Chapter 5.1, Problem 51PPSE

Solution

To show:The third differences of a polynomial function of degree 3 are nonzero and constant.

Given information:

Given function is $f\left(x\right)=x^3-3x^2-2x-6$ .

Calculation:

Substitute different values of $x$ and find corresponding values of $f\left(x\right)$ .

Substitute $x=3$ in $f\left(x\right)=x^3-3x^2-2x-6$ gives:

  $\begin{array}{l}f\left(3\right)=3^3-\left(3\right)3^2-2\cdot 3-6\\ f\left(3\right)=27-27-6-6\\ f\left(3\right)=-12\end{array}$

Now Substitute $x=2$ in $f\left(x\right)=x^3-3x^2-2x-6$ gives:

  $\begin{array}{l}f\left(2\right)=2^3-\left(3\right)2^2-2\cdot 2-6\\ f\left(2\right)=8-12-4-6\\ f\left(2\right)=-14\end{array}$

Similarly, more values can be substituted, and the corresponding value will be found. The table for different values is given below:

$x$	$f\left(x\right)$
3	-12
2	-14
1	-10
0	-6
-1	-8
-2	-22
-3	-54

Now first difference can be found by subtracting 2 digits from first digit and so on.

This gives:

$x$	$f\left(x\right)$	First difference	Second difference	Third difference
3	-12	$-14-\left(-12\right)=-2$	4-2=2	0-2=-2
2	-14	$\begin{array}{l}-10-\left(-14\right)=4\hfill \end{array}$	4-4=0	$-6-0=-6$
1	-10	$-6-\left(-10\right)=4$	-2-4=-6	$-12-\left(-6\right)=-6$
0	-6	$-8-\left(-6\right)=-2$	$\begin{array}{l}-14-\left(-2\right)=-12\hfill \end{array}$	$\begin{array}{l}-18-\left(-12\right)=-6\hfill \end{array}$
-1	-8	$-22-\left(-8\right)=-14$	$-32-\left(-14\right)=-18$
-2	-22	$-54-\left(-22\right)=-32$
-3	-54

It is clear from the table that third difference is nonzero and constant.

Now take the polynomial $a{x}^3+b{x}^2+cx+d$

Substitute different values of $x$ and find corresponding values of $f\left(x\right)$ .

Substitute $x=3$ in $a{x}^3+b{x}^2+cx+d$ gives:

  $\begin{array}{l}f\left(3\right)=a{(}^3+b{(}^3+c\cdot 3+d\\ f\left(3\right)=27a+9b+3c+d\end{array}$

Now Substitute $x=2$ in $a{x}^3+b{x}^2+cx+d$ gives:

  $\begin{array}{l}f\left(3\right)=a{(}^2+b{(}^2+c\cdot 2+d\\ f\left(3\right)=8a+4b+2c+d\end{array}$

Similarly, more values can be substituted, and the corresponding value will be found. The table for different values is given below:

$x$	$f\left(x\right)$
3	$27a+9b+3c+d$
2	$8a+4b+2c+d$
1	$a+b+c+d$
0	$d$
-1	$\text{-a+b-c+d}$
-2	$\text{-8a+4b-2c+d}$
-3	$-27a+9b-3c+d$

Now first difference can be found by subtracting 2 digits from first digit and so on. This gives:

$x$	$f\left(x\right)$	First difference	Second difference	Third difference
3	$27a+9b+3c+d$	$-19a-5b-c$	$12a+2b$	$-6a$
2	$8a+4b+2c+d$	$-7a-3b-c$	$6a+2b$	$-6a$
1	$a+b+c+d$	$-a-b-c$	$2b$	$-6a$
0	$d$	$-a+b-c$	$-6a+2b$	$-6a$
-1	$-a+b-c+d$	$-7a+3b-c$	$-12a+2b$
-2	$-8a+4b-2c+d$	$-19a+5b-c$
-3	$-27a+9b-3c+d$

It is clear from the table that third difference is nonzero and constant.