F(x) = x^3 / (x-16) is concave up and down on what intervals?

Question

I got the derivative to be f' = (x^4 - 48) / (x^2-16)^2 and I found the local min and max but now I need to find where it concaves up and down.

Answer 1

Now you need to take the second derivative.

Find where the second derivative is 0 or undefined, and make a sign line around those values.

When f''(x)<0, f(x) is concave down; when f''(x)>0, f(x) is concave up.