Need help with concavity
suppose f(x)=(3x)/(x^2+49)
first derivative is f'(x)=(147-3x^2)/((x^2+49)^2)
I know that concavity has to do with the second derivative but I'm not sure what i'm really doing. some help clearing this up would be great.

Question

Need help with concavity 
suppose f(x)=(3x)/(x^2+49)
first derivative is f'(x)=(147-3x^2)/((x^2+49)^2)
I know that concavity has to do with the second derivative but I'm not sure what i'm really doing. some help clearing this up would be great.

anonymous · Answer

I assume you know how to find the critical points, take the second deriv﻿ative, the same way you took the first, then find the critical points of the second derivative make a table, then insert your numbers in between the points of the critical numbers. Then if you get negative you get a frown, and positive goes up. Then you just insert the concavity into the graph based on where it is in the domain.
so from -infinity to zero the function would be increasing and from 2 to infinity it would be decreasing. from 0 to two it would look like a frown

anonymous · Answer

You take the critical numbers you get and you sub them into your original function check to see if you get a positive or negative number