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