Calculus / Unit 6: Applications of Derivatives II
6.4 Absolute Extrema on a Closed Interval
Notes
The Extreme Value Theorem: a continuous function on $[a, b]$ has an absolute maximum and an absolute minimum.
- Method: find critical numbers inside $(a, b)$, evaluate $f$ at them and at both endpoints, then compare.
- Example: $f(x) = x^3 - 3x$ on $[0, 2]$. Critical number $x = 1$. Values: $f(0) = 0$, $f(1) = -2$, $f(2) = 2$. The absolute max is 2 and the absolute min is $-2$.
Practice
What does the Extreme Value Theorem guarantee?
It requires continuity on a closed interval.
Find the absolute maximum value of $f(x) = x^3 - 3x$ on $[0, 2]$.
Compare $f(0) = 0$, $f(1) = -2$, $f(2) = 2$.
Find the absolute minimum value of $f(x) = x^2 - 4x + 3$ on $[0, 5]$.
The critical number is $x = 2$. Values: $f(0) = 3$, $f(2) = -1$, $f(5) = 8$.
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