Interval related to increasing/decreasing and concavity/convexity Why do some people use closed intervals when describing the intervals where a function is increasing/decreasing or concave/convex?
For example, given the function $f(x)= x^2-5x+6$, it says the interval of increase is $[5/2, \infty)$. Why is this written as a closed interval, and not an open one?
Concavity, on the other hand, uses open intervals.
 A: It just depends on what points you are interested in. Sometimes closed intervals are the ones you're making statements about, sometimes it's open intervals. The concavity property only applies to the open interval, so that's what is used.
A: In an open interval the domain is unbounded when concavity is considered. The particular example function is symmetrical about $x= \frac52 $ so includes full range $ +\infty , -\infty. $
A: Increasing or decreasing compares the function value at 2 points in the interval. 
If $f'\gt 0$ on $(a,b)$ and $f$ is continuous on $[a,b]$ then if $x$ is such that $a \lt x \lt b$ then $f(x)\lt f(b)$ so $f$ IS increasing on $[a,b]$.  
On the other hand, concavity is generally an attribute at a specific point, so if $f ' ' \gt 0$ for $x \lt a$ and $f ' ' \lt 0$ for $x \gt a$, we wouldn't say $f$ is concave up or down at $x = a$. 
The reason I put the modifier "generally" in there is in the case where $f ' '(a) = 0$ but $f' ' \gt 0$ for $x \lt a$ AND $f' ' \lt 0$ for $x \gt a$. Then we still say $f$ is concave up at $a$. (Think of $f(x) = x^4$.)
