Relative Extrema, Absolute Maxima and Minima

Relative Extrema, Absolute Maxima, and Minima

1. Relative (Local) Extrema

A function $f(x)$ has a relative maximum at $x = c$ if $f(c)$ is greater than all nearby function values. Similarly, $f(x)$ has a relative minimum at $x = c$ if $f(c)$ is less than all nearby values.

First Derivative Test for Relative Extrema

To find relative extrema:

1. Find the first derivative $f'(x)$.
2. Set $f'(x) = 0$ or find where $f'(x)$ is undefined (critical points).
3. Use sign changes in $f'(x)$ to determine extrema:
   • If $f'(x)$ changes from positive to negative, $f(x)$ has a relative maximum.
   • If $f'(x)$ changes from negative to positive, $f(x)$ has a relative minimum.

Example

Find the relative extrema of $f(x) = x^3 - 3x^2 + 4$.

1. Compute the derivative:

   $$f'(x) = 3x^2 - 6x$$

2. Set $f'(x) = 0$:

   $$3x(x - 2) = 0$$ $$x = 0, \quad x = 2$$

3. Use test points to check sign changes:

   • For $x < 0$, pick $x = -1$: $f'(-1) = 3(1) - 6(-1) = 9$ (positive, increasing).
   • For $0 < x < 2$, pick $x = 1$: $f'(1) = 3(1) - 6(1) = -3$ (negative, decreasing).
   • For $x > 2$, pick $x = 3$: $f'(3) = 3(9) - 6(3) = 9$ (positive, increasing).

   Conclusion:

   • $x = 0$ is a relative maximum because $f'(x)$ changes from positive to negative.
   • $x = 2$ is a relative minimum because $f'(x)$ changes from negative to positive.

2. Absolute (Global) Maximum and Minimum

A function has an absolute maximum at $x = c$ if $f(c)$ is the largest value over the entire domain. It has an absolute minimum if $f(c)$ is the smallest value.

Steps to Find Absolute Extrema

1. Find critical points by setting $f'(x) = 0$.
2. Evaluate the function at critical points and endpoints (for closed intervals).
3. Compare function values to determine the highest and lowest points.

Example on a Closed Interval

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

1. Critical points: From the previous example, we found $x = 0, 2$.

2. Evaluate $f(x)$ at critical points and endpoints:

   • $f(-1) = (-1)^3 - 3(-1)^2 + 4 = -1 - 3 + 4 = 0$
   • $f(0) = (0)^3 - 3(0)^2 + 4 = 4$
   • $f(2) = (2)^3 - 3(2)^2 + 4 = 8 - 12 + 4 = 0$
   • $f(3) = (3)^3 - 3(3)^2 + 4 = 27 - 27 + 4 = 4$

3. Compare values:

   • Maximum value: $f(0) = f(3) = 4$ → Absolute maximum at $x = 0$ and $x = 3$.
   • Minimum value: $f(-1) = f(2) = 0$ → Absolute minimum at $x = -1$ and $x = 2$.

Summary

• Relative extrema (local max/min) occur where $f'(x) = 0$ and the derivative changes sign.
• Absolute extrema (global max/min) occur at critical points or endpoints in a closed interval.
• The First Derivative Test helps determine relative extrema.
• The Absolute Extrema Test compares function values at critical points and endpoints.

These concepts are fundamental in optimization problems, physics, economics, and engineering.