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!
Thanks for your replay.
f''(x) > 0 ... which indicates f(x) is concave up for x > -3.
Thats what the solution said too.
But I do not understand how I can make this conclusion..
Why is f''(x) >0 ?
Do I have to plug in the critical points which are not max or min and see if the y-value turns out positive?
That's what I do not get?
I know, once I have figured out if f''(x) is positive or negative, what concavity the function has.
I thought you have to set f''(x)=0
For example, I did that when I had f''(x)= 12x(x-2)
So, I got x=0 and x=2
These numbers formed the intervals I was looking at.
Now I have f''(x)= 3/4 * [ (x+4) / (x+3)^(3/2) ]
So, I thought I need to set it zero
which would be only the numerator 3*(x+4)=0 --> x+4=0 and therefore x=-4
Since, this is not in the domain, I do not have a inflection point. Is that right
But how do they get that the rest is convave upwards?
Thank you.
The problem was, that I have to draw the graph.
So, I pluged in a number out of the interval (-3, infinite) to see it f''(x) gets positive and therefore is concave upward?
Like, x=0
so, f''(0) ~ 0.58 which is greater than zero ->
f''(x) > 0 for the interval (-3, infinite)
I guess I understood it now. Thanks again!