Q$6-f(x)=(x-a{)}^3(x-b)$

$a=$

$-f^{\text{'}}(x)=(x-a{)}^2(4x-3b-a)$

$b=$

$-f^{\text{'}\text{'}}(x)=6(x-a)(2x-(a+b))$

fhas $2$ cntricalpoints $x=$a and $x=-\frac{9}{4}$

fhas $2$ inflection points $x=$a and $x=-\frac{1}{2}$