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anonymous
 5 years ago
f(t)=t^412t^3+16t^2
anonymous
 5 years ago
f(t)=t^412t^3+16t^2

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0find the critical numbers

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0oh goody.... the critical numbers are our max min and inflection points. So off to derivatives we go.... the first derivative tells us where the bends are. and the second derivative tells us what those bends are doing.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y' = 4t^3 36t^2 +32t make that equal to 0 to find the bends.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y'' = 12t^2 72t +32 when this is negative we have a maximum, where it is positive we have a minimum (might have those backwards) and where it is 0 are inflection points

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y'' = 4(3t^2 18t +8) [18 +sqrt((18^2)  4(8)(3))] / 2(3) 3 +sqrt(324  96)/6 3 (+) 2sqrt(57)/3 2 inflections: x=(3 2sqrt(57)/3) and (3 +2sqrt(57)/3) theres the two critical points of inflection... whew this is exhausting :)
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