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anonymous

  • 5 years ago

f'(x)=2(1-x^2)-2(x^2+10) find interval of increasing, decreasing,concave up and down...local max and min, inflection point

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  1. dumbcow
    • 5 years ago
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    For inflection point: find second derivative and set it equal to 0 For max and min: set f'(x) = 0 solve for x and then plug it back in to f

  2. dumbcow
    • 5 years ago
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    a function is concave up when f''(x) >0 concave down when f''(x) < 0

  3. anonymous
    • 5 years ago
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    i know this but i don't how to find the critical numbers with this function maybe my algebra is rusty...lol

  4. dumbcow
    • 5 years ago
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    hmm..when setting f' = 0 i get x^2 = -9/2 so there are no real solutions so the slope is never zero. is this what you get?

  5. anonymous
    • 5 years ago
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    yea thats what i got that why i didnt know if there was a max or min and how would i find any other solutions

  6. anonymous
    • 5 years ago
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    I don't think that's the right derivative.

  7. anonymous
    • 5 years ago
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    I just redid your derivative for f and came up with something slightly different.

  8. anonymous
    • 5 years ago
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    I have \(-2x(2x^2 + 9)\) for f'

  9. anonymous
    • 5 years ago
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    hmmm...yea i don't know what i did wrong...

  10. anonymous
    • 5 years ago
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    You were missing a factor of x from each term. I showed all my steps.

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