# Problems from Inverse interpolation

We know the following points of the function $$f (x)$$:

 $$x$$ $$0$$ $$1$$ $$2$$ $$3$$ $$f(x)$$ $$0$$ $$1$$ $$4$$ $$9$$

Find the value of x $$x$$ such that $$f (x) = 2$$.

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### Development:

We have a case of inverse interpolation. Therefore we write the points in the table but exchanging the columns.

 $$0$$ $$0$$ $$\dfrac{1-0}{1-0}=1$$ $$1$$ $$1$$ $$\dfrac{\dfrac{1}{3}-1}{4-0}=\dfrac{\dfrac{-2}{3}}{4}=-\dfrac{1}{6}$$ $$\dfrac{2-1}{4-1}=\dfrac{1}{3}$$ $$\dfrac{-\dfrac{1}{60}+\dfrac{1}{6}}{9-0}=\dfrac{\dfrac{3}{20}}{9}= \dfrac{1}{60}$$ $$4$$ $$2$$ $$\dfrac{\dfrac{1}{5}-\dfrac{1}{3}}{9-1}=\dfrac{\dfrac{-2}{15}}{8}= -\dfrac{1}{60}$$ $$\dfrac{3-2}{9-4}=\dfrac{1}{5}$$ $$9$$ $$3$$

$$\begin{array}{rl} P_3(y)=& 0+1\cdot (y-0)-\dfrac{1}{6}(y-0)(y-1)+\dfrac{1}{60}(y-0)(y-1)(y-4)\\ =&y-\dfrac{1}{6}y^2+\dfrac{1}{6}y+\dfrac{1}{60}y^3-\dfrac{1}{60}y^2-\dfrac{1}{15}y\\ =& \dfrac{1}{60}y^3-\dfrac{1}{4}y^2+\dfrac{37}{30}y \end{array}$$\$

We evaluate in $$2$$:

$$x\approx P_3(2)=\dfrac{1}{60}\cdot8-\dfrac{1}{4}\cdot4+\dfrac{37}{30}\cdot2= \dfrac{8}{5}=1.6$$

### Solution:

$$x=1.6$$

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