Chapter 7.4, Problem 87E

To determine

To calculate:

To find the cubic function $f\left(x\right)=a{x}^3+b{x}^2+cx+d$ that satisfies the equations.

Answer to Problem 87E

  $f\left(x\right)=x^3-2x^2-4x+1$ is the cubic function.

Explanation of Solution

Given information:

The general cubic function is $f\left(x\right)=a{x}^3+b{x}^2+cx+d$ and the given points are

  $\begin{array}{l}f\left(-2\right)=-7\\ f\left(-1\right)=2\\ f\left(1\right)=-4\\ f\left(2\right)=-7\end{array}$

Calculation:

Consider, the general form of the cubic function is $f\left(x\right)=a{x}^3+b{x}^2+cx+d$ and the given points are

  $\begin{array}{l}f\left(-2\right)=-7\\ f\left(-1\right)=2\\ f\left(1\right)=-4\\ f\left(2\right)=-7\end{array}$

Since the points lie on the cubic function, they satisfy when we substitute them.

Substituting each of the four points in the equations gives us four equations with four variables.

  $\begin{array}{l}f\left(-2\right)=-8a+4b-2c+d=-7\to \left(1\right)\\ f\left(-1\right)=-a+b-c+d=2\to \left(2\right)\\ f\left(1\right)=a+b+c+d=-4\to \left(3\right)\\ f\left(2\right)=8a+4b+2c+d=-7\to \left(4\right)\end{array}$

The given system of equations can be written in matrix notation as shown below,

  $\left(\begin{array}{cccc}-8& 4& -2& 1\\ -1& 1& -1& 1\\ 1& 1& 1& 1\\ 8& 4& 2& 1\end{array}\right)\left(\begin{array}{c}a\\ b\\ c\\ d\end{array}\right)=\left(\begin{array}{c}-7\\ 2\\ -4\\ -7\end{array}\right)$

Multiply both sides by the inverse of the coefficient matrix.

Note that the product of a matrix with its inverse is an identity matrix of the same order.

  $\left(\begin{array}{c}a\\ b\\ c\\ d\end{array}\right)={\left(\begin{array}{cccc}-8& 4& -2& 1\\ -1& 1& -1& 1\\ 1& 1& 1& 1\\ 8& 4& 2& 1\end{array}\right)}^{-1}\left(\begin{array}{c}-7\\ 2\\ -4\\ -7\end{array}\right)$

Consider, ${\left(\begin{array}{cccc}-8& 4& -2& 1\\ -1& 1& -1& 1\\ 1& 1& 1& 1\\ 8& 4& 2& 1\end{array}\right)}^{-1}$

Augment with a $4\times 4$ identity matrix,

  $=\left[\begin{array}{ccccccccc}-8& 4& -2& 1& \mid & 1& 0& 0& 0\\ -1& 1& -1& 1& \mid & 0& 1& 0& 0\\ 1& 1& 1& 1& \mid & 0& 0& 1& 0\\ 8& 4& 2& 1& \mid & 0& 0& 0& 1\end{array}\right]$

Reduce the matrix to row echelon form,

  $=\left[\begin{array}{ccccccccc}-8& 4& -2& 1& \mid & 1& 0& 0& 0\\ 0& 8& 0& 2& \mid & 1& 0& 0& 1\\ 0& 0& \frac{3}{4}& \frac{3}{4}& \mid & -\frac{1}{16}& 0& 1& -\frac{3}{16}\\ 0& 0& 0& \frac{3}{2}& \mid & -\frac{1}{4}& 1& 1& -\frac{1}{4}\end{array}\right]$

Reduce further,

  $=\left[\begin{array}{ccccccccc}1& 0& 0& 0& \mid & -\frac{1}{12}& \frac{1}{6}& -\frac{1}{6}& \frac{1}{12}\\ 0& 1& 0& 0& \mid & \frac{1}{6}& -\frac{1}{6}& -\frac{1}{6}& \frac{1}{6}\\ 0& 0& 1& 0& \mid & \frac{1}{12}& -\frac{2}{3}& \frac{2}{3}& -\frac{1}{12}\\ 0& 0& 0& 1& \mid & -\frac{1}{6}& \frac{2}{3}& \frac{2}{3}& -\frac{1}{6}\end{array}\right]$

Hence, we get

  $=\left(\begin{array}{cccc}-\frac{1}{12}& \frac{1}{6}& -\frac{1}{6}& \frac{1}{12}\\ \frac{1}{6}& -\frac{1}{6}& -\frac{1}{6}& \frac{1}{6}\\ \frac{1}{12}& -\frac{2}{3}& \frac{2}{3}& -\frac{1}{12}\\ -\frac{1}{6}& \frac{2}{3}& \frac{2}{3}& -\frac{1}{6}\end{array}\right)$

Now, $\left(\begin{array}{c}a\\ b\\ c\\ d\end{array}\right)={\left(\begin{array}{cccc}-8& 4& -2& 1\\ -1& 1& -1& 1\\ 1& 1& 1& 1\\ 8& 4& 2& 7\end{array}\right)}^{-1}\left(\begin{array}{c}-7\\ 2\\ -4\\ -7\end{array}\right)$

  $=\left(\begin{array}{cccc}-\frac{1}{12}& \frac{1}{6}& -\frac{1}{6}& \frac{1}{12}\\ \frac{1}{6}& -\frac{1}{6}& -\frac{1}{6}& \frac{1}{6}\\ \frac{1}{12}& -\frac{2}{3}& \frac{2}{3}& -\frac{1}{12}\\ -\frac{1}{6}& \frac{2}{3}& \frac{2}{3}& -\frac{1}{6}\end{array}\right)\left(\begin{array}{c}-7\\ 2\\ -4\\ -7\end{array}\right)$

  $=\left(\begin{array}{c}1\\ -2\\ -4\\ 1\end{array}\right)$

Therefore,

  $a=1,b=-2,c=-4,d=1$

Hence, the equation of the cubic function is $f\left(x\right)=x^3-2x^2-4x+1$