Critical Points and Increasing/Decreasing Behavior for Optimization

1) Critical Numbers, Stationary Points, and Where the Derivative Fails

Optimization questions often reduce to: where does a quantity switch from increasing to decreasing (or vice versa)? The first derivative f'(x) is the main tool because its sign tells you whether f is rising or falling.

Key definitions

• Critical number: a number c in the domain of f such that either f'(c)=0 or f'(c) does not exist.
• Stationary point: a point (c, f(c)) where f'(c)=0. (Every stationary point comes from a critical number, but not every critical number is stationary.)
• Nondifferentiable critical point: a point in the domain where f'(c) does not exist (common causes: corners/cusps, vertical tangents, piecewise “kinks”).

Important: a critical number must be in the domain of f. Values where the formula breaks but the function is not defined are not candidates for extrema of f.

Quick examples of critical numbers

• Stationary: f(x)=x^3-3x. Then f'(x)=3x^2-3, so f'(x)=0 at x=±1. Both are critical numbers.
• Nondifferentiable: f(x)=|x-2|. The derivative fails at x=2, and 2 is in the domain, so x=2 is a critical number.
• Not in domain: f(x)=1/(x-1). The derivative also fails at x=1, but f is not defined there, so x=1 is not a critical number of f.

2) Sign Charts for f'(x) and Increasing/Decreasing Intervals

The sign of f'(x) controls monotonic behavior:

• If f'(x)>0 on an interval, then f is increasing on that interval.
• If f'(x)<0 on an interval, then f is decreasing on that interval.
• If f'(x)=0 throughout an interval, then f is constant there.

How to build a sign chart (first-derivative sign analysis)

Step-by-step:

• Compute f'(x).
• Find critical numbers: solve f'(x)=0 and locate where f'(x) does not exist (but f does).
• Use these numbers (and any domain breaks) to split the domain into test intervals.
• Pick a test point in each interval and evaluate the sign of f'(x) there.
• Record + or - on each interval; translate to increasing/decreasing behavior.

Example: sign chart and monotonicity

Let f(x)=x^3-3x. Then f'(x)=3x^2-3=3(x-1)(x+1). Critical numbers: x=-1, 1.

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So f increases on (-∞,-1), decreases on (-1,1), and increases on (1,∞).

Connecting sign changes to turning behavior

• If f' changes from + to - at c, then f has a local maximum at c.
• If f' changes from - to + at c, then f has a local minimum at c.
• If f' does not change sign at c, there may be no local extremum there (it could be a “flat spot” or an inflection-type behavior).

3) Local vs Absolute Extrema and Why Endpoints Matter

Local (relative) extrema

Local maximum at x=c: f(c) is the largest value in some open interval around c. Local minimum is defined similarly.

Local extrema are detected by sign changes in f' (or other tests), and they typically occur at critical numbers.

Absolute (global) extrema

Absolute maximum on a domain: the largest value of f over the entire domain. Absolute minimum is the smallest value.

On a closed interval [a,b], absolute extrema (if they exist) occur at:

• critical numbers in (a,b), and/or
• endpoints a or b.

Endpoints matter because the “best” value can occur at the boundary even when there is no turning point. For example, if a function is increasing on [a,b], then the absolute minimum is at a and the absolute maximum is at b.

When the domain is not a closed interval

If the domain is open (like (0,5)) or unbounded (like [0,∞)), an absolute maximum/minimum may fail to exist even if the function has local extrema. In applied problems, the domain is often restricted by physical constraints, which can create endpoints that must be checked.

4) Workflow: Candidate Points and Comparing Values

For many optimization tasks, the most reliable method is the “candidates + compare values” workflow.

Step-by-step workflow

1. Write the domain clearly (include real constraints such as x≥0, capacity limits, geometry constraints, or piecewise conditions). Convert words into inequalities.
2. Find critical numbers inside the domain:
   • Solve f'(x)=0.
   • Find where f'(x) does not exist (corners, cusps, piecewise joins), but keep only those x values where f is defined.
3. Include endpoints of the domain if they exist (for example, [a,b], or a constraint like 0≤x≤10).
4. Evaluate f at every candidate point (critical points + endpoints). Put the results in a small table.
5. Choose the largest/smallest value depending on the question (maximize or minimize).
6. Write the final answer with units and include a short reasonableness check.

Template for final answers (use this format)

• State the optimum: “The maximum/minimum value of f is ....”
• State where it occurs: “It occurs at x=... (within the allowed domain ...).”
• Include units: “x is in meters; f(x) is in square meters,” etc.
• Reasonableness check (1–2 sentences): confirm the point is in the domain, compare to endpoint values, and/or confirm the sign change of f' matches max/min behavior.

5) Practice Problems (with Piecewise Domains and Real Constraints)

Work each problem using the workflow: domain → critical points → endpoints → compare values. Then write a final answer with units and a reasonableness check.

Problem 1: Closed interval (endpoints included)

Let f(x)=x^3-6x^2+9x+1 on [0,4]. Find the absolute maximum and absolute minimum values.

Solution outline:

• Domain: [0,4].
• f'(x)=3x^2-12x+9=3(x^2-4x+3)=3(x-1)(x-3).
• Critical numbers in (0,4): x=1,3.
• Candidates: x=0,1,3,4.

Final answer (model): The absolute maximum value is 5 units of f, occurring at x=1 and x=4 (both in [0,4]). The absolute minimum value is 1 units of f, occurring at x=0 and x=3. Reasonableness check: all candidates are in the domain and the compared values show ties at both the max and min.

Problem 2: Constraint x≥0 (endpoint at 0, unbounded to the right)

A toy model for cost is C(x)=x^2-8x+20, where x is the number of items produced (in hundreds), and C is in thousands of dollars. For x≥0, find the minimum cost and where it occurs.

Solution outline:

• Domain: [0,∞).
• C'(x)=2x-8. Set to zero: 2x-8=0 gives x=4 (allowed).
• Endpoint candidate: x=0.
• Compare values: C(0)=20, C(4)=16-32+20=4.

Final answer (model): The minimum cost is 4 thousand dollars, occurring at x=4 hundred items (i.e., 400 items). Reasonableness check: x=4≥0 is allowed, and since the quadratic opens upward, the critical point gives the global minimum on [0,∞).

Problem 3: Piecewise function (derivative may fail at the join)

Let f(x)=

f(x) = { x^2 + 2x,    0 ≤ x < 2  { 6x - 4,     2 ≤ x ≤ 5

Find where f is increasing/decreasing on [0,5], and find absolute extrema on [0,5].

Solution outline:

• Domain: [0,5]. Note the join at x=2 is in the domain and is a candidate point.
• For 0≤x<2: f'(x)=2x+2, which is positive for all x≥0. So increasing on [0,2).
• For 2≤x≤5: f'(x)=6, positive. So increasing on [2,5].
• At x=2, check continuity: left value 2^2+2·2=8, right value 6·2-4=8 (continuous). Derivative changes from 2x+2 (which equals 6 at x=2) to 6, so it is differentiable here as well; regardless, x=2 is still a key checkpoint.
• Since increasing everywhere on [0,5], absolute min at x=0, absolute max at x=5.
• Compute: f(0)=0. f(5)=6·5-4=26.

Final answer (model): f is increasing on [0,2) and on [2,5] (so overall increasing on [0,5]), with no decreasing intervals. The absolute minimum is 0 units at x=0, and the absolute maximum is 26 units at x=5. Reasonableness check: a positive derivative on each piece means the function must rise as x increases, so endpoints should give min/max.

Problem 4: Nondifferentiable critical point (corner) with a constraint

Let g(x)=|x-3|+1 for 0≤x≤8. Find the absolute minimum and maximum.

Solution outline:

• Domain: [0,8].
• g'(x) does not exist at x=3 (corner), and 3 is in the domain, so x=3 is a critical number.
• Candidates: x=0,3,8.
• Evaluate: g(0)=| -3 |+1=4, g(3)=0+1=1, g(8)=|5|+1=6.

Final answer (model): The absolute minimum value is 1 unit at x=3. The absolute maximum value is 6 units at x=8. Reasonableness check: the V-shape of an absolute value function has its lowest point at the corner, and values grow as you move away; the farthest endpoint from 3 is 8, giving the largest value.

Problem 5: Rational function with restricted domain (exclude non-domain points)

Let h(x)=x + 4/x with the real constraint x>0 (so x is positive). Find the minimum value of h on (0,∞).

Solution outline:

• Domain: (0,∞). Note x=0 is not allowed and is not an endpoint to test.
• h'(x)=1 - 4/x^2. Set to zero: 1-4/x^2=0 gives x^2=4, so x=2 (only positive root allowed).
• Sign chart: for 0<x<2, 4/x^2>1 so h'(x)<0 (decreasing). For x>2, 4/x^2<1 so h'(x)>0 (increasing). Therefore x=2 is a local minimum, and because the function decreases then increases on the whole domain, it is an absolute minimum.
• Compute: h(2)=2+4/2=4.

Final answer (model): The minimum value of h on (0,∞) is 4 units, occurring at x=2 (positive and allowed). Reasonableness check: the sign change of h' from negative to positive at 2 confirms a minimum, and there are no endpoints to beat it because the domain is open and unbounded.

Instructions for submitting your solutions

• Start by writing the domain as an inequality or interval (include constraints like x≥0, 0≤x≤L, or piecewise ranges).
• List candidate points explicitly: “critical points: …; endpoints: …”.
• Show a small comparison table of f(x) values at candidates.
• Write the final answer in a sentence with units (both for x and for the optimized quantity).
• Add a brief reasonableness check: confirm the point is in the domain and mention either endpoint comparison or the sign change of f'.