Question

Let $p\left(x\right)$ be a real polynomial of least degree which has a local maximum at $x=1$ and a local minimum at $x=3$. if $p\left(1\right)=6,p\left(3\right)=2$, then $p^{\prime }\left(0\right)$ is

Answer 1

$p\left(x\right)$ have local maxima and minima at $x=1$ and $x=3$
So, $p^{\prime }=\lambda (x-1)(x-3)=\lambda (x^2-4x+3)$
$p\left(x\right)=\lambda (x^3/3-2x^2+3x)+\mu$
$p\left(1\right)=6$
$6=\lambda (\frac{1}{3}-2+3)+\mu$
$18=4\lambda +3\mu ...$(i)
$p\left(3\right)=2$
$2=\lambda (\frac{27}{3}-2\times 9+9)+\mu$
$\therefore \mu =2\Rightarrow \lambda =3$
$p^{\prime }\left(x\right)=3(x-1)(x-3)$
$p^{\prime }\left(0\right)=3(-1)(-3)$
$=9$