Ta có:
\(\eqalign{
& f(x) = 3x + {{60} \over x} - 64.{x^{ - 3}} + 5 \cr
& \Rightarrow f'(x) = 3 - {{60} \over {{x^2}}} + 192{x^{ - 4}} \cr&= 3 - {{60} \over {{x^2}}} + {{192} \over {{x^4}}} = {{3{x^4} - 60{x^2} + 192} \over {{x^4}}} \cr} \)
Vậy:
\(\eqalign{
& f'(x) = 0 \Leftrightarrow 3{x^4} - 60{x^2} + 192 = 0(x \ne 0) \cr
& \Leftrightarrow \left[ \matrix{
{x^2} = 16 \hfill \cr
{x^2} = 4 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x = \pm 4 \hfill \cr
x = \pm 2 \hfill \cr} \right.\text{ thỏa mãn } \cr}\)
