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 $$f (x) = 2$$.
See development and solution
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:
$$x=1.6$$