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

Let the cubic polynomial be $y=f\left(x\right)$
$f$ is maximum at $x=-1$
$f$ is minimum at $x=1$
$\frac{dy}{dx}=k(x+1)(x-1)$
$dy=k(x^2-1)dx$
$\int dy=\int k(x^2-1)dx\Rightarrow y=k(\frac{x^3}{3}-x)+c$
when $x=-1$, $y=4$
$4=k(\frac{-1}{3}+1)+c$
$\Rightarrow 4=\frac{2k}{3}+c$
$\Rightarrow 12=2k+3c$ ...........(1)
when $x=1$, $y=0$
$0=k(\frac{1}{3}-1)+c$
$\Rightarrow 0=\frac{2k}{3}+c$
$0=-2k+3c$ .........(2)
$12=2k+3c$ ..........(1)
............................
$12=6c$
$\Rightarrow c=2$
Substitute $c=2$ in (1),
$12=2k+6$
$\Rightarrow 2k+6$
$\Rightarrow k=3$
The required polynomial is $=3(\frac{x^3}{3}-x)+2$

$\Rightarrow y=x^3-3x+2$