4.3 Maxima and Minima

Maxima and minima help you find the highest and lowest values a function can reach. These are the foundation of optimization, which shows up constantly in later calculus, physics, and engineering courses. To find them reliably, you need to understand critical points, the Extreme Value Theorem, and the closed interval method.

Maxima and Minima

Absolute vs. Local Extrema

Absolute extrema are the single highest or lowest function values across an entire domain. If $f(c) \geq f(x)$ for every $x$ in the domain, then $f(c)$ is the absolute maximum. If $f(c) \leq f(x)$ for every $x$ in the domain, then $f(c)$ is the absolute minimum.

Local extrema are high or low points relative to nearby values only. A local maximum at $x = c$ means $f(c) \geq f(x)$ for all $x$ in some small open interval around $c$. A local minimum is the reverse.

A few things to keep straight:

• Every absolute extremum on an open interval is also a local extremum, but a local extremum is not necessarily an absolute one. A function can have several local peaks and valleys, yet only one absolute max and one absolute min on a given domain.
• On a closed interval, absolute extrema can occur at endpoints. Endpoints are not considered local extrema (since you can't look at both sides of an endpoint).

Critical Points

A critical point of $f$ occurs at $x = c$ (where $c$ is in the domain of $f$) if either:

• $f'(c) = 0$ (the tangent line is horizontal), or
• $f'(c)$ does not exist (corners, cusps, or vertical tangents)

To find critical points:

1. Compute $f'(x)$.
2. Set $f'(x) = 0$ and solve for $x$.
3. Identify any $x$-values in the domain where $f'(x)$ is undefined.
4. Confirm each candidate is actually in the domain of $f$.

Not every critical point is an extremum. For example, $f(x) = x^3$ has $f'(0) = 0$, but $x = 0$ is neither a max nor a min. Critical points are candidates for extrema; you still need to test them.

Closed Interval Method for Finding Absolute Extrema

The Extreme Value Theorem guarantees that if $f$ is continuous on a closed interval $[a, b]$, then $f$ must attain both an absolute maximum and an absolute minimum somewhere on that interval. This theorem is what makes the closed interval method work.

Here's the procedure:

1. Verify that $f$ is continuous on $[a, b]$.
2. Find all critical points of $f$ in the open interval $(a, b)$.
3. Evaluate $f$ at each critical point.
4. Evaluate $f$ at the endpoints $x = a$ and $x = b$.
5. Compare all the values from steps 3 and 4. The largest is the absolute maximum; the smallest is the absolute minimum.

Example: Find the absolute extrema of $f(x) = 2x^3 - 3x^2 - 12x + 5$ on $[-2, 3]$.

1. $f$ is a polynomial, so it's continuous everywhere.

2. $f'(x) = 6x^2 - 6x - 12 = 6(x^2 - x - 2) = 6(x - 2)(x + 1)$. Setting $f'(x) = 0$ gives $x = 2$ and $x = -1$. Both are in $(-2, 3)$.

3. $f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) + 5 = -2 - 3 + 12 + 5 = 12$

4. $f(2) = 2(8) - 3(4) - 12(2) + 5 = 16 - 12 - 24 + 5 = -15$

5. $f(-2) = 2(-8) - 3(4) - 12(-2) + 5 = -16 - 12 + 24 + 5 = 1$

6. $f(3) = 2(27) - 3(9) - 12(3) + 5 = 54 - 27 - 36 + 5 = -4$

Comparing: $f(-1) = 12$, $f(2) = -15$, $f(-2) = 1$, $f(3) = -4$. The absolute maximum is $12$ at $x = -1$, and the absolute minimum is $-15$ at $x = 2$.

Real-World Significance of Extrema

Optimization problems are one of the biggest applications of derivatives. A few classic examples:

• Maximizing: profit, area enclosed by a fence, volume of a container
• Minimizing: production cost, material used, travel time

For instance, finding the maximum height of a projectile means finding the absolute maximum of its height function on a time interval. Determining the cheapest way to build a container means minimizing a cost function. In each case, you're setting up a function and then using critical points (and often the closed interval method) to locate the extreme values.

Related Concepts

• Inflection points are where the concavity of $f$ changes (from concave up to concave down, or vice versa). These are found where $f''(x) = 0$ or $f''(x)$ is undefined, but you must verify the concavity actually switches. Inflection points are not extrema, though they often appear between a local max and a local min.
• Monotonicity refers to whether $f$ is increasing or decreasing on an interval. If $f'(x) > 0$ on an interval, $f$ is increasing there; if $f'(x) < 0$, it's decreasing. A sign change in $f'$ at a critical point signals a local extremum, which connects directly to the First Derivative Test (covered in the next section).