Question

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

f(x)=6x+(24/x^sqr)+3, x>0

Answer 1

`f(x)=6x+(24)/(x^2)+3\\ f'(x)=6-(48)/(x^3)\\ 6-(48)/(x^3)=0\\ 6x^3-48=0\\ 6x^3=48\\ x^3=8\\ x=2\\`

For
`x<2 \wedge x\\ot=0` the derivative is negative.
For
`x>2` the derivative is positive.
Therefore at
`x=2` there's a minimum.


`f_(min)=6\cdot2+(24)/(2^2)+3=12+6+3=21`