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Maximum and Minimum
Absolute Maximum or Minimum
A function f has an absolute maximum at c
if f(c)≥f(x) for all x in the domain. The
number f(c) is called the maximum value
of f
 A function f has an absolute minimum at c
if f(c)≤f(x) for all x in the domain. The
number f(c) is called the minimum value
of f
 These absolute maximum or minimum
values are called the extreme values of f

Local (Relative) Maximum or
Minimum
Maximum or minimum in a smaller area
 Occur at any “hill” or “valley” on a
function

Examples:
Tell all max and min values for each
1) f(x) = cos x
2) f(x) = x2
3) f(x) = x3
4) f(x) = 3x4 – 16x3+ 18x2 on interval [-1, 4]
Extreme Value Theorem

If f is continuous on a closed interval [ a,
b], then f attains an absolute maximum
f(c) and an absolute minimum f(d) at
some number c and d in [a, b]
- more than one extreme may exist
- if f is not continuous, may not have an
extreme
- if the interval is not closed, there may
not be an extreme
Fermat’s Theorm
If f has a local maximum or minimum at c,
and if f’(c) exists, then f’(c) = 0
 Be careful, at every local max or min the
tangent is horizontal, but not every
horizontal tangent is a local max or min
 Also may be a max or min at locations
where f’ does not exist

Critical Number

A number where c in the domain of f
such that either f’(c) = 0 or f’(c) does not
exist
Example

Find the critical numbers of
f ( x )  x (4  x )
3/5
Finding Absolute Max and Min on a
Closed Interval
1)
2)
3)
4)
Find the critical numbers
Evaluate f at each critical number in the
interval
Evaluate each endpoint of the interval
The least of these values is the minimum,
the most is the maximum
Example
Find the absolute maximum and minimum
3
2
of f ( x)  x  3x  1 on the interval
[ -1/2, 4]
Practice: Find Max and Min
1) f ( x)  x 2 ( x  3)
0,4
2) f ( x )   x 2  3x
2,2
3) f ( x )  2(3  x )
1,2
4) g (t )  t 4  t
0,4
5) y  3x 2/3  2 x
1,1
t2
6) g (t )  2
t 3
7)h( x)  sin 2 x  cos x
1,1
0, 
The Hubble Space Telescope was deployed
by the Space Shuttle. A model for the
velocity of the shuttle during the mission
from liftoff at t=0 to when the boosters
were jettisoned at t=126s is given by
v(t )  0.001302t 3  0.09029t 2  23.61t  3.083
Use this model to estimate the maximum
and minimum values of acceleration.