Maximum of the function F(X)=X*(1+X)^0.5

[Mike s]

Hello,
Is x=-1 local maximum of $F(x)=x\sqrt{1+x}$?
On the one hand, $F(-1+\delta)<F(-1)$ for $0<\delta<1$.
However, $F(-1-\delta)$ is not defined.
As far a I know, the point x=a is considered local maximum, if there exists small neighborhood $\delta$ such that $$F(a)>F(x)$$ for every $a-\delta<x<a+\delta$.

So is x=-1 a local maximum or not?

 

[Fightfish]

I believe that the term "local maximum" is not applicable because the concept of neighbourhood does not exist about the point x=-1 as you have pointed out.

Rather this is what we call a "boundary maximum". If you look at the gradient of F(x) at x=-1, it is undefined, as opposed to 0 gradient for stationary maxima/minima.

It is worth noting that boundary points are always "maximum" or "minimum" (except when the curve is level about that point).

 

[Ray Vickson]

Hello,
Is x=-1 local maximum of $F(x)=x\sqrt{1+x}$?
On the one hand, $F(-1+\delta)<F(-1)$ for $0<\delta<1$.
However, $F(-1-\delta)$ is not defined.
As far a I know, the point x=a is considered local maximum, if there exists small neighborhood $\delta$ such that $$F(a)>F(x)$$ for every $a-\delta<x<a+\delta$.

So is x=-1 a local maximum or not?

Yes, it is. It satisfies F(-1) > F(x) for all sufficiently small |x+1| with x > -1. In this problem it is impossible to compare F(-1) to F(x) for x < -1, but that is not really the reason that x = -1 is a local max. In the presence of constraints a local max or min need not have derivative = 0, and whether or not the function could be defined outside the interval is not particularly important; for example, for the problem min/max f(x) = x, subject to 0 <= x <= 1, x = 0 is a min and x = 1 is a max. These two statements are true even though f(x) is perfectly well defined outside the interval.

Note: you may find that some writers disagree with these statements. However, they are commonly asserted in modern treatments of optimization theory and practice; they are really matters of definition, not of substance.

RGV

 

[Fightfish]

Note: you may find that some writers disagree with these statements. However, they are commonly asserted in modern treatments of optimization theory and practice; they are really matters of definition, not of substance.

Agreed. Most texts do define local maxima/minima as being on an open interval. In a closed interval, it is trivial as the endpoints certainly must be minima/maxima.

For all practical purposes though, it doesn't matter if its "local" or "not local". It's not going to affect your calculations or results, so don't fret too much over it.

 

[Mike s]

I understand, thanks for both of you.