Question

Find the cubic polynomial in $x$ which attains its maximum value 4 and minimum value 0 at $x=-1$ and $x=1$ respectively.

Answer 1

Hint: For solving this question first we will assume a general cubic polynomial in $x$ then we will use conditions of maxima and minima as per the given data and then we will get the required cubic polynomial expression.

Complete step-by-step answer:
Given:
A cubic polynomial in $x$ , whose maximum value 4 occurs at $x=-1$ and minimum value 0 occurs at $x=1$ .
Now, let the cubic polynomial is $f\left( x \right)=a{{x}^{3}}+b{{x}^{2}}+cx+d$ .
Before we proceed we should know how the derivative of the function $f\left( x \right)$ helps us in determining the maximum and minimum value of the function. If the maximum and minimum value of $f\left( x \right)$ occurs at $x={{x}_{1}}$ and $x={{x}_{2}}$ respectively . Then,
1. ${f}'\left( {{x}_{1}} \right)={{\left[ \dfrac{d\left( f\left( x \right) \right)}{dx} \right]}_{x={{x}_{1}}}}=0$ and ${f}''\left( {{x}_{1}} \right)={{\left[ \dfrac{d\left( {f}'\left( x \right) \right)}{dx} \right]}_{x={{x}_{1}}}}<0$ .
2. ${f}'\left( {{x}_{2}} \right)={{\left[ \dfrac{d\left( f\left( x \right) \right)}{dx} \right]}_{x={{x}_{2}}}}=0$ and ${f}''\left( {{x}_{2}} \right)={{\left[ \dfrac{d\left( {f}'\left( x \right) \right)}{dx} \right]}_{x={{x}_{2}}}}>0$ .
Now, in this question, the maximum value of $f\left( x \right)=a{{x}^{3}}+b{{x}^{2}}+cx+d$ is 4 and it occurs at $x=-1$ . Then,
$\begin{align}
  & f\left( -1 \right)=4 \\
 & \Rightarrow -a+b-c+d=4..............\left( 1 \right) \\
\end{align}$
$\begin{align}
  & {f}'\left( -1 \right)={{\left[ \dfrac{d\left( f\left( x \right) \right)}{dx} \right]}_{x=-1}}=0 \\
 & \Rightarrow {{\left[ \dfrac{d\left( a{{x}^{3}}+b{{x}^{2}}+cx+d \right)}{dx} \right]}_{x=-1}}=0 \\
\end{align}$
$\begin{align}
  & \Rightarrow {{\left[ 3a{{x}^{2}}+2bx+c \right]}_{x=-1}}=0 \\
 & \Rightarrow 3a-2b+c=0...................\left( 2 \right) \\
\end{align}$
Now, in the question, it is given that the minimum value of $f\left( x \right)=a{{x}^{3}}+b{{x}^{2}}+cx+d$ is 0 and it occurs at $x=1$ . Then,
$\begin{align}
  & f\left( 1 \right)=0 \\
 & \Rightarrow a+b+c+d=0..............\left( 3 \right) \\
\end{align}$
$\begin{align}
  & {f}'\left( 1 \right)={{\left[ \dfrac{d\left( f\left( x \right) \right)}{dx} \right]}_{x=1}}=0 \\
 & \Rightarrow {{\left[ \dfrac{d\left( a{{x}^{3}}+b{{x}^{2}}+cx+d \right)}{dx} \right]}_{x=1}}=0 \\
 & \Rightarrow {{\left[ 3a{{x}^{2}}+2bx+c \right]}_{x=1}}=0 \\
 & \Rightarrow 3a+2b+c=0...................\left( 4 \right) \\
\end{align}$
Now, subtract equation (2) from equation (4). Then,
$\begin{align}
  & \left( 3a+2b+c \right)-\left( 3a-2b+c \right)=0 \\
 & \Rightarrow 4b=0 \\
 & \Rightarrow b=0..........\left( 5 \right) \\
\end{align}$
Now, substituting the value of $b=0$ from equation (5) into equation (2). Then,
$\begin{align}
  & 3a-2b+c=0 \\
 & \Rightarrow c=-3a.............\left( 6 \right) \\
\end{align}$
Now, substituting the value of $c=-3a$ from equation (6) and value of $b=0$ from equation (5) into equation (3). Then,
$\begin{align}
  & a+b+c+d=0 \\
 & \Rightarrow a+0-3a+d=0 \\
 & \Rightarrow d=2a..............\left( 7 \right) \\
\end{align}$
Now, substituting the value of $b=0$ from equation (5), the value of $c=-3a$ from equation (6) and value of $d=2a$ from equation (7) into equation (1). Then,
$\begin{align}
  & -a+b-c+d=4 \\
 & \Rightarrow -a+0+3a+2a=4 \\
 & \Rightarrow 4a=4 \\
 & \Rightarrow a=1..............\left( 8 \right) \\
\end{align}$
Now, substitute $a=1$ from equation (8) into equation (6) and equation (7). Then,
$\begin{align}
  & c=-3a=-3............\left( 9 \right) \\
 & d=2a=2................\left( 10 \right) \\
\end{align}$
Now, from equation (8), (5), (9) and (10). We have the value of $a=1$ , $b=0$ , $c=-3$ and $d=2$ . Then,
$\begin{align}
  & f\left( x \right)=a{{x}^{3}}+b{{x}^{2}}+cx+d \\
 & \Rightarrow f\left( x \right)={{x}^{3}}-3x+2 \\
\end{align}$
Thus, ${{x}^{3}}-3x+2$ is our required polynomial.

Note: Although the question is very simple to understand, the student should apply the concept of maxima and minima accurately and don’t confuse the value of the first derivative and second derivative for each case. Moreover, equations should be solved stepwise to get the correct answer.