Technicality of describing where sqrt(x) is increasing

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SUMMARY

The square root function, sqrt(x), is monotonically increasing from zero to infinity, represented as [0, ∞). Unlike x^2, which switches from decreasing to increasing at zero, sqrt(x) does not switch behavior at this point, allowing for the inclusion of zero in its domain. The definition of a monotonically increasing function confirms that for any x < y, f(x) < f(y) holds true for sqrt(x) across the interval [0, ∞). This discussion clarifies the proper notation for indicating the increasing nature of sqrt(x).

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When describing where $sqrt{x}$ (square root of x) is increasing, it's from zero to infinity. But, do you say (0,inf) or [0, inf) ?

(I'm tutoring a student in pre-calc, and this came up. They don't know any calculus.)

In a situation like where is $x^2$ inc/dec, we'd say inc: (0, inf) and dec (-inf, 0). We wouldn't include the zero because that's the point where the function switches from dec to inc.

But on square root of x, it's not "switching" from dec to inc, so can we include that zero?

Thanks so much.
 
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A function f is monotonically increasing over some set S if for any [itex]x<y \Rightarrow f(x) < f(y) \, \forall x,y \in S[/itex]. No need for derivatives here! With this definition, you could say [tex]\sqrt x[/itex] is a monitonically increasing function over the set [itex]S=[0,\infty)[/itex].[/tex]
 
D H said:
A function f is monotonically increasing over some set S if for any [itex]x<y \Rightarrow f(x) < f(y) \, \forall x,y \in S[/itex]. No need for derivatives here! With this definition, you could say [tex]\sqrt x[/itex] is a monitonically increasing function over the set [itex]S=[0,\infty)[/itex].[/tex]
[tex] And also, note that [itex]x^2[/itex] is strictly increasing on [itex][0, +\infty)[/itex]![/tex]
 

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