Chapter 7, Problem 46RE

Solution

To calculate: The coefficient of the function so that its graph goes through the given points.

The function is $f\left(x\right)=\frac{18}{108}{x}^4-\frac{29}{18}{x}^3+\frac{59}{36}{x}^2+\frac{505}{54}x-\frac{68}{9}$ .

Given Information:

The points for the function are $\left(-2,-4\right),\left(1,2\right),\left(3,6\right),\left(4,-2\right),\left(7,8\right)$ .

Calculation:

Consider the given general function,

  $f\left(x\right)=a{x}^4+b{x}^3+c{x}^2+dx+f$

Substitute the given point $\left(-2,-4\right)$ in the function $f\left(x\right)=a{x}^4+b{x}^3+c{x}^2+dx+f$ .

  $\begin{array}{c}-4=a{\left(-2\right)}^4+b{\left(-2\right)}^3+c{\left(-2\right)}^2-2d+e\\ 16a-8b+4c-2d+e=-4\end{array}$

Substitute the given point $\left(1,2\right)$ in the function $f\left(x\right)=a{x}^4+b{x}^3+c{x}^2+dx+f$ .

  $a+b+c+d+e=2$

Substitute the given point $\left(3,6\right)$ in the function $f\left(x\right)=a{x}^4+b{x}^3+c{x}^2+dx+f$ .

  $\begin{array}{c}6=a{\left(3\right)}^4+b{\left(3\right)}^3+c{\left(3\right)}^2+3d+e\\ 81a+27b+9c+3d+e=6\end{array}$

Substitute the given point $\left(4,-2\right)$ in the function $f\left(x\right)=a{x}^4+b{x}^3+c{x}^2+dx+f$ .

  $\begin{array}{c}-2=a{\left(4\right)}^4+b{\left(4\right)}^3+c{\left(4\right)}^2+4d+e\\ 256a+64b+16c+4d+e=-2\end{array}$

Substitute the given point $\left(7,8\right)$ in the function $f\left(x\right)=a{x}^4+b{x}^3+c{x}^2+dx+f$ .

  $\begin{array}{c}8=a{\left(7\right)}^4+b{\left(7\right)}^3+c{\left(7\right)}^2+7d+e\\ 2401a+343b+49c+7d+e=8\end{array}$

Write the all obtained system as $AX=B$ matrix form.

  $A=\left[\begin{array}{ccccc}16& -8& 4& -2& 1\\ 1& 1& 1& 1& 1\\ 81& 27& 9& 3& 1\\ 256& 64& 16& 4& 1\\ 2401& 343& 49& 7& 1\end{array}\right]$ , $X=\left[\begin{array}{c}a\\ b\\ c\\ d\\ e\end{array}\right]$ , and $B=\left[\begin{array}{c}-4\\ 2\\ 6\\ -2\\ 8\end{array}\right]$

Solve the equation $AX=B$ .

  $\begin{array}{c}{A}^{-1}AX=A^{-1}B\\ X=A^{-1}B\\ ={\left[\begin{array}{ccccc}16& -8& 4& -2& 1\\ 1& 1& 1& 1& 1\\ 81& 27& 9& 3& 1\\ 256& 64& 16& 4& 1\\ 2401& 343& 49& 7& 1\end{array}\right]}^{-1}\cdot \left[\begin{array}{c}-4\\ 2\\ 6\\ -2\\ 8\end{array}\right]\left[\text{using calculator}\right]\\ =\left[\begin{array}{c}19/108\\ -174/108\\ 177/108\\ 1010/108\\ -816/108\end{array}\right]\\ =\left[\begin{array}{c}18/108\\ -29/18\\ 59/36\\ 505/54\\ -68/9\end{array}\right]\end{array}$

Substitute the obtained values in the function.

  $f\left(x\right)=\frac{18}{108}{x}^4-\frac{29}{18}{x}^3+\frac{59}{36}{x}^2+\frac{505}{54}x-\frac{68}{9}$

Hence, the obtained function is $f\left(x\right)=\frac{18}{108}{x}^4-\frac{29}{18}{x}^3+\frac{59}{36}{x}^2+\frac{505}{54}x-\frac{68}{9}$ .