Posted by Desperado on Monday, January 11, 2010 at 2:55am.
CORRECTION!
I read the problem wrong
..
its actually
g(x) = x√(8 - x^2)
the second x is the only thing that is squared not the whole (8-x).
sorry!
The maxima/minima are part of the important ingredients when we try to sketch the graph of a function without having to plot every possible point.
A local maximum of a function occurs at an interior point c (i.e. f(c+) and f(c-) exist) of its domain if f(x) ≤ f(c) for all x in some open interval containing c. The definition of local minimum is similar.
It is possible to have multiple local maxima/minima.
A global (absolute) maximum of a function occurs at a point c if f(x) ≤ f(c) for all x on its domain. the definition for global minimum is similar.
For example, the function y=x² has a local minimum at x=0. We note that y(0)=0, and at both x=0+ and at x=x-, the value of the function is greater than 0. Therefore x=0 is a local minimum for y=x².
However, there is no local (nor global) maximum for y=x², since for any value of y(x) corresponding to a particular value of x, we can find a greater value of y(x). Since x=±∞ is not in the domain of y=x², there is no local nor global maximum.
To find the local and global maxima/minima, it is necessary to follow these steps:
1. find the domain of the function.
2. Calculate all the critical points in the domain. A critical point is a point where the derivative f'(x) becomes zero, or or where f'(x) is undefined.
3. Since a local extremum can only occur at the following points:
a. interior point where f'(x)=0.
b. endpoints of the domain of f(x).
c. interior points where f'(x) is undefined.
So calculate the values of f(x) at all the critical points and at endpoints of the domain. From the calculated values of f(x), determine the local/global extrema.
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