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!

- 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.

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

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.

## Answers (1)

You did the first question correctly, all thats left is the interpretation of your answers. A negative 2nd derivative means concave down. The 2nd derivative is negative for all negative numbers. So from (-infinity, 0) everything is concave down. Same thing but opposite for (0, infinity).

2.

f(x) = xe^(-x)

use the product rule to find the derivative

f '(x) = x * ((d/dx) e^(-x)) + e^(-x) * (d/dx) x

Apply the chain rule when doing ((d/dx) e^(-x))

f '(x) = xe^(-x)*((d/dx) -x) + e^(-x)*1

f '(x) = -xe^(-x) + e^(-x)

Now take the 2nd derivative. I'm just going to give the answer this time.

f ''(x) = xe^(-x) - 2e^(-x)

then you set this equal to zero to find the points of inflection (to do your sign chart).

0 = xe^(-x) - 2e^(-x)

factor out an e^(-x)

0 = e^(-x) (x - 2)

x = 2

Technically it will also equal 0 as x approaches infinity, but that's useless here.

From (-infinity, 2) the 2nd derivative is negative, so it's concave down. From (2, infinity) the 2nd derivative is positive, so it's concave up.

The sin graph switches signs every pi radians and starts off positive. So your original function will be concave up from (o, pi) concave down from (pi, 2pi) and concave up again from (2pi, 3pi).

When a problem is restricted to a certain domain, it usually means that there are infinite answers (here you could keep listing forever) and they want you to give them for just that domain. In other words, just list concavity changes on the given domain.

I'll help with the rest later if I have time.