# Find a Polynomial Satisfying the Given Conditions on Derivatives

## Problem 87

Find a cubic polynomial

\[p(x)=a+bx+cx^2+dx^3\]
such that $p(1)=1, p'(1)=5, p(-1)=3$, and $ p'(-1)=1$.

## Solution.

By differentiating $p(x)$, we obtain

\[p'(x)=b+2cx+3dx^2.\]
Thus the given conditions are

\begin{align*}

p(1)&=a+b+c+d=1\\

p'(1)&=b+2c+3d=5\\

p(-1)&=a-b+c-d=3\\

p'(-1)&=b-2c+3d=1.

\end{align*}

We want to solve these four equations for $a,b,c,d$. So $a,b,c,d$ are unknowns and we regard the conditions as a system of four linear equations.

We solve the system using Gauss-Jordan elimination.

The augmented matrix for the system is

\[\left[\begin{array}{rrrr|r}

1 & 1 & 1 & 1 &1 \\

0 & 1 & 2 & 3 & 5 \\

1 & -1 & 1 & -1 & 3 \\

0 & 1 & -2 & 3 & 1 \\

\end{array}\right] \]

We apply the elementary row operations as follows.

\begin{align*}

&\left[\begin{array}{rrrr|r}

1 & 1 & 1 & 1 &1 \\

0 & 1 & 2 & 3 & 5 \\

1 & -1 & 1 & -1 & 3 \\

0 & 1 & -2 & 3 & 1 \\

\end{array}\right]
\xrightarrow{R_3-R_1}

\left[\begin{array}{rrrr|r}

1 & 1 & 1 & 1 &1 \\

0 & 1 & 2 & 3 & 5 \\

0 & -2 & 0 & -2 & 2 \\

0 & 1 & -2 & 3 & 1 \\

\end{array}\right]
\xrightarrow{\substack{R_1-R_2 \\ R_3+2R_2\\R_4-R_2}}\\[6pt]
&\left[\begin{array}{rrrr|r}

1 & 0& -1 & -2 &-4 \\

0 & 1 & 2 & 3 & 5 \\

0 & 0 & 4 & 4 & 12 \\

0 & 0 & -4 & 0 & -4 \\

\end{array}\right]
\xrightarrow[\frac{-1}{4}R_4]{\frac{1}{4}R_3}

\left[\begin{array}{rrrr|r}

1 & 0& -1 & -2 &-4 \\

0 & 1 & 2 & 3 & 5 \\

0 & 0 & 1 & 1 & 3 \\

0 & 0 & 1 & 0& 1 \\

\end{array}\right]
\xrightarrow{\substack{R_1+R_4 \\ R_2-2R_4\\R_3-R_4}}\\[6pt]
&\left[\begin{array}{rrrr|r}

1 & 0& 0 & -2 &-3 \\

0 & 1 & 0 & 3 & 3 \\

0 & 0 & 0 & 1 & 2 \\

0 & 0 & 1 & 0& 1 \\

\end{array}\right]
\xrightarrow{R_3 \leftrightarrow R_4}

\left[\begin{array}{rrrr|r}

1 & 0& 0 & -2 &-3 \\

0 & 1 & 0 & 3 & 3 \\

0 & 0 & 1 & 0& 1 \\

0 & 0 & 0 & 1 & 2 \\

\end{array}\right]\\[6pt]
& \xrightarrow{\substack{R_1+2R_4 \\ R_2-3R_4}}

\left[\begin{array}{rrrr|r}

1 & 0& 0 & 0 & 1 \\

0 & 1 & 0 & 0 & -3 \\

0 & 0 & 1 & 0& 1 \\

0 & 0 & 0 & 1 & 2 \\

\end{array}\right].

\end{align*}

The last matrix is in reduced row echelon form. Thus the solution to the system is

\[a=1, b=-3, c=1, d=2.\]
Therefore, the polynomial is

\[p(x)=1-3x+x^2+2x^3.\]

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