Closed Interval Method

---

Concept Explanation

1. The 5-Step Closed Interval Method Pipeline

When a function satisfies the preconditions of the Extreme Value Theorem, execute the following 5-step pipeline:

Step 1: Check Preconditions (Continuity on [a, b])
                  |
Step 2: Differentiate (Compute f'(x))
                  |
Step 3: Find & Filter Interior Critical Numbers (c in (a, b) where f'(c)=0 or DNE)
                  |
Step 4: Evaluate Original Function f(x) at:
        - Surviving interior critical numbers: f(c_1), f(c_2), ...
        - Left boundary endpoint: f(a)
        - Right boundary endpoint: f(b)
                  |
Step 5: Compare Heights:
        - Highest value  --> Absolute Maximum Value
        - Lowest value   --> Absolute Minimum Value

2. Common Pitfalls to Avoid

• Pitfall 1: The Derivative Evaluation Trap: Never substitute candidate points back into the derivative $f'(x)$. Substituting critical numbers into $f'(x)$ yields $0$ or undefined. Always evaluate the original function $f(x)$ to find the physical heights ($y$-values).
• Pitfall 2: Forgetting Endpoints: Absolute extrema frequently occur at the boundaries $x = a$ or $x = b$, where $f'(x) \neq 0$. Neglecting endpoint evaluations can cause you to miss the true global maximum or minimum.
• Pitfall 3: Failing to Filter Candidates: Only critical numbers located strictly inside the open interval $(a, b)$ are valid candidates. Any solutions to $f'(x) = 0$ outside $(a, b)$ must be discarded.
• Pitfall 4: Overlooking Singular Points ($f'(x) \text{ DNE}$): Always check whether the denominator of $f'(x)$ can equal zero within $(a, b)$ (e.g., fractional exponents such as $x^{2/3}$ or $x^{1/3}$).

---

Engineering Applications

• Civil Structural Beam Bending Moments: Reinforced concrete bridge beams spanning $x \in [0, L]$ experience internal bending moment distributions $M(x)$. Civil engineers use the Closed Interval Method to compare moments at support piers ($x = 0, L$) against internal critical moment sections ($M'(x) = 0$), ensuring rebar steel reinforcement is placed at the absolute peak bending stress location.
• Transistor Thermal Duty-Cycle Analysis: Power semiconductor devices (such as GaN MOSFETs) experience thermal junction heating $T(t)$ over a periodic switching cycle $[0, t_{\text{cycle}}]$. Electronic systems engineers apply the Closed Interval Method to verify that the absolute peak junction temperature does not breach the silicon thermal damage ceiling.
• Subsea Pipeline Pressure Cycling: Marine petroleum pipelines spanning a seabed elevation profile $z(s)$ over total pipeline length $s \in [0, L_{\text{pipe}}]$ must withstand hydrostatic burst pressure. Engineers evaluate pipeline wall stresses at intermediate high/low points and boundary manifolds using closed-interval optimization.

---

Example Problems

Example 1: Polynomial on a Closed Interval

Find the absolute maximum and minimum values of $f(x) = x^3 - 3x^2 + 1$ on the closed interval $[-1, 4]$:

Solution

1. Check Continuity: $f(x)$ is a polynomial, so it is continuous on $[-1, 4]$. The EVT applies.
2. Differentiate:
   $$f'(x) = 3x^2 - 6x = 3x(x - 2)$$
3. Locate and Filter Critical Numbers: Set $f'(x) = 0$:
   $$3x(x - 2) = 0 \implies x = 0 \quad \text{and} \quad x = 2$$
   Filter: Both $x = 0$ and $x = 2$ lie strictly inside the interior interval $(-1, 4)$. Both are retained.
4. Evaluate Original Function $f(x)$ at All Candidates:
   • Left Endpoint ($x = -1$):
     $$f(-1) = (-1)^3 - 3(-1)^2 + 1 = -1 - 3 + 1 = -3$$
   • First Critical Number ($x = 0$):
     $$f(0) = (0)^3 - 3(0)^2 + 1 = 1$$
   • Second Critical Number ($x = 2$):
     $$f(2) = (2)^3 - 3(2)^2 + 1 = 8 - 12 + 1 = -3$$
   • Right Endpoint ($x = 4$):
     $$f(4) = (4)^3 - 3(4)^2 + 1 = 64 - 48 + 1 = 17$$
5. Compare and Conclude:
   • Absolute Maximum Value: $17$ at location $x = 4$.
   • Absolute Minimum Value: $-3$ at locations $x = -1$ and $x = 2$.

Example 2: Function with a Cusp on a Closed Interval

Find the absolute extrema of $g(x) = 2x - 3x^{2/3}$ on $[-1, 8]$:

Solution

1. Check Continuity: $g(x) = 2x - 3(\sqrt[3]{x})^2$ is continuous everywhere, including on $[-1, 8]$.
2. Differentiate:
   $$g'(x) = 2 - 3\left(\frac{2}{3}x^{-1/3}\right) = 2 - \frac{2}{x^{1/3}} = \frac{2x^{1/3} - 2}{x^{1/3}}$$
3. Find Critical Numbers:
   • Stationary Points ($g'(x) = 0$): Numerator equals zero:
     $$2x^{1/3} - 2 = 0 \implies x^{1/3} = 1 \implies x = 1$$
   • Singular Points ($g'(x) \text{ DNE}$): Denominator equals zero:
     $$x^{1/3} = 0 \implies x = 0$$
   • Filter: Both $x = 0$ and $x = 1$ lie in $(-1, 8)$.
4. Evaluate Original Function $g(x)$:
   • Left Endpoint ($x = -1$): $g(-1) = 2(-1) - 3(-1)^{2/3} = -2 - 3(1) = -5$
   • Singular Point ($x = 0$): $g(0) = 2(0) - 3(0)^{2/3} = 0$
   • Stationary Point ($x = 1$): $g(1) = 2(1) - 3(1)^{2/3} = 2 - 3 = -1$
   • Right Endpoint ($x = 8$): $g(8) = 2(8) - 3(8)^{2/3} = 16 - 3(4) = 16 - 12 = 4$
5. Compare and Conclude:
   • Absolute Maximum Value: $4$ at location $x = 8$.
   • Absolute Minimum Value: $-5$ at location $x = -1$.

---

Connections & References

• Theoretical Basis:
  • Extreme Value Theorem
  • Critical Numbers and Critical Points
• Alternative/Related Tests:
  • First Derivative Test
  • Absolute and Relative Extrema
• Course Sessions:
  • Calc 1 Session 13 - Critical Points and Extrema
  • Calc 1 Session 14 - Optimizing Areas and Volumes
  • Calc 1 Session 15 - Optimizing Distance and Time