champ215 champ215

08-04-2015
Mathematics

Answered

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

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

Verified Answer :

konrad509 konrad509

08-04-2015

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

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

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

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