Thread: Find the intervals of concavity and inflection point

1. Find the intervals of concavity and inflection point

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.
Thanks!

2. Originally Posted by DBA

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.
Thanks!
the domain of the original function is x > -3

$f''(x) = \frac{3(x+4)}{4(x+3)^{3/2}}
$

the only critical value in the domain is x = -3 ... for all values of x > -3, f''(x) > 0 ... which indicates f(x) is concave up for x > -3.

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?

4. $f''(x) = \frac{3(x+4)}{4(x+3)^{3/2}} = 0
$

x = -4 is the only solution, but is is not in the domain of f(x).

x = -3 is the only critical value, because it makes f''(x) undefined.

f''(x) > 0 for any value of x > -3 ... f(x) is concave up on the interval $(-3, \infty)$

look at the graph ...

5. 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!

6. you're welcome.