# Local Vs Global Max/Min by HC11120608751

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```									                               Local Vs Global Max/Min
Local Max/Min
A local maximum/minimum is a point (p, f(p)) on the graph of that has the
largest/smallest function value around. Formally we say there is a local max/min at x = p
if f(p) is the largest/smallest function value in a neighborhood or set of points around p.
Local maxima and minima occur at points where f   x   0 or where f   x  is
undefined. It’s important to remember that not all points where f   x   0 or is
undefined are local extrema but that all local extrema are at points where f   x   0 or is
undefined. The requirement that f   x   0 or is undefined is known as a necessary but
not sufficient condition for local extrema. It also gives us a good way to look for them.
a.) Find f '(x)
b.) Find all points where f '(x) = 0 or is undefined. These are known as
the critical points (CPs) of f(x).
c.) Test each CP to see if it is a local max/min or neither.
d.) The first derivative test says that if f '(x) changes sign at a CP then it
is a local max/min. If f '(x) changes from positive to negative at the CP
then the CP is a local max, if f '(x) changes from negative to positive at the
CP then the CP is a local min and if f '(x) doesn’t change sign at the CP
then the CP is neither a local max nor a local min.
e.) The second derivative test says that if f "(x) is positive at the CP then
the CP is a local min, if f "(x) is negative at the CP then the CP is a local
max and if f "(x) is zero at the CP the second derivative test is
inconclusive.
x5 16 x3
Example: f  x   
5      3
Find the CPs:
         
f   x   x 4  16 x 2  x 2 x 2  16  x 2  x  4  x  4   0
when x  0 or x  4 or x  4
Test the CPs:
Using the first derivative test, when x = 4 f   x   0 for
values to the left of 4 and f   x   0 for values to the right
of 4 (try evaluating f  3.9 and f   4.1 to see what
happens). This tells us that there is a local minimum at
x = 4.

Using the second derivative test, when x = -4
f   x   4x3  32x and
f   4  4  4  32  4  256  128  128  0 .
3

This tells us that there is a local maximum at x = -4.
When x = 0 the second derivative test is inconclusive
because f   0  0 . Using the first derivative test we see
that f   x   0 both to the right of 0 and to the left of 0.
(try evaluating f   .1 and f  .1 to see what happens)
This tells us that x = 0 is neither a local max nor a local
min.

Global Max/Min
A global max/min is a point (p, f(p)) on the graph of f(x) that has the
largest/smallest function value on the domain of f(x). Formally we say there is a global
max at x = p if f  p   f  x  for all x in the domain and a global minimum at x = p if
f  p   f  x  for all x in the domain.
Any function that is continuous on a closed, bounded interval [a, b] will have both
a global max and a global min on that interval. To find them
a.) Find all the CPs in [a, b]
b.) Find f(a) , f(b) and f(CP) for all CPs in [a, b]
c.) The largest of these function values will be the global max and the
smallest will be the global min. It’s important to remember that the global
max/min might be at one of the endpoints or it might be at one of the CPs
but it will exist.

x5 16 x3
Example: Find the global max and min of the function f  x               on the
5      3
interval 7,5 . We already know there is a local max at x = -4 and a
local min at x = 4. We want to find the function values at thest CPs and
also at the endpoints:
f  7   1532
f  4   136.53
f  4   136.53
f  5   41.67
We can see that on the interval 7,5 the global max occurs at x = -4 and
the global min at x = -7.

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