20.1 First & Second Derivative Tests, Extrema, Concavity & Optimization

20.1 First & Second Derivative Tests, Extrema, Concavity & Optimization

Critical Numbers & The Extreme Value Theorem

Differential calculus provides the analytical framework for uncovering the qualitative and quantitative geometry of functions. At the core of curve analysis is Fermat's Theorem on local extrema: if a function $f$ attains a local extremum (maximum or minimum) at an interior point $c$, and if $f'(c)$ exists, then $f'(c) = 0$. This motivates the formal definition of a critical number: a number $c$ in the domain of $f$ such that either $f'(c) = 0$ or $f'(c)$ does not exist (is undefined).

Crucially, an $x$-value where $f'$ is undefined is a critical number only if $x$ belongs to the domain of $f$. For instance, for $f(x) = x^{2/3}$, $f'(x) = \frac{2}{3x^{1/3}}$ is undefined at $x = 0$; because $0$ is in the domain of $f$, $x = 0$ is a critical number (corresponding to a vertical cusp). Conversely, for $g(x) = \frac{1}{x}$, $g'(x) = -\frac{1}{x^2}$ is undefined at $x = 0$, but $x = 0$ is not in the domain of $g$, so $x = 0$ is a vertical asymptote rather than a critical number.

The Extreme Value Theorem (EVT) states that if a real-valued function $f$ is continuous on a closed, bounded interval $[a, b]$, then $f$ must attain both an absolute (global) maximum and an absolute (global) minimum at least once on $[a, b]$. These absolute extrema must occur either at an interior critical number $c \in (a, b)$ or at one of the boundary endpoints $x = a$ or $x = b$.

The Candidates Test (or Closed Interval Method) operationalizes the EVT:

1. Confirm $f$ is continuous on $[a, b]$.
2. Calculate $f'(x)$ and find all critical numbers $c_1, c_2, \dots$ lying strictly inside the open interval $(a, b)$.
3. Evaluate the original function $f$ at each interior critical number: $f(c_1), f(c_2), \dots$
4. Evaluate $f$ at the boundary endpoints: $f(a)$ and $f(b)$.
5. Identify the largest evaluated value as the absolute maximum and the smallest evaluated value as the absolute minimum on $[a, b]$.

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Monotonicity & The First Derivative Test

The sign of the first derivative characterizes the direction of a function's graph over an interval:

• If $f'(x) > 0$ for all $x \in (a, b)$, then $f$ is strictly increasing on $[a, b]$.
• If $f'(x) < 0$ for all $x \in (a, b)$, then $f$ is strictly decreasing on $[a, b]$.
• If $f'(x) = 0$ for all $x \in (a, b)$, then $f$ is constant on $[a, b]$.

The First Derivative Test classifies local extrema at a continuous critical number $c$:

1. Local Maximum: If $f'(x)$ changes sign from positive to negative as $x$ increases through $c$ (i.e., $f$ transitions from increasing to decreasing), then $f(c)$ is a local maximum.
2. Local Minimum: If $f'(x)$ changes sign from negative to positive as $x$ increases through $c$ (i.e., $f$ transitions from decreasing to increasing), then $f(c)$ is a local minimum.
3. No Local Extremum: If $f'(x)$ maintains the same sign on both sides of $c$ (positive to positive, or negative to negative), then $f(c)$ is neither a local maximum nor a local minimum. For example, $f(x) = x^3$ has $f'(0) = 0$, but $f'(x) = 3x^2 > 0$ for all $x \neq 0$, producing a stationary point of inflection rather than an extremum.

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Concavity & Points of Inflection

While the first derivative dictates the slope of the tangent line, the second derivative dictates the rate of change of that slope, which defines the concavity of the curve:

• A differentiable function $f$ is concave upward on an open interval $I$ if $f'$ is strictly increasing on $I$. If $f''(x) > 0$ on $I$, the graph is concave upward, curving like an open cup $\cup$. Tangent lines lie strictly below the curve.
• A differentiable function $f$ is concave downward on an open interval $I$ if $f'$ is strictly decreasing on $I$. If $f''(x) < 0$ on $I$, the graph is concave downward, curving like an inverted cup $\cap$. Tangent lines lie strictly above the curve.

A point of inflection is a point $(c, f(c))$ on the curve where the concavity strictly changes sign (from upward to downward, or downward to upward) and where the function is continuous with a well-defined tangent line.

• A necessary condition for an inflection point on a twice-differentiable function is $f''(c) = 0$ or $f''(c)$ does not exist.
• However, $f''(c) = 0$ alone is not sufficient. For $f(x) = x^4$, $f''(x) = 12x^2$, so $f''(0) = 0$. Yet $f''(x) > 0$ for all $x \neq 0$; because the concavity does not change sign across $x = 0$, $(0, 0)$ is a local minimum, not a point of inflection.

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The Second Derivative Test for Local Extrema

The Second Derivative Test uses curvature at a stationary critical point where $f'(c) = 0$:

1. If $f'(c) = 0$ and $f''(c) < 0$, then $f$ is concave downward at $c$, meaning the horizontal tangent sits atop the curve. Thus, $f(c)$ is a local maximum.
2. If $f'(c) = 0$ and $f''(c) > 0$, then $f$ is concave upward at $c$, meaning the horizontal tangent sits below the curve. Thus, $f(c)$ is a local minimum.
3. If $f'(c) = 0$ and $f''(c) = 0$ (or $f''(c)$ does not exist), the test is inconclusive. The point may be a local maximum ($f(x) = -x^4$ at $0$), a local minimum ($f(x) = x^4$ at $0$), or an inflection point ($f(x) = x^3$ at $0$). When this occurs, one must revert to the First Derivative Test.

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Curve Sketching & Derivative Behavior Summary

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Applied Optimization: Systematic Analytical Methodology

Applied optimization problems require finding the absolute maximum or absolute minimum of a quantity modeled by a mathematical function subject to physical or economic constraints.

1. Formulate Variables and Objective Function: Identify the target quantity to optimize (e.g., area, volume, cost, material surface area) and write an algebraic expression for it.
2. Identify Constraint Equations: State the equations relating the auxiliary variables based on geometric, physical, or monetary conditions.
3. Reduce to a Single Variable: Substitute the constraint into the objective function so that it depends on exactly one independent variable.
4. Establish the Feasible Domain: Determine the physically and mathematically valid interval for the independent variable, taking note of whether the domain is closed $[a, b]$ or open $(a, b)$.
5. Differentiate & Locate Critical Numbers: Compute the first derivative of the single-variable objective function, set it equal to zero, and solve for critical values within the feasible domain.
6. Verify Optimality: If the domain is closed, apply the Candidates Test. If the domain is an open interval with exactly one critical number, apply the Second Derivative Test or First Derivative Test to prove the critical value yields a global extremum.

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Worked Exemplar: Optimization of a Cylindrical Container

Problem: A manufacturing engineer must design a closed right circular cylindrical container with a fixed volume of $54\pi\text{ cm}^3$. Determine the radius $r$ and height $h$ that minimize the total surface area of the container, and compute this minimum surface area.

Step 1: Define the Objective Function and Constraint. Let $r$ denote the base radius and $h$ denote the height in centimeters.

• Volume constraint: $V = \pi r^2 h = 54\pi\text{ cm}^3$.
• Objective: Minimize the total surface area $A$ consisting of the top, bottom, and lateral surface: $A = 2\pi r^2 + 2\pi r h$

Step 2: Express the Objective Function in a Single Variable. Solve the volume constraint for the secondary variable $h$: $h = \frac{54\pi}{\pi r^2} = \frac{54}{r^2}$ Substitute $h$ into the surface area equation: $A(r) = 2\pi r^2 + 2\pi r \left(\frac{54}{r^2}\right) = 2\pi r^2 + \frac{108\pi}{r}$ Because $r$ represents a physical length, the feasible domain is the open interval $r \in (0, \infty)$.

Step 3: Differentiate and Solve for Critical Numbers. Compute the first derivative with respect to $r$: $A'(r) = \frac{d}{dr}\left[2\pi r^2 + 108\pi r^{-1}\right] = 4\pi r - \frac{108\pi}{r^2}$ Set $A'(r) = 0$ to locate stationary critical numbers: $4\pi r - \frac{108\pi}{r^2} = 0 \implies 4\pi r = \frac{108\pi}{r^2}$ Multiply through by $r^2$ and divide by $4\pi$: $r^3 = \frac{108\pi}{4\pi} = 27 \implies r = \sqrt[3]{27} = 3\text{ cm}$

Step 4: Verify the Global Minimum via the Second Derivative Test. Differentiate $A'(r)$ to compute the second derivative: $A''(r) = \frac{d}{dr}\left[4\pi r - 108\pi r^{-2}\right] = 4\pi + \frac{216\pi}{r^3}$ Evaluate at the critical radius $r = 3$: $A''(3) = 4\pi + \frac{216\pi}{3^3} = 4\pi + \frac{216\pi}{27} = 4\pi + 8\pi = 12\pi > 0$ Because $A''(r) > 0$ for all $r > 0$, the function $A(r)$ is strictly concave upward across its entire domain. Therefore, $r = 3\text{ cm}$ is the unique absolute minimum.

Step 5: Determine Dimensions and Minimum Surface Area. Calculate the optimal container height: $h = \frac{54}{(3)^2} = \frac{54}{9} = 6\text{ cm}$ Notice that $h = 2r$, confirming the classic geometric result that a cylinder minimizing surface area for a fixed volume has a height equal to its diameter. Calculate the minimum surface area: $A(3) = 2\pi(3)^2 + \frac{108\pi}{3} = 18\pi + 36\pi = 54\pi\text{ cm}^2$