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anonymous

  • 5 years ago

Given f(x)=x^3-4x^2-2. State the intervals on which f is increasing and on which f is decreasing.

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  1. amistre64
    • 5 years ago
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    derive and set to 0

  2. amistre64
    • 5 years ago
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    \[f' = 3x^2 -8x =0\] \[x(3x-8)=0\] \(x=0,8/3 \) right?

  3. amistre64
    • 5 years ago
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    then maybe this; but we gotta check to make sure that x=0 aint an inflectionin disguise <.........0...........8/3...........> - + + - - + ------------------------ + - +

  4. anonymous
    • 5 years ago
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    Wait, where are you getting the equation from?

  5. amistre64
    • 5 years ago
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    its called a derivative; the first derivative tells you how the function is moving.... its slope at any given point

  6. amistre64
    • 5 years ago
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    when the derivative =0 that means we are bending back on itself and changing direction ..usually. but, an inflection in a graph can also show a 0 in the 1st derivative

  7. amistre64
    • 5 years ago
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    but; 6x-8 aint =0 at x=0 so its a bender

  8. amistre64
    • 5 years ago
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    i gues the long way would be to factor the equation if we can

  9. amistre64
    • 5 years ago
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    i just assumed by the nature of the question that it was a derivative type question lol

  10. amistre64
    • 5 years ago
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    http://www.wolframalpha.com/input/?i=x^3-4x^2-2 might help you see whats happenin with it

  11. anonymous
    • 5 years ago
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    I'm not up to calc yet :/ sorry...

  12. anonymous
    • 5 years ago
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    Helppppp....

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