Question

6)                 If f(x) takes the values -21, 15, 12 and 3 respectively when x assumes the values -1, 1, 2 and 3, find the polynomial which approximates f(x).

Answer 1

Solution.

Each of the points (-1,-21), (1,15),(2,12),(3,3​) satisfies the equation $p=ax^3+bx^2+cx+d$  for some unknown $a,b,c,d.$  Substitute each point in the equation and make a matrix equation. Let be $A=\begin{pmatrix} -1& 1 &-1&1 \\ 1 & 1 & 1&1\\ 8 & 4 &2& 1\\27&9&3&1 \end{pmatrix},$ $X=\begin{pmatrix} a \\ b\\ c\\d \end{pmatrix} \text{and }B=\begin{pmatrix} -21 \\ 15\\ 12\\3 \end{pmatrix}.$ We will have eqution

$$AX=B, \text{or} \newline\begin{pmatrix} -1& 1 &-1&1 \\ 1 & 1 & 1&1\\ 8 & 4 &2& 1\\27&9&3&1 \end{pmatrix}\cdot \begin{pmatrix} a \\ b\\ c\\d \end{pmatrix} =\begin{pmatrix} -21 \\ 15\\ 12\\3 \end{pmatrix}.$$

Solve matrix equation: $X = A^{-1} · B.$

$\det A=\begin{vmatrix} -1& 1 &-1&1 \\ 1 & 1 & 1&1\\ 8 & 4 &2& 1\\27&9&3&1 \end{vmatrix}=48.$

$A^{-1}=\begin{pmatrix} -\frac{1}{6} & -\frac{1}{2} &\frac{1}{6}&0\\ \frac{3}{4} & 2&-1&\frac{1}{4}\\ -\frac{7}{12}&-\frac{1}{2}&\frac{5}{6}&-\frac{1}{4}\\ -\frac{1}{2}&-3&-2&-\frac{1}{2} \end{pmatrix}.$

$X=\begin{pmatrix} 1 \\ -9\\ 17\\6 \end{pmatrix}, \text{or } a=1, b=-9, c=17,d=6.$

So, $p=x^3-9x^2+17x+6.$  It is the polynomial function.