-1.936$$ | $$0.4$$ |

Thus the interpolating polynomial is:

$$$\begin{array}{rl} P_3(y)=&0.2+0.0675\cdot(y-1.008)+0.00028963\cdot(y-1.008)(y-0.473) \\ =&0.2679019090-0.067654y+0.00028963y^2 \end{array}$$$

So an approximation of the zero of the function is:

$$$P_3(0)=0.2679019090$$$