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mad011

  • 2 years ago

Determine all the critical points for the function f(x)= x^2 + (16/x)

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  1. freckles
    • 2 years ago
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    You found f' yet?

  2. mad011
    • 2 years ago
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    f '= 2x-(16/x^2)???

  3. freckles
    • 2 years ago
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    Yep! :)

  4. freckles
    • 2 years ago
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    Now combine fractions! :)

  5. freckles
    • 2 years ago
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    \[f'=\frac{2x(x^2)-16}{x^2}\] right? :)

  6. mad011
    • 2 years ago
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    Thats what I got! :)

  7. freckles
    • 2 years ago
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    Now set Mr.top=0! :) and solve for x We would look at where the bottom is 0 except our function is not even define for x=0 So we are only look for when the top of f' is 0

  8. freckles
    • 2 years ago
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    \[2x(x^2)-16=0\]

  9. freckles
    • 2 years ago
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    Can you solve this for x?

  10. mad011
    • 2 years ago
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    x=2 ??

  11. freckles
    • 2 years ago
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    \[2x^3-16=0\] Yes that is the only real critical number \[x^3-8=0 \] \[(x-2)(x^2+2x+4)=0\] x=2 --- \[b^2-4ac=2^2-4(1)(4)=4-16=-12 <0 => x=2 \text{ is the only critical number}\] :) So good job! :)

  12. mad011
    • 2 years ago
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    Oh great! thank you

  13. freckles
    • 2 years ago
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    You deserve a medal! Seriously great job! :)

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