A function ${\displaystyle f(x)})$ defined for ${\displaystyle x>a}$, where ${\displaystyle a}$ is a constant, and a quartic function ${\displaystyle g(x)}$ whose leading coefficient is ${\displaystyle -1}$ satisfy the three conditions below:
A) For all real numbers ${\displaystyle x}$, such that ${\displaystyle x>a}$, ${\displaystyle (x-a)f(x)=g(x)}$
B) For two different real numbers ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ , ${\displaystyle f(x)}$ has the same local maximum ${\displaystyle M}$ at ${\displaystyle x=\alpha }$ and ${\displaystyle x=\beta }$ . $({\displaystyle M>0}$
C) ${\displaystyle f(x)}$ has more local extrema than ${\displaystyle g(x)}$ does. ${\displaystyle \beta -\alpha =6{\sqrt {3}}}$. Find the minimum of ${\displaystyle M}$
My solution is as follow
$x-a>0$ and $(x-a)f(x)=g(x)$
Hence $f(x)=-x^3+bx^2+cx+d$
Given $f'(\alpha)=0$ and $f'(\beta)=0$ and $f''(\alpha)<0$ and $f''(\beta)<0$ and also $f(\alpha)=f(\beta)$
How do we proceed from here. For reference the I got the question from the following website https://en.wikipedia.org/wiki/College_Scholastic_Ability_Test#Mathematics_2
