Question

Identify the absolute extrema of the function and the x-values where they occur.

Answer 1

$f(x)=6x+\dfrac{24}{x^2}+3\\ f'(x)=6-\dfrac{48}{x^3}\\ 6-\dfrac{48}{x^3}=0\\ 6x^3-48=0\\ 6x^3=48\\ x^3=8\\ x=2$

For $x$ the derivative is negative.
For $x>2$ the derivative is positive.
Therefore at $x=2$ there's a minimum.

$f_{min}=6\cdot2+\dfrac{24}{2^2}+3=12+6+3=21$