ONE MORE STEP TOWARD BETTER TOMORROW : Global Extrema

Tuesday, 26 March 2013

Global Extrema

to find local extrema; one is often more interested in finding global extrema:
We say that the function f(x) has a global maximum at x=x₀ on the interval I, if $f(x_0)\geq f(x)$ for all $x\in I$ . Similarly, the function f(x) has a global minimum at x=x₀ on the interval I, if $f(x_0)\leq f(x)$ for all $x\in I$ .
If f(x) is a continuous function on a closed bounded interval [a,b], then f(x) will have a global maximum and a global minimum on [a,b]! (This is not easy to prove, though).
On the other hand, if the interval is not bounded or closed, then there is no guarantee that a continuous function f(x) will have global extrema. Examples: f(x)=x² does not have a global maximum on the interval $[0,\infty)$ , the function $f(x)=-\frac{1}{x}$ does not have a global minimum on the interval (0,1).
How can we find global extrema? Unfortunately, not every global extremum is also a local extremum:
Example. Consider the function f(x) = (x-1)², for $x \in [0,3]$ . The only critical point is x=1. And the first or second derivative test will imply that x=1 is a local minimum. Looking at the graph (see below) we see that the right endpoint of the interval [0,3] is the global maximum.

This leads us to introduce the new concept of endpoint extrema. Indeed, if c is an endpoint of the domain of f(x), then f(x) is said to have an endpoint maximum at c iff $f(x) \leq f(c)$ for all x in the domain close to c. Similarly one can define the concept of an endpoint minimum.
The news is not too bad, though. If f(x) is differentiable on the interval I, then:

Every global extremum is a local extremum or an endpoint extremum.

This suggests the following strategy to find global extrema:

Find the critical points.
List the endpoints of the interval under consideration.
The global extrema of f(x) can only occur at these points! Evaluate f(x) at these points to check where the global maxima and minima are located.

Example. Let us find the global extrema of the function f(x)=x e^-x on the interval [0.1,3.5]. The function f(x)is differentiable everywhere, its derivative f'(x)=e^-x-xe^-x=(1-x)e^-x is zero only at x=1. Thus x=1 is the only critical point. Throw in the endpoints of the interval x=0.1 and x=3.5, and evaluate f(x):