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

1. May 13, 2012

### 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?

2. May 13, 2012

### 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).

3. May 13, 2012

### Ray Vickson

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

4. May 14, 2012

### Fightfish

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.

5. May 14, 2012

### Mike s

I understand, thanks for both of you.