7.2 Absolute and relative extrema

Absolute and Relative Extrema

Defining Absolute and Relative Extrema

Understanding the difference between absolute and relative extrema comes down to scope. Absolute extrema look at the entire domain, while relative extrema only care about what's happening in a small neighborhood around a point.

• Absolute maximum: the largest value of $f(x)$ over its entire domain. This is the single highest point on the graph across the whole interval you're considering. For $f(x) = -x^2 + 4x - 3$ on $[0, 4]$, the absolute maximum is 1, occurring at $x = 2$.
• Absolute minimum: the smallest value of $f(x)$ over its entire domain. For $f(x) = x^2 - 4x + 5$ on $[-1, 3]$, the absolute minimum is 1 at $x = 2$.
• Relative (local) maximum: the largest value of $f(x)$ within some neighborhood of a point. The function rises to this point and then falls away. For $f(x) = x^3 - 3x^2 - 9x + 7$, a relative maximum occurs at $x = -1$.
• Relative (local) minimum: the smallest value of $f(x)$ within some neighborhood of a point. The function decreases to this point and then increases. For the same function, a relative minimum occurs at $x = 3$.

A relative extremum doesn't have to be an absolute extremum. A function can have several relative maxima, but only one value can be the absolute maximum on a given domain. Conversely, an absolute extremum on a closed interval might occur at an endpoint, where it wouldn't be considered a relative extremum by most definitions.

Finding Absolute and Relative Extrema

Finding absolute extrema on a closed interval follows a direct procedure:

1. Compute $f'(x)$ and find all critical points where $f'(x) = 0$ or $f'(x)$ is undefined within the interval.
2. Evaluate $f(x)$ at each critical point and at both endpoints of the interval.
3. The largest value is the absolute maximum; the smallest is the absolute minimum.

Classifying relative extrema requires one of two tests:

First Derivative Test: Examine the sign of $f'(x)$ on either side of a critical point.

• If $f'(x)$ changes from positive to negative, you have a relative maximum (the function was increasing, then starts decreasing).
• If $f'(x)$ changes from negative to positive, you have a relative minimum.
• If the sign doesn't change, the critical point is neither a max nor a min (think of $f(x) = x^3$ at the origin).

Second Derivative Test: Evaluate $f''(x)$ at the critical point.

• $f''(c) < 0$ → concave down → relative maximum at $x = c$
• $f''(c) > 0$ → concave up → relative minimum at $x = c$
• $f''(c) = 0$ → the test is inconclusive, and you need to fall back on the First Derivative Test or higher-order derivatives

Extreme Value Theorem

Closed and Bounded Sets

The Extreme Value Theorem has specific conditions, so you need to know what "closed" and "bounded" actually mean.

• A set is closed if it contains all of its boundary points. For intervals, square brackets denote closed: $[a, b]$ includes both $a$ and $b$, while $(a, b)$ excludes them.
• A set is bounded if it fits within some finite range, meaning it has both a finite upper bound and a finite lower bound. The interval $[2, 7]$ is bounded; the interval $[2, \infty)$ is not.

For the theorem to apply, you need both properties. The interval $[a, b]$ with finite $a$ and $b$ is the standard example of a closed and bounded set in $\mathbb{R}$.

Extreme Value Theorem

This theorem guarantees existence. It tells you the extrema are out there, and it narrows your search: they must occur at critical points or at the boundary points $a$ and $b$.

Both hypotheses are essential. If continuity fails (say, the function has a vertical asymptote inside the interval), the conclusion can break down. Similarly, on an open interval like $(0, 1)$, a continuous function might approach but never reach its supremum.

Boundary Points and Extrema

Boundary points are the endpoints of a closed interval. For $[2, 6]$, those are $x = 2$ and $x = 6$. A common mistake is to find all the critical points and forget to check the endpoints. Absolute extrema frequently occur at boundary points, not at critical points.

Worked example: Find the absolute extrema of $f(x) = x^3 - 9x^2 + 24x - 16$ on $[0, 4]$.

1. Compute $f'(x) = 3x^2 - 18x + 24 = 3(x^2 - 6x + 8) = 3(x - 2)(x - 4)$.

2. Set $f'(x) = 0$: critical points at $x = 2$ and $x = 4$. Both lie in $[0, 4]$.

3. Evaluate $f$ at critical points and endpoints:

   • $f(0) = 0 - 0 + 0 - 16 = -16$
   • $f(2) = 8 - 36 + 48 - 16 = 4$
   • $f(4) = 64 - 144 + 96 - 16 = 0$

4. The absolute maximum is $4$ at $x = 2$, and the absolute minimum is $-16$ at $x = 0$ (a boundary point).

If a function is not continuous on the interval, or the domain isn't closed and bounded, the Extreme Value Theorem doesn't apply. In those cases, you'll need to analyze limits and asymptotic behavior to determine whether absolute extrema exist at all.