91Ó°ÊÓ

We have been given:

$f(-2)=-10,f(-1)=3,f\left(1\right)=5$and$f\left(3\right)=15$

We have to find role="math" localid="1647455224381" $f\left(x\right)=a{x}^3+b{x}^2+cx+d$.

$$\begin{gathered} aâ‹\dots (−2{)}^3+bâ‹\dots (−2{)}^2+câ‹\dots (−2)+d=−10 \\ −8a+4b−2c+d=−10 \end{gathered}$$

$$\begin{gathered} aâ‹\dots (−1{)}^3+bâ‹\dots (−1{)}^2+câ‹\dots (−1)+d=3 \\ −a+b−c+d=3 \end{gathered}$$

$$\begin{gathered} aâ‹\dots (1{)}^3+bâ‹\dots (1{)}^2+câ‹\dots \left(1\right)+d=5 \\ a+b+c+d=5 \end{gathered}$$

$$\begin{gathered} aâ‹\dots (3{)}^3+bâ‹\dots (3{)}^2+câ‹\dots \left(3\right)+d=15 \\ 27a+9b+3c+d=15 \end{gathered}$$

$$\begin{gathered} \left\{\begin{array}{l}−8a+4b−2c+dâ€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots =−10\\ −a+b−c+dâ€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots =3\\ a+b+c+dâ€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots =5\\ 27a+9b+3c+dâ€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots =15\end{array}\right. \\ \mathrm{The}\mathrm{corresponding}\mathrm{augmented}\mathrm{matrix}\mathrm{is}: \\ \left[\begin{array}{ccccc}−8& 4& −2& 1& −10\\ −1& 1& −1& 1& 3\\ 1& 1& 1& 1& 5\\ 27& 9& 3& 1& 15\end{array}\right] \end{gathered}$$

$$\begin{gathered} R_1↔R_3 \\ \left[\begin{array}{ccccc}1& 1& 1& 1& 5\\ −1& 1& −1& 1& 3\\ −8& 4& −2& 1& −10\\ 27& 9& 3& 1& 15\end{array}\right] \\ {R}_2→R_2+R_1; \\ {R}_3→R_3+8R_1; \\ {R}_4→R_4−27R_1 \\ \left[\begin{array}{ccccc}1& 1& 1& 1& 5\\ 0& 2& 0& 2& 8\\ 0& 12& 6& 9& 30\\ 0& −18& −24& −26& −120\end{array}\right] \\ {R}_3→R_3−6R_2; \\ {R}_4→R_4+9R_2 \\ \begin{array}{r}:\left[\begin{array}{ccccc}1& 1& 1& 1& 5\\ 0& 2& 0& 2& 8\\ 0& 0& 6& −3& −18\\ 0& 0& −24& −8& −48\end{array}\right]\end{array} \end{gathered}$$

$$\begin{gathered} R_2→\frac{1}{2}{R}_2; \\ {R}_4→R_4+4R_3 \\ \begin{array}{r}:\left[\begin{array}{ccccc}1& 1& 1& 1& 5\\ 0& 1& 0& 1& 4\\ 0& 0& 6& −3& −18\\ 0& 0& 0& −20& −120\end{array}\right]\end{array} \\ {R}_3→\frac{1}{3}{R}_3; \\ {R}_4→−\frac{1}{20}{R}_4 \\ :\left[\begin{array}{ccccc}1& 1& 1& 1& 5\\ 0& 1& 0& 1& 4\\ 0& 0& 2& −1& −6\\ 0& 0& 0& 1& 6\end{array}\right] \end{gathered}$$

$$\begin{gathered} \left\{\begin{array}{l}a+b+c+dâ€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots =5\\ b+dâ€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots =4\\ 2c−dâ€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots =−6\\ dâ€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots =6\end{array}\right. \\ d=6 \\ \left(3\right):2c−6=−6 \\ c=0 \\ \left(2\right):b+6=4 \\ b=−2 \\ \left(1\right):a−2+6=5 \\ a=1 \end{gathered}$$

Therefore, the function will be $f\left(x\right)=x^3-2x^2+6$.