Math 130 §6.1
Nguyen 1 of 3
Section 6.1 Absolute Extrema
The largest possible value of a function is called the absolute maximum, and the smallest
possible value of a function is called the absolute minimum.
On a graph of a function, absolute extrema often coincide with relative extrema. But watch
out: although a function may have several relative maxima or relative minima, it never has more
than one absolute maximum or absolute minimum. Also, if the absolute maximum or minimum
of a function does exist, this absolute extremum value may occur at many points. How?
* A function may have no absolute
maximum or absolute minimum or both.
Let’s look at some graphs:
2
Graph f ( x)  x 2 e x here:
Absolute max at 2 locations!
No absolute minimum!
Absolute min at 1, 5 
No absolute maximum!
No absolute extrema!
One of the main reasons for the importance of absolute extrema is given by the following theorem:
Extreme Value Theorem
A function f that is continuous on a closed interval  a, b will have both an absolute
maximum and an absolute minimum on the interval.
For instance, consider y  f ( x ) on the closed interval c1 , c3  . Then:
y
y f x
f has an absolute maximum at …x = c1…
y  f ( x)
f has an absolute minimum at …x = c2….
c1
c2
c3
x
But, if on the Domain of f = (, ) , then:
f has NO absolute min, and an absolute
max at x = c1.
Math 130 §6.1
Nguyen 2 of 3
To find Absolute Extrema for a function f Continuous on a Closed Interval  a, b
Procedure:
1) Find all critical numbers for f in the open interval (a, b) .
2) Evaluate f for all such critical numbers in (a, b) .
3) Evaluate f for the endpoints a and b of the interval.
4) The largest value found in steps 2 & 3 is the absolute maximum for f on  a, b . The
smallest value found in steps 2 & 3 is the absolute minimum for f on  a, b .
* Remember: an absolute extremum is a y-value, not an x-value.
f ( x)  x 3  3 x  5
Example 1:
Find the absolute extrema for
Solution:
Since f ( x)  3 x 2  3 , f  is never undefined.
over  2,3
Solving f ( x)  0 yields x = ±1, both are inside the OPEN interval (−2,3). (Check?)
Then, let’s evaluate the ORIGINAL function f at the 4 values of x = −2, −1, 1, and 3:
f (2)  3; f (1)  7; f (1)  3; f (3)  23
(Check?)
Thus, f has Absolute Max value of 23 at x = 3, and Absolute Min value of 3 at x = −2 and 1
g ( x)  4 x  x
over 0,16
Example 2:
Find the absolute extrema for
Solution:
1
2
Since f ( x)  4  x 1 2  1 
 1 , f  is undefined at x = 0.
2
x
(Check?)
0 is NOT inside the OPEN interval (0,16), but 0 is an endpoint of the GIVEN interval, so
we’ll use it later!!!
Solving f ( x)  0 yields x = 4, which is inside the OPEN interval (0,16). (Check?)
Then, let’s evaluate the ORIGINAL function g at the 3 values of x = 0, 4, and 16:
g (0)  0; g (4)  4; g (16)  0
(Check?)
Thus, g has Absolute Max value of 4 at x = 4, and Absolute Min value of 0 at x = 0 and 16
Math 130 §6.1
Example 3:
Solution:
Nguyen 3 of 3
Find the absolute extrema for
h( x)  2 x 2  ln x
1 
over  ,2
4 
1
Since h( x)  4 x  , h is undefined at x = 0, which is NOT inside the open
x
interval (1/4, 2), so we IGNORE it. (Check?)
Solving h( x)  0 yields x = ±1/2, ONLY 1/2 is inside the open interval (1/4, 2). Check?)
Then, let’s evaluate h at the 3 values of x = 1/4, 1/2, and 2:
h(1 4)  1.5113; h(1 2)  1.1931; h(2)  7.3069
(Check?)
Thus, h has Absolute Max value of 7.3069 at x = 2, and Absolute Min value of 1.1931 at x  1 4
p( x) 
x 1
( x  1) 2
Example 4:
Find the absolute extrema for
Solution:
Use Quotient Rule to find and simplify p to be:
over 0,4
p( x) 
3 x
( x  1)3
(Check?)
p is undefined when x = −1, which is NOT inside (0,4), so we IGNORE it. [By the
way, even if x = −1 were inside (0,4), we would still ignore it because it’s NOT a critical number
– remember that Domain of p is x  1 . Aha!!!]
Solving p( x)  0 yields x = 3, which is inside the open interval (0,4).
(Check?)
Then, let’s evaluate p at the 3 values of x = 0, 3 and 4:
p (0)  1; p (3)  0.125; p (4)  0.12
(Check?)
Thus, p has Absolute Max value of 0.125 at x = 3, and Absolute Min value of −1 at x = 0