Chapter 4.3, Problem 59E

To determine

To find: The cubic equation $f\left(x\right)=a{x}^3+b{x}^2+cx+d$ .

Answer to Problem 59E

The cubic equation is $f\left(x\right)=\left(\frac{1}{9}\right)\left[2x^3+3x^2+12x+7d\right]$ .

Explanation of Solution

Given:

The given equation is $3$

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

The local maximum value of 3 at $x=-2$ and a local minimum value of $0$ at $x=1$ .

Calculation:

Consider the given expression is,

  $f\left(x\right)=a{x}^3+b{x}^2+cx+d$

The local maximum value of 3 at $x=-2$ and a local minimum value of $0$ at $x=1$ .

Then,

  $\begin{array}{l}f\left(-2\right)=a{\left(-2\right)}^3+b{\left(-2\right)}^2+c\left(-2\right)+d\\ a{\left(-2\right)}^3+b{\left(-2\right)}^2+c\left(-2\right)+d=3\\ -8a+4b-2c+d=3\end{array}$ ……. (I)

For local maximum of $0$ at $x=1$ .

  $\begin{array}{l}a{\left(1\right)}^3+b{\left(1\right)}^2+c\left(1\right)+d=0\\ a+b+c+d=0\end{array}$ …… (II)

Derive the given function,

  $\begin{array}{l}f\left(x\right)=a{x}^3+b{x}^2+cx+d\\ f^{\prime }\left(-2\right)=3a{\left(-2\right)}^2+2b\left(-2\right)+c\\ 12a-4b+c=0\end{array}$ …… (III)

Also,

  $\begin{array}{l}{f}^{\prime }\left(1\right)=3a+2b+c\\ 0=3a+2b+c\end{array}$ …… (IV)

Subtract equation (II) from equation (I).

  $\begin{array}{l}-9a+3b-3c=3\\ -3a+b-c=1\end{array}$ ……. (V)

Subtract equation (IV) from equation (III).

  $9a-6b=0$ ……. (VI)

From equation (IV) and (V).

  $\begin{array}{l}3b=1\\ b=\frac{1}{3}\end{array}$

Then, the $9a-6b=0$ becomes,

  $\begin{array}{l}9a-6\left(\frac{1}{3}\right)=0\\ a=\frac{2}{9}\end{array}$

The equation $0=3a+2b+c$ is,

  $\begin{array}{l}0=3\left(\frac{2}{9}\right)+2\left(\frac{1}{3}\right)+c\\ c=\frac{4}{3}\end{array}$

For equation (II).

  $\begin{array}{l}\frac{2}{9}+\frac{1}{3}-\frac{4}{3}+d=0\\ d=\frac{7}{9}\end{array}$

Then the given expression is,

  $f\left(x\right)=\left(\frac{1}{9}\right)\left[2x^3+3x^2+12x+7d\right]$