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F(x) = 1/(x^24), such that F''(x) = (6x^28)/(x^24)^3
I am asked to find the intervals of concavity and possible points of inflection. But....
 one year ago
 one year ago
F(x) = 1/(x^24), such that F''(x) = (6x^28)/(x^24)^3 I am asked to find the intervals of concavity and possible points of inflection. But....
 one year ago
 one year ago

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EdutopiaBest ResponseYou've already chosen the best response.0
When i set F''(x)=0 then i get a value (+2/sqrt(3)) and WOLF says that this graph has no points of inflection (because it is asymptotic). What am i misunderstanding?
 one year ago

NoelGrecoBest ResponseYou've already chosen the best response.0
The minus sign in the numerator of F'' should be a plus sign. That makes the numerator always >0.
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.0
Let me know if you're on line at he same time I am and I'll explain.
 one year ago
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