The fact that $f'(x) = 2(2x + 3)^2(4 - x)^3(6 - 7x)$ combined with $2x + 3 > 0$ and $4 - x > 0$ (since ${\textstyle -\frac{3}{2} < x < 4}$) implies ${\textstyle f(x)_{\max} = f(\frac{6}{7}) = (\frac{33}{7})^3 \cdot (\frac{22}{7})^4 = \frac{3^3 \cdot 11^3 \cdot 2^4 \cdot 11^4}{7^7} = 432 \cdot (\frac{11}{7})^7}$.

By AM-GM $$\frac{11}{14}=\frac{3\cdot\frac{3+2x}{6}+4\cdot\frac{4-x}{4}}{7}\ge\sqrt[7]{\left(\frac{3+2x}{6}\right)^3\cdot \left(\frac{4-x}{4}\right)^4}$$

WolfusA wrote: By AM-GM $$\frac{11}{14}=\frac{3\cdot\frac{3+2x}{6}+4\cdot\frac{4-x}{4}}{7}\ge\sqrt[7]{\left(\frac{3+2x}{6}\right)^3\cdot \left(\frac{4-x}{4}\right)^4}$$ How do you now complete the b part of the problem

Criterion for equality in AM-GM: all variables are equal. In our case $$\frac{3+2x}{6}=\frac{4-x}{4}\iff x=\frac67$$