How To Find Critical Numbers Subject To

Finding critical numbers is a fundamental skill in calculus, serving as the gateway to solving optimization problems, sketching accurate graphs, and understanding the behavior of functions. While the basic process involves setting the derivative equal to zero, the phrase "subject to" introduces a layer of complexity that appears frequently in advanced coursework and real-world applications. It signals that the search for critical numbers is restricted by a specific condition—be it a closed interval, a geometric boundary, or an explicit constraint equation.

The official docs gloss over this. That's a mistake.

This full breakdown walks you through the hierarchy of finding critical numbers, starting from the standard definition and progressing to constrained optimization using Lagrange Multipliers Small thing, real impact. And it works..

What Are Critical Numbers? The Standard Definition

Before tackling constraints, we must solidify the baseline definition. A critical number (or critical point) of a function $f(x)$ is a value $c$ in the domain of $f$ such that either:

1. $f'(c) = 0$ (Stationary Point): The tangent line is horizontal. This indicates potential local maxima, minima, or saddle points (inflection points with horizontal tangents).
2. $f'(c)$ Does Not Exist (DNE): The function has a sharp corner, a cusp, a vertical tangent, or a discontinuity at $c$.

Key Rule: The number $c$ must be in the domain of the original function $f$. If $f(c)$ is undefined (e.g., division by zero, square root of a negative), $c$ is not a critical number, even if the derivative is zero or undefined there.

Step-by-Step: Standard Procedure (Unconstrained)

1. Find the Domain: Identify all $x$ values for which $f(x)$ is defined.
2. Differentiate: Compute $f'(x)$ using derivative rules (Power, Product, Quotient, Chain).
3. Solve $f'(x) = 0$: Find all solutions. Check each against the domain from Step 1. Keep only valid ones.
4. Find Where $f'(x)$ is Undefined: Look for division by zero in $f'(x)$, even roots of negative numbers, or logarithms of non-positive numbers. Check these $x$-values against the domain of $f$. Keep valid ones.
5. List Critical Numbers: The union of valid results from Steps 3 and 4.

Example: $f(x) = x^{2/3}(x - 4)$ Domain: All real numbers. Which means > $f'(x) = \frac{2}{3}x^{-1/3}(x-4) + x^{2/3} = \frac{2(x-4)}{3\sqrt[3]{x}} + \frac{3x}{3\sqrt[3]{x}} = \frac{5x - 8}{3\sqrt[3]{x}}$ Set numerator to 0: $5x - 8 = 0 \Rightarrow x = 8/5$. Consider this: (Valid). Think about it: > Denominator 0: $x = 0$. This leads to $f(0)$ is defined (equals 0). Valid. Critical Numbers: $x = 0, \frac{8}{5}$ Small thing, real impact. No workaround needed..

Critical Numbers "Subject to" a Closed Interval: The Closed Interval Method

In Calculus I, "subject to" often appears as: "Find the absolute extrema of $f(x)$ subject to the constraint $a \le x \le b$." This is the Closed Interval Method (or Candidate's Test).

When a function is restricted to a closed interval $[a, b]$, the endpoints $a$ and $b$ become critical candidates for absolute extrema, even if the derivative is not zero there. The Extreme Value Theorem guarantees absolute max/min exist for continuous functions on closed intervals Still holds up..

This is the bit that actually matters in practice Most people skip this — try not to..

Modified Procedure for Closed Intervals $[a, b]$

1. Verify Continuity: Ensure $f$ is continuous on $[a, b]$.
2. Find Standard Critical Numbers: Use the standard procedure above, but only keep critical numbers that lie strictly inside the open interval $(a, b)$. Discard any $c \le a$ or $c \ge b$.
3. Evaluate Candidates: Calculate $f(x)$ for:
   • The endpoints $a$ and $b$.
   • All valid critical numbers from Step 2.
4. Compare: The largest value is the Absolute Maximum; the smallest is the Absolute Minimum.

Why endpoints matter: On $f(x) = x^2$ subject to $[-1, 2]$, $f'(x)=2x=0 \Rightarrow x=0$. But the absolute max is at the endpoint $x=2$ ($f=4$), not at the critical number $x=0$ ($f=0$).

Critical Numbers "Subject to" a Constraint Equation: Lagrange Multipliers

In Multivariable Calculus (Calculus III), "subject to" takes on its most rigorous meaning: Constrained Optimization. ) = k$ (or $g(x, y, z...You are asked to find critical points (max/min) of a function $f(x, y, z...)$ subject to a constraint $g(x, y, z...) = 0$) Took long enough..

Here, we are not just looking for where $\nabla f = \vec{0}$. We are looking for points on the level curve/surface $g=k$ where $f$ stops increasing/decreasing. Geometrically, this happens when the gradient of $f$ is parallel to the gradient of $g$ Took long enough..

The Lagrange Multiplier Method

Theorem: If $f$ has a local extremum at point $P$ on the constraint curve $g = k$, and $\nabla g(P) \neq \vec{0}$, then there exists a scalar $\lambda$ (lambda) such that: $ \nabla f(P) = \lambda \nabla g(P) $ $ g(P) = k $

Step-by-Step Procedure

1. Set Up the System: Write the gradient equations component-wise.
   • $f_x = \lambda g_x$
   • $f_y = \lambda g_y$
   • $(f_z = \lambda g_z)$
   • Constraint: $g(x, y, z) = k$
2. Solve for $\lambda$ (or eliminate it): Solve two of the gradient equations for $\lambda$ and set them equal to each other. This removes $\lambda$ and gives a relationship between variables (e.g., $x = 2y$).
3. Substitute into Constraint: Plug the relationship from Step 2 into the constraint equation $g = k$.
4. Solve for Coordinates: Find the specific points $(x, y, z)$.
5. Check Edge Cases (Crucial): The theorem requires $\nabla g \neq \vec{0}$. You must separately check points where $\nabla g = \vec{0}$ (if they satisfy the constraint), as the Lagrange equations fail there.
6. Evaluate $f$: Plug all candidate points into $f(x, y, z)$. The largest is the constrained max; the smallest is the constrained min.

Example: Maximize $f(x,