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Denise

  • 5 years ago

Find the critical vales of 12x^3 -24x

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  1. anonymous
    • 5 years ago
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    to find the critical numbers, you have to derive the function once and take 2 conditions for f'(x): 1) f'(x) = 0 2 f'(x) = UND (undefined) in your case, it'll be, f'(x) = 0 <-- to find the critical numbers ^_^ give it a try now

  2. anonymous
    • 5 years ago
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    set f'(x) to zero after deriving and find the zeros ^_^

  3. Denise
    • 5 years ago
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    The original equation was f(x)=3x^4-12x^2+4

  4. anonymous
    • 5 years ago
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    yep, now derive it

  5. Denise
    • 5 years ago
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    12x(x^2-12)

  6. anonymous
    • 5 years ago
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    now set this equal to 0 and find the zeros normally ^_^

  7. anonymous
    • 5 years ago
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    the zeros that you'll find = will be your critical values :)

  8. Denise
    • 5 years ago
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    that is where I get stuck.

  9. anonymous
    • 5 years ago
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    my dear: f'(x) = 12x^3 - 24x = 12x(x^2-2)

  10. anonymous
    • 5 years ago
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    so your critical values are: x = 0 and \[x = \pm \sqrt(2)\] ^_^

  11. anonymous
    • 5 years ago
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    you had a lil mistake in finding the derivative ^_^

  12. Denise
    • 5 years ago
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    yep

  13. Denise
    • 5 years ago
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    so to find the relative min and max I need to do ???

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