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

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dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0For 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

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0a function is concave up when f''(x) >0 concave down when f''(x) < 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i know this but i don't how to find the critical numbers with this function maybe my algebra is rusty...lol

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0hmm..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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea thats what i got that why i didnt know if there was a max or min and how would i find any other solutions

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't think that's the right derivative.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I just redid your derivative for f and came up with something slightly different.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have \(2x(2x^2 + 9)\) for f'

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmmm...yea i don't know what i did wrong...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You were missing a factor of x from each term. I showed all my steps.
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