Originally Posted by

**DBA** **I already know:**

f(x)= x* sqrt(x+3)

f'(x) = [3(x+2)] / [2*sqrt(x+3)]

Critical numbers at x=-2 and x= -3

(-3,-2) --> function f is decreasing

(-2, infinite) --> function f is increasing

Local minimum is at x=-2

**Find the intervals of concavity and the inflection points:**

I found f''(x) to be

f''(x)= 3/4 * [ (x+4) / (x+3)^(3/2) ]

I know that

f''(x) > 0 for all x in the interval --> the graph is concave upwards

f''(x) < 0 for all x in the interval --> the graph is concave downwards

The solution says that

f''(x) > 0 for the interval (-3, infinite).

I do not know how they got this information. **So, how do I get from the f''(x) to the point I know, at what interval it is positive or negative?**

I understand that I will not have an inflection point then.

Please provide an example.

Thanks!