Absolute Extremum

Tags: #◇Computer Science #◇CSMATH1 #◇Problem Solving

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◇Overview

An absolute extremum refers to either the absolute maximum or absolute minimum value of a function over a specific interval.

Absolute Maximum: A function $y = f(x)$ has an absolute maximum on an interval $I$ if there is a number $c$ in $I$ such that $f(x) \le f(c)$ for all $x \in I$. The value $f(c)$ is the absolute maximum value, and $(c, f(c))$ is the absolute maximum point.
Absolute Minimum: A function $y = f(x)$ has an absolute minimum on an interval $I$ if there is a number $c$ in $I$ such that $f(x) \ge f(c)$ for all $x \in I$. The value $f(c)$ is the absolute minimum value, and $(c, f(c))$ is the absolute minimum point.

◇Important Formulas

Extreme Value Theorem

If $y = f(x)$ is a function continuous on a closed interval $[a, b]$, then $f$ has both an absolute maximum value and an absolute minimum value on $[a, b]$.

Procedure for Finding Absolute Extrema on $[a, b]$

1. Find $f'(x)$ and solve for the critical numbers of $f$ in the interval $[a, b]$ (i.e., where $f'(x) = 0$ or $f'(x)$ is undefined).
2. Compute the function value $f(c)$ for each critical number $c$ found in step 1.
3. Compute the function values at the endpoints, $f(a)$ and $f(b)$.
4. Compare all the computed values. The largest value is the absolute maximum, and the smallest value is the absolute minimum of $f$ on the interval $[a, b]$.

◇Worked Examples

Example 1: Find the absolute extrema of $f(x) = x^3 + 3x^2 - 9x$ on $[-4, 4]$.

1. Since $f(x)$ is a polynomial function, it is continuous everywhere, including on $[-4, 4]$. By the Extreme Value Theorem, it has an absolute maximum and minimum on this interval.
2. Find critical numbers: $f'(x) = 3x^2 + 6x - 9$. Setting $f'(x) = 0$, we have:
   $3(x^2 + 2x - 3) = 0 \implies 3(x+3)(x-1) = 0$
   The critical numbers are $x = -3$ and $x = 1$. Both are in $[-4, 4]$.
3. Evaluate $f$ at the critical numbers:
   $f(-3) = (-3)^3 + 3(-3)^2 - 9(-3) = -27 + 27 + 27 = 27$
   $f(1) = (1)^3 + 3(1)^2 - 9(1) = 1 + 3 - 9 = -5$
4. Evaluate $f$ at the endpoints:
   $f(-4) = (-4)^3 + 3(-4)^2 - 9(-4) = -64 + 48 + 36 = 20$
   $f(4) = (4)^3 + 3(4)^2 - 9(4) = 64 + 48 - 36 = 76$
5. Comparing the values: $\{27, -5, 20, 76\}$.
   Absolute maximum value is $76$ at $x = 4$.
   Absolute minimum value is $-5$ at $x = 1$.

Therefore, $(4, 76)$ is the absolute maximum point and $(1, -5)$ is the absolute minimum point of $f$ on $[-4, 4]$.

◇References & Navigation

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• Related Notes: ◇Maximum and Minimum Function Values with Applications, ◇Functions and Their Graphs