Precalculus (6th Edition) Blitzer

The cubic function is $f\left( x \right)=-{{x}^{3}}+4{{x}^{2}}-2$.
To obtain the cubic function, at first we will find the coefficients a, b, c, and d. Since $f\left( x \right)=a{{x}^{3}}+b{{x}^{2}}+cx+d$ So, $f\left( -1 \right)=3$ gives, \begin{align} & f\left( -1 \right)=a{{\left( -1 \right)}^{3}}+b{{\left( -1 \right)}^{2}}+c\left( -1 \right)+d \\ & 3=-a+b-c+d \\ & -a+b-c+d=3 \end{align} (I) $f\left( 1 \right)=1$ gives, \begin{align} & f\left( 1 \right)=a{{\left( 1 \right)}^{3}}+b{{\left( 1 \right)}^{2}}+c\left( 1 \right)+d \\ & 1=a+b+c+d \\ & a+b+c+d=1 \end{align} (II) $f\left( 2 \right)=6$ gives, \begin{align} & f\left( 2 \right)=a{{\left( 2 \right)}^{3}}+b{{\left( 2 \right)}^{2}}+c\left( 2 \right)+d \\ & 6=8a+4b+2c+d \\ & 8a+4b+2c+d=6 \end{align} (III) $f\left( 3 \right)=7$ gives, \begin{align} & f\left( 3 \right)=a{{\left( 3 \right)}^{3}}+b{{\left( 3 \right)}^{2}}+c\left( 3 \right)+d \\ & 7=27a+9b+3c+d \\ & 27a+9b+3c+d=7 \end{align} (IV) Equations (I)–(IV) give, \begin{align} & -a+b-c+d=3 \\ & a+b+c+d=1 \\ & 8a+4b+2c+d=6 \\ & 27a+9b+3c+d=7 \end{align} Use the Gauss elimination method to obtain the solution of the above system of equations. The matrix corresponding to the system of equations is as follows: $\left[ \begin{matrix} -1 & 1 & -1 & 1 & 3 \\ 1 & 1 & 1 & 1 & 1 \\ 8 & 4 & 2 & 1 & 6 \\ 27 & 9 & 3 & 1 & 7 \\ \end{matrix} \right]$ Find the echelon form of the matrix by using the elementary row transformation. ${{R}_{2}}\to {{R}_{2}}+{{R}_{1}},{{R}_{3}}\to {{R}_{3}}+\left( 8 \right){{R}_{1}},{{R}_{4}}\to {{R}_{4}}+\left( 27 \right){{R}_{1}}$ gives, $\left[ \begin{matrix} -1 & 1 & -1 & 1 & 3 \\ -1+1 & 1+1 & -1+1 & 1+1 & 3+1 \\ 8+(-8) & 4+8 & 2-8 & 1+8 & 6+24 \\ 27-27 & 9+27 & 3+27 & 1+27 & 7+81 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & -1 & 1 & -1 & -3 \\ 0 & 2 & 0 & 2 & 4 \\ 0 & 12 & -6 & 9 & 30 \\ 0 & 36 & -24 & 28 & 88 \\ \end{matrix} \right]$ ${{R}_{1}}\to -{{R}_{1}}$ gives, $\left[ \begin{matrix} -(-1) & -(1) & -(-1) & -(1) & -(3) \\ 0 & 2 & 0 & 2 & 4 \\ 0 & 12 & -6 & 9 & 30 \\ 0 & 36 & -24 & 28 & 88 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & -1 & 1 & -1 & -3 \\ 0 & 2 & 0 & 2 & 4 \\ 0 & 12 & -6 & 9 & 30 \\ 0 & 36 & -24 & 28 & 88 \\ \end{matrix} \right]$ ${{R}_{3}}\to {{R}_{3}}+\left( -6 \right){{R}_{2}},{{R}_{4}}\to {{R}_{4}}+\left( -18 \right){{R}_{2}}$ gives, $\left[ \begin{matrix} 1 & -1 & 1 & -1 & -3 \\ 0 & 2 & 0 & 2 & 4 \\ 0 & 12-12 & -6-0 & 9-12 & 30-24 \\ 0 & 36-36 & -24-0 & 28-36 & 88-72 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & -1 & 1 & -1 & -3 \\ 0 & 2 & 0 & 2 & 4 \\ 0 & 0 & -6 & -3 & 6 \\ 0 & 0 & -24 & -8 & 16 \\ \end{matrix} \right]$ ${{R}_{2}}\to \frac{1}{2}{{R}_{2}}$ gives, $\left[ \begin{matrix} 1 & -1 & 1 & -1 & -3 \\ 0 & 2 & 0 & 2 & 4 \\ 0 & 0 & -6 & -3 & 6 \\ 0 & 0 & -24 & -8 & 16 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & -1 & 1 & -1 & -3 \\ 0 & 1 & 0 & 1 & 2 \\ 0 & 0 & -6 & -3 & 6 \\ 0 & 0 & -24 & -8 & 16 \\ \end{matrix} \right]$ ${{R}_{1}}\to {{R}_{1}}+{{R}_{2}}$ gives, $\left[ \begin{matrix} 1 & -1+1 & 1+0 & -1+1 & -3+2 \\ 0 & 1 & 0 & 1 & 2 \\ 0 & 0 & -6 & -3 & 6 \\ 0 & 0 & -24 & -8 & 16 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & 1 & 0 & -1 \\ 0 & 1 & 0 & 1 & 2 \\ 0 & 0 & -6 & -3 & 6 \\ 0 & 0 & -24 & -8 & 16 \\ \end{matrix} \right]$ ${{R}_{4}}\to {{R}_{4}}-4{{R}_{3}}$ gives, $\left[ \begin{matrix} 1 & 0 & 1 & 0 & -1 \\ 0 & 1 & 0 & 1 & 2 \\ 0 & 0 & -6 & -3 & 6 \\ 0 & 0 & -24-(-24) & -8-(-12) & 16-24 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & 1 & 0 & -1 \\ 0 & 1 & 0 & 1 & 2 \\ 0 & 0 & -6 & -3 & 6 \\ 0 & 0 & 0 & 4 & -8 \\ \end{matrix} \right]$ ${{R}_{3}}\to -\frac{1}{6}{{R}_{3}}$ gives, $\left[ \begin{matrix} 1 & 0 & 1 & 0 & -1 \\ 0 & 1 & 0 & 1 & 2 \\ 0 & 0 & -\frac{6}{-6} & -\frac{3}{-6} & 6 \\ 0 & 0 & 0 & 4 & -8 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & 1 & 0 & -1 \\ 0 & 1 & 0 & 1 & 2 \\ 0 & 0 & 1 & \frac{1}{2} & -1 \\ 0 & 0 & 0 & 4 & -8 \\ \end{matrix} \right]$ ${{R}_{1}}\to {{R}_{1}}-{{R}_{3}}$ gives, $\left[ \begin{matrix} 1 & 0 & 1-1 & 0-\frac{1}{2} & -1-(-1) \\ 0 & 1 & 0 & 1 & 2 \\ 0 & 0 & 1 & \frac{1}{2} & -1 \\ 0 & 0 & 0 & 4 & -8 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 1 & 0 & 1 & 2 \\ 0 & 0 & 1 & \frac{1}{2} & -1 \\ 0 & 0 & 0 & 4 & -8 \\ \end{matrix} \right]$ ${{R}_{3}}\to \frac{1}{4}{{R}_{3}}$ gives, $\left[ \begin{matrix} 1 & 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 1 & 0 & 1 & 2 \\ 0 & 0 & 1 & \frac{1}{2} & -1 \\ 0 & 0 & 0 & \frac{4}{4} & \frac{-8}{4} \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 1 & 0 & 1 & 2 \\ 0 & 0 & 1 & \frac{1}{2} & -1 \\ 0 & 0 & 0 & 1 & -2 \\ \end{matrix} \right]$ ${{R}_{1}}\to {{R}_{1}}+\frac{1}{2}{{R}_{4}},\,{{R}_{2}}\to {{R}_{2}}-{{R}_{4}}\,\text{ and }\,{{R}_{3}}\to {{R}_{3}}-\frac{1}{2}{{R}_{4}}$ gives, $\left[ \begin{matrix} 1 & 0 & 0 & -\frac{1}{2}+\frac{1}{2} & 0-\frac{2}{2} \\ 0 & 1 & 0 & 1-1 & 2-(-2) \\ 0 & 0 & 1 & \frac{1}{2}-\frac{1}{2} & -1-(-\frac{1}{2}) \\ 0 & 0 & 0 & 1 & -2 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & 0 & 0 & -1 \\ 0 & 1 & 0 & 0 & 4 \\ 0 & 0 & 1 & \frac{1}{2} & 0 \\ 0 & 0 & 0 & 1 & -2 \\ \end{matrix} \right]$ So, $a=-1,\,b=4,\,c=0,\,\text{ and }\,d=-2$ Now, put the values of a, b, c, and d in function $f\left( x \right)=a{{x}^{3}}+b{{x}^{2}}+cx+d$. This gives: \begin{align} & f\left( x \right)=\left( -1 \right){{x}^{3}}+\left( 4 \right){{x}^{2}}+\left( 0 \right)x-2 \\ & =-{{x}^{3}}+4{{x}^{2}}-2 \end{align} Hence, the required cubic function is: $f\left( x \right)=-{{x}^{3}}+4{{x}^{2}}-2$