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# Concavity and the Second Deriv Test...having some serious issues. HELP!?

Alright, so I totally understand that you need to take the second derivative to find where it's concave up and concave down. Here are the problems and here is what I did. Please give me your input and help me out. Thank you!

1. F(x) = x^3 --> f'(x) = 3x^2 --> f"(x) = 6x = 0 so x = 0

concave up: (0, infinity) and concave down (-infinity, 0)

I did a sign chart and found that when x = any number less than zero, f"(x) was negative and when x = any number greater than zero, f"(x) was positive.

1. F(x) = xe^(-x) Unfortunetly, I keep screwing up to find the derivative when I chain rule it...

2. F(x) = x-2sin x restricted to 0 - 3π; f'(x) = 1 - 2 cosx; f"(x) = 2sin(x)

how do i explain where its concave up and down? when you graph it, you can see my dilemma. and it says "restricted to 0-3pi." what!?!

Any help would be amazing...and please, no spam. 