Решите неравенство f'(x)<0 f(x)=4x^4-4x^3-9

$\left \{ {{f(x) = 4 x^{4} - 4x ^ {3} - 9} \atop {f'(x) \ \textless \ 0}} \right. =\ \textgreater \ 16x^{3}-12x^{2} \ \textless \ 0 \ \textless \ =\ \textgreater \ 4x^{2}(4x - 3) \ \textless \ 0 \\ \ \textless \ =\ \textgreater \ \left \{ {{x^{2} \neq 0} \atop {4x - 3 \ \textless \ 0}} \right \ \textless \ =\ \textgreater \ \left \{ {{x \ \textless \ \frac{3}{4}} \atop {x \neq 0}} \right. \\$
x ∈ (-∞; 0) ∪ (0; 3/4)