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?

Bavikhove8h 2022-10-01 Answered
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 5 x + 6, it says the interval of increase is [ 5 / 2 , ). 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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Answers (2)

falwsay
Answered 2022-10-02 Author has 8 answers
Explanation:
In an open interval the domain is unbounded when concavity is considered. The particular example function is symmetrical about x = 5 2 so includes full range + , ..
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shadyufog0
Answered 2022-10-03 Author has 1 answers
Step 1
Increasing or decreasing compares the function value at 2 points in the interval.
If f > 0 on (a,b) and f is continuous on [a,b] then if x is such that a < x < b then f ( x ) < f ( b ) 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 < a 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 ( a ) = 0 but f > 0 for x < a and f < 0 for x > a. Then we still say f is concave up at a. (Think of f ( x ) = x 4 .)
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