[tex]f(x)=6x+\dfrac{24}{x^2}+3\\
f'(x)=6-\dfrac{48}{x^3}\\
6-\dfrac{48}{x^3}=0\\
6x^3-48=0\\
6x^3=48\\
x^3=8\\
x=2\\
[/tex]
For [tex]x<2 \wedge x\not=0[/tex] the derivative is negative.
For [tex]x>2[/tex] the derivative is positive.
Therefore at [tex]x=2[/tex] there's a minimum.
[tex]f_{min}=6\cdot2+\dfrac{24}{2^2}+3=12+6+3=21[/tex]
