Module 6 the extreme value theorem
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A continuous function on a closed interval will always have both a maximum and minimum value. The maximum value is the largest value the function takes on within the interval, while the minimum value is the smallest value. To find the extreme values, take the derivative to find any critical values, then evaluate the original function at the critical values and endpoints to determine which yields the maximum and minimum.
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![THE EXTREME VALUE THEOREM
If 𝑓(𝑥) is a continuous on a closed
interval [𝑎, 𝑏], then 𝑓(𝑥) has both a
maximum and minimum value on the
interval. In other words, it must have at
least two extreme values.
The MAXIMUM VALUE of a function
on an interval is the largest value the
function takes on within the interval.
The MINIMUM VALUE of a function on
an interval is the smallest value the
function takes on within the interval.
STEPS IN SOLVING THE EXTREME
VALUES
Solve for the extreme values of the
function: 𝒇(𝒙) = 𝟑𝒙𝟐
− 𝟏𝟐𝒙 + 𝟓; [𝟏, 𝟒]
STEP 1 Find the critical value (find the derivative)
𝒇(𝒙) = 𝟑𝒙𝟐
− 𝟏𝟐𝒙 + 𝟓
𝒇′(𝒙) = 𝟔𝒙 − 𝟏𝟐
𝟔𝒙 − 𝟏𝟐 = 𝟎
𝟔𝒙 = 𝟏𝟐
𝟔𝒙
𝟔
=
𝟏𝟐
𝟔
𝒙 = 𝟐
STEP 2 Find the ‘y’ values
• 𝒇(𝟏) = 𝟑(𝟏)𝟐
− 𝟏𝟐(𝟏) + 𝟓 = −𝟒
• 𝒇(𝟒) = 𝟑(𝟒)𝟐
− 𝟏𝟐(𝟒) + 𝟓 = 𝟓
• 𝒇(𝟐) = 𝟑(𝟐)𝟐
− 𝟏𝟐(𝟐) + 𝟓 = −𝟕
STEP 3 Find the extreme values
• 𝒇(𝟏) = −𝟒
• 𝒇(𝟒) = 𝟓 → 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒗𝒂𝒍𝒖𝒆
• 𝒇(𝟐) = −𝟕 → 𝒎𝒊𝒏𝒊𝒎𝒖𝒎 𝒗𝒂𝒍𝒖𝒆](https://image.slidesharecdn.com/module6theextremevaluetheorem-210225010502/85/Module-6-the-extreme-value-theorem-1-320.jpg)
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Module 6 the extreme value theorem
- 1. THE EXTREME VALUE THEOREM If 𝑓(𝑥) is a continuous on a closed interval [𝑎, 𝑏], then 𝑓(𝑥) has both a maximum and minimum value on the interval. In other words, it must have at least two extreme values. The MAXIMUM VALUE of a function on an interval is the largest value the function takes on within the interval. The MINIMUM VALUE of a function on an interval is the smallest value the function takes on within the interval. STEPS IN SOLVING THE EXTREME VALUES Solve for the extreme values of the function: 𝒇(𝒙) = 𝟑𝒙𝟐 − 𝟏𝟐𝒙 + 𝟓; [𝟏, 𝟒] STEP 1 Find the critical value (find the derivative) 𝒇(𝒙) = 𝟑𝒙𝟐 − 𝟏𝟐𝒙 + 𝟓 𝒇′(𝒙) = 𝟔𝒙 − 𝟏𝟐 𝟔𝒙 − 𝟏𝟐 = 𝟎 𝟔𝒙 = 𝟏𝟐 𝟔𝒙 𝟔 = 𝟏𝟐 𝟔 𝒙 = 𝟐 STEP 2 Find the ‘y’ values • 𝒇(𝟏) = 𝟑(𝟏)𝟐 − 𝟏𝟐(𝟏) + 𝟓 = −𝟒 • 𝒇(𝟒) = 𝟑(𝟒)𝟐 − 𝟏𝟐(𝟒) + 𝟓 = 𝟓 • 𝒇(𝟐) = 𝟑(𝟐)𝟐 − 𝟏𝟐(𝟐) + 𝟓 = −𝟕 STEP 3 Find the extreme values • 𝒇(𝟏) = −𝟒 • 𝒇(𝟒) = 𝟓 → 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒗𝒂𝒍𝒖𝒆 • 𝒇(𝟐) = −𝟕 → 𝒎𝒊𝒏𝒊𝒎𝒖𝒎 𝒗𝒂𝒍𝒖𝒆