Extract all possible values of $b$ for which the function $f(x)=\{\begin{array}{ll}7x-x^3+\log (b^2-4b+4),& 0\le x<3\\ x-9,& x\ge 3\end{array}$ has local maxima at $x=3$:

For a function to have a local maximum at $x=a$, the value $f(a)$ must be greater than or equal to the values in its immediate neighborhood.At $x=3$, $f(3)=3-9=-6$.For a local maximum at $x=3$, we must have $f(3)\ge f(3^{-})$.$f(3^{-})={\lim }_{x\to 3^{-}}(7x-x^3+\log (b^2-4b+4))$$f(3^{-})=7(3)-(3{)}^3+\log (b^2-4b+4)=21-27+\log (b^2-4b+4)=-6+\log (b^2-4b+4)$.Condition: $-6\ge -6+\log (b^2-4b+4)$$\log (b^2-4b+4)\le 0$.