# 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?

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 $f\left(x\right)={x}^{2}-5x+6$, it says the interval of increase is $\left[5/2,\mathrm{\infty }\right)$. Why is this written as a closed interval, and not an open one?
Concavity, on the other hand, uses open intervals.
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falwsay
Explanation:
In an open interval the domain is unbounded when concavity is considered. The particular example function is symmetrical about $x=\frac{5}{2}$ so includes full range $+\mathrm{\infty },-\mathrm{\infty }.$.
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Step 1
Increasing or decreasing compares the function value at 2 points in the interval.
If ${f}^{\prime }>0$ on (a,b) and f is continuous on [a,b] then if x is such that $a then $f\left(x\right) so f IS increasing on [a,b].
Step 2
On the other hand, concavity is generally an attribute at a specific point, so if ${f}^{″}>0$ for $x and ${f}^{″}<0$ for $x>a$, 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 ${f}^{″}\left(a\right)=0$ but ${f}^{″}>0$ for $x and ${f}^{″}<0$ for $x>a$. Then we still say f is concave up at a. (Think of $f\left(x\right)={x}^{4}$.)