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Math Help - Find the intervals of concavity and inflection point

  1. #1
    DBA
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    Find the intervals of concavity and inflection point

    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!
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  2. #2
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    Quote Originally Posted by DBA View Post
    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!
    the domain of the original function is x > -3

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

    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.
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  3. #3
    DBA
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    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?
    Last edited by DBA; October 24th 2009 at 02:58 PM. Reason: additional information
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  4. #4
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    f''(x) = \frac{3(x+4)}{4(x+3)^{3/2}} = 0<br />

    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 ...
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  5. #5
    DBA
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    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!
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  6. #6
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    you're welcome.

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