SOLUTION: Find the open intervals where the function is concave up and concave down. f(x)= x^3 -3x^2 -1 can you also explain the inflection point and how it is related to the concavity s

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Question 1100519: Find the open intervals where the function is concave up and concave down.
f(x)= x^3 -3x^2 -1
can you also explain the inflection point and how it is related to the concavity since i dont really understand the whole concept
Thanks for helping out.
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
Graph it
graph%28300%2C300%2C-10%2C10%2C-10%2C10%2Cx%5E3-3x%5E2-1%29
Set first derivative equal to 0
3x^2-6x=0
x(2x-6)=0
x=0, 3 critical points.
Second derivative is 6x-6=0, x=1.
At x=1, function is negative, so this is a local maximum
At x=3, function is positive, so this is a local minimum
Inflection point is at x=1, where the curve is changing from forming a concave up/down to a concave down/up
The function is concave down from (-oo, 1) and concave up from (1,+oo)