Interpreting the graph of $g(x) = \int_0^x f(t)dt$ Disclaimer: I am not a lazy student trying to get free internet homework help. I am an adult who is learning Calculus from a textbook.  I am deeply grateful to the members of this community for their time. 
I made up a graph and wanted to make up questions about it.
I just wanted to clarify 2 answers I got.
Note that $g(x) = \int_0^x f(t)dt$
So, $g'(x) = f(x) $
and $g''(x) = f'(x)$
1) x=3 is indeed a relative max, correct?  Since it's one of the bounds, the derivative doesn't change sign at 3, but g(x) is increasing all the way to x=3.
2) Inflection point is where $g''(x) = f'(x)$ changes sign.  Well, this happens where $f(x)$ changes slope (reverses direction).  This happens at $x=-2.$  But, what is the deal with the flat part of the graph at [0,3]?  It technically reverses direction but only after a flat part from [1,2].  Is my below conclusion correct?  Is there a better way to explain this?


1) Yes. $g(x)$ has a relative max at $x=3$ for that interval. This is because the area from $x=0$ to $x=3$ is greater than the area from $x=0$ to $x=-3$ minus the rest of the area there (recall: $\int_0^{-3}=-\int_{-3}^0=$ positive for negative area). I wouldn't really look at derivatives to figure that one out because, if $f(3)$ was at a point higher than 0, the value of $g(3)$ would be even higher. Slopes are more of to determine extrema, which are not necessarily maximum/minimum values of an interval.
2) As far as I know (and correct me if I'm wrong), you're right - there is no inflection point because there isn't a point where the second derivatives of the points to the left and right sides have different signs. And $x=-2$ is indeed an inflection point even though $f'(-2)$ is undefined.
