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

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

Sagot :

konrad509
[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]

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