Chapter 7.3, Problem 86E

Solution

To state: The value of $f$ so that the graph contains the points $\left(-1,-6\right),\left(0,-1\right),\left(1,2\right)$ and the family of curves is $a{x}^3+b{x}^2+cx+d$ .

The resultant answer is $f\left(x\right)=\left(-c+4\right)x^3-x^2+cx-1$ .

Given information:

The given points $\left(-1,-6\right),\left(0,-1\right),\left(1,2\right)$ and the family of curves is $a{x}^3+b{x}^2+cx+d$ .

Explanation:

Consider the given points:

  $\left(-1,-6\right),\left(0,-1\right),\left(1,2\right)$

Now set up a system of equations using given points and the family of curves is $a{x}^3+b{x}^2+cx+d$ .

Point $\left(-1,-6\right)$ ,

  $\begin{array}{l}-6=a{\left(-1\right)}^3+b{\left(-1\right)}^2+c\left(-1\right)+d\\ -6=-a+b-c+d\cdots \left(1\right)\end{array}$

Point $\left(0,-1\right)$ ,

  $\begin{array}{l}-1=a{\left(0\right)}^3+b{\left(0\right)}^2+c\left(0\right)+d\\ -1=d\cdots \left(2\right)\end{array}$

Point $\left(1,2\right)$ ,

  $\begin{array}{l}2=a{\left(1\right)}^3+b{\left(1\right)}^2+c\left(1\right)+d\\ 2=a+b+c+d\cdots \left(3\right)\end{array}$

It can be seen that there are 4 variables but only 3 equations, it means there are infinitely many solutions.

Write the above system of equation in matrix and solve the system of equations using matrix:

  $\left[\begin{array}{ccc}-1& 1& -1\\ 0& 0& 0\\ 1& 1& 1\end{array}\begin{array}{c}1\\ 1\\ 1\end{array}\begin{array}{c}-6\\ -1\\ 2\end{array}\right]$

Apply the operation $R_{13}$ ,

  $\left[\begin{array}{ccc}1& 1& 1\\ 0& 0& 0\\ -1& 1& -1\end{array}\begin{array}{c}1\\ 1\\ 1\end{array}\begin{array}{c}2\\ -1\\ -6\end{array}\right]$

Apply the operation $R_3\to R_1+R_3$ ,

  $\left[\begin{array}{ccc}1& 1& 1\\ 0& 0& 0\\ 0& 2& 0\end{array}\begin{array}{c}1\\ 1\\ 2\end{array}\begin{array}{c}2\\ -1\\ -4\end{array}\right]$

Apply the operation $R_{23}$ ,

  $\left[\begin{array}{ccc}1& 1& 1\\ 0& 2& 0\\ 0& 0& 0\end{array}\begin{array}{c}1\\ 2\\ 1\end{array}\begin{array}{c}2\\ -4\\ -1\end{array}\right]$

Apply the operation $R_2\to \frac{1}{2}{R}_2$ ,

  $\left[\begin{array}{ccc}1& 1& 1\\ 0& 1& 0\\ 0& 0& 0\end{array}\begin{array}{c}1\\ 1\\ 1\end{array}\begin{array}{c}2\\ -2\\ -1\end{array}\right]$

Apply the operation $R_1\to \left(-1\right)R_2+R_1$ ,

  $\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 0& 0& 0\end{array}\begin{array}{c}0\\ 1\\ 1\end{array}\begin{array}{c}4\\ -2\\ -1\end{array}\right]$

Apply the operation $R_2\to \left(-1\right)R_3+R_2$ ,

  $\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 0& 0& 0\end{array}\begin{array}{c}0\\ 1\\ 1\end{array}\begin{array}{c}4\\ -1\\ -1\end{array}\right]$

This correspond the system: $a+c=4,b=-1,\text{and}d=-1$

The solution of the system is:

  $a=-c+4,b=-1,c=c\text{and}d=-1$ and $f\left(x\right)=\left(-c+4\right)x^3-x^2+cx-1$ .