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.