Bavikhove8h

2022-10-01

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 $[5/2,\mathrm{\infty})$. Why is this written as a closed interval, and not an open one?

Concavity, on the other hand, uses open intervals.

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 $[5/2,\mathrm{\infty})$. Why is this written as a closed interval, and not an open one?

Concavity, on the other hand, uses open intervals.

falwsay

Beginner2022-10-02Added 8 answers

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 $+\mathrm{\infty},-\mathrm{\infty}.$.

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 $+\mathrm{\infty},-\mathrm{\infty}.$.

shadyufog0

Beginner2022-10-03Added 1 answers

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 $a<x<b$ 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 ${f}^{\u2033}>0$ for $x<a$ and ${f}^{\u2033}<0$ 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 ${f}^{\u2033}(a)=0$ but ${f}^{\u2033}>0$ for $x<a$ and ${f}^{\u2033}<0$ for $x>a$. Then we still say f is concave up at a. (Think of $f(x)={x}^{4}$.)

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 $a<x<b$ 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 ${f}^{\u2033}>0$ for $x<a$ and ${f}^{\u2033}<0$ 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 ${f}^{\u2033}(a)=0$ but ${f}^{\u2033}>0$ for $x<a$ and ${f}^{\u2033}<0$ for $x>a$. Then we still say f is concave up at a. (Think of $f(x)={x}^{4}$.)

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