Question

Curve Fitting Find the function $f\left(x\right)=a{x}^3+b{x}^2+cx+d$for which $f(-3)=-112,f(-1)=-2,f\left(1\right)=4$, and $f\left(2\right)=13$.

Short answer

$f\left(x\right)=3x^3-4x^2+5$

Step-by-step solution

Step 1. Given information 

We have been given:

$f(-3)=-112,f(-1)=-2,f\left(1\right)=4$and $f\left(2\right)=13$

We have to find $f\left(x\right)=a{x}^3+b{x}^2+cx+d$.

Step 2. Put x=-3 in the function f(x) = ax3 + bx2 + cx + d.

$$\begin{gathered} a\cdot (-3{)}^3+b\cdot (-3{)}^2+c\cdot (-3)+d=-112 \\ -27a+9b-3c+d=-112 \end{gathered}$$

Step 3. Put x=-1 in the function f(x) = ax3 + bx2 + cx + d.

$$\begin{gathered} a\cdot (-1{)}^3+b\cdot (-1{)}^2+c\cdot (-1)+d=-112 \\ -a+b-c+d=-2 \end{gathered}$$

Step 4. Put x=1 in the function f(x) = ax3 + bx2 + cx + d.

$$\begin{gathered} a\cdot (1{)}^3+b\cdot (1{)}^2+c\cdot \left(1\right)+d=-112 \\ a+b+c+d=4 \end{gathered}$$

Step 5. Put x=2 in the function f(x) = ax3 + bx2 + cx + d.

$$\begin{gathered} a\cdot (2{)}^3+b\cdot (2{)}^2+c\cdot \left(2\right)+d=13 \\ 8a+4b+2c+d=13 \end{gathered}$$

Step 6. Write the system of equations and corresponding augmented matrix

$\left\{\begin{array}{l}-27a+9b-3c+d\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=-112\\ -a+b-c+d\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=-2\\ a+b+c+d\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=4\\ 8a+4b+2c+d\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=13\end{array}\right.$

The corresponding augmented matrix is:

$\left[\begin{array}{ccccc}-27& 9& -3& 1& -112\\ -1& 1& -1& 1& -2\\ 1& 1& 1& 1& 4\\ 8& 4& 2& 1& 13\end{array}\right]$

Step 7. Perform elementary row operations

$$\begin{gathered} R_1\leftrightarrow R_3 \\ \left[\begin{array}{ccccc}1& 1& 1& 1& 4\\ -1& 1& -1& 1& -2\\ -27& 9& -3& 1& -112\\ 8& 4& 2& 1& 13\end{array}\right] \\ {R}_2\to R_2+R_1 \\ {R}_3\to R_3+27R_1 \\ {R}_4\to R_4-8R_1 \\ \left[\begin{array}{ccccc}1& 1& 1& 1& 4\\ 0& 2& 0& 2& 2\\ 0& 36& 24& 28& -4\\ 0& -4& -6& -7& -19\end{array}\right] \\ {R}_3\to R_3-18R_2 \\ {R}_4\to R_4+2R_2 \\ \left[\begin{array}{ccccc}1& 1& 1& 1& 4\\ 0& 2& 0& 2& 2\\ 0& 0& 24& -8& -40\\ 0& 0& -6& -3& -15\end{array}\right] \end{gathered}$$

Step 8. Perform more row operations

$$\begin{gathered} R_2\to \frac{1}{2}{R}_2 \\ {R}_4\to R_4+\frac{6}{24}{R}_3 \\ \left[\begin{array}{ccccc}1& 1& 1& 1& 4\\ 0& 1& 0& 1& 1\\ 0& 0& 24& -8& -40\\ 0& 0& 0& -5& -25\end{array}\right] \\ {R}_3\to \frac{{\displaystyle 1}}{{\displaystyle 8}}{R}_3 \\ {R}_4\to -\frac{{\displaystyle 1}}{{\displaystyle 5}}{R}_4 \\ :\left[\begin{array}{ccccc}1& 1& 1& 1& 4\\ 0& 1& 0& 1& 1\\ 0& 0& 3& -1& -5\\ 0& 0& 0& 1& 5\end{array}\right] \end{gathered}$$

Step 9. Solve for a,b,c and d

$$\begin{gathered} a+b+c+d\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=4\left(1\right) \\ b+d\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=1\left(2\right) \\ 3c-d\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=-5\left(3\right) \\ d\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=5\left(4\right) \\ d\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=5 \\ From\left(3\right):3c-5\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=-5 \\ c\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=0 \\ From\left(2\right):b+5\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=1 \\ b\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=-4 \\ From\left(1\right):a-4+5\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=4 \\ a\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=3 \end{gathered}$$

$\mathrm{Therefore},\mathrm{the}\mathrm{function}\mathrm{will}\mathrm{be}\mathrm{f}\left(\mathrm{x}\right)=3{\mathrm{x}}^3-4{\mathrm{x}}^2+5.$