Asked by Z32

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

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.

Answers

Answered by Marth
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.
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