Answered

Discover the answers to your questions at Westonci.ca, where experts share their knowledge and insights with you. Discover comprehensive answers to your questions from knowledgeable professionals on our user-friendly platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

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]

We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.