How to Determine Extrema using Candidates Test

What You Need to Know

What this is (and why AP loves it)

The Candidates Test (often called the Closed Interval Method) is your go-to process for finding absolute maximum and absolute minimum values of a function on a closed interval $[a,b]$.

AP Calculus AB loves this because it’s:

• Algorithmic (clear steps)
• Based on big theorems (especially the Extreme Value Theorem)
• Full of common traps (endpoints, undefined derivative, domain issues)

Core idea (the rule you’re using)

If $f$ is **continuous** on a **closed interval** $[a,b]$, then:

• $f$ **must** attain an **absolute maximum** and **absolute minimum** somewhere on $[a,b]$.
• Those absolute extrema can only occur at:
  1. Endpoints: $x=a$ or $x=b$
  2. Critical numbers in $(a,b)$ where:
  • $f'(x)=0$, or
  • $f'(x)$ **does not exist** (but $f(x)$ is defined)

That list (endpoints + critical numbers) gives you the candidates.

When to use it

Use the Candidates Test when a problem asks for:

• Absolute max/min on an interval (especially a closed interval $[a,b]$)
• “Find the maximum value” / “minimum value” on a given domain
• Optimization problems after you create an objective function and an interval

Critical reminder: The guarantee (that absolute extrema exist) relies on continuity on $[a,b]$. You can still try the method when discontinuous, but you must check whether an absolute extremum actually exists.

Step-by-Step Breakdown

The Candidates Test algorithm

1. Confirm the interval and domain

   • Identify the interval: usually $[a,b]$.
   • Ensure $f$ is defined on the entire interval.
   • If the problem implies “absolute extrema,” check continuity (or be cautious if not continuous).

2. Compute the derivative

   • Find $f'(x)$.

3. Find critical numbers in the open interval $(a,b)$

   • Solve $f'(x)=0$.
   • Also find where $f'(x)$ does not exist.
   • Keep only points where $f$ is defined and $x \in (a,b)$.

4. Create the candidate list

   • Candidates = {endpoints $a,b$} ∪ {critical numbers in $(a,b)$}.

5. Evaluate $f$ at every candidate

   • Compute $f(a)$, $f(b)$, and $f(c)$ for each critical number $c$.

6. Compare values and conclude

   • Largest value = absolute maximum value.
   • Smallest value = absolute minimum value.
   • State both the value and the location (where it occurs).

Mini worked walkthrough (template)

Suppose you’re asked: “Find absolute extrema of $f(x)$ on $[a,b]$.”

• Step 1: Domain/continuity check on $[a,b]$.
• Step 2: Compute $f'(x)$.
• Step 3: Solve $f'(x)=0$ and include derivative-undefined points.
• Step 4: Candidates = endpoints + interior critical numbers.
• Step 5: Evaluate $f$ at candidates.
• Step 6: Compare values → pick biggest/smallest.

Decision point: If the interval is open like $(a,b)$, you **do not** automatically include endpoints. Instead, you may need to analyze limits as $x \to a^+$ and $x \to b^-$ to see if extrema exist.

Key Formulas, Rules & Facts

Must-know facts (high yield)

• Absolute maximum/minimum: compares to all values in the interval.
• Local maximum/minimum: compares only to nearby values.
• Critical number $c$ (interior point): $f'(c)=0$ or $f'(c)$ DNE, with $f(c)$ defined.

Theorems you’re implicitly using

Candidates Test summary table

Important edge-case rules

• If $f$ is **not continuous** on $[a,b]$, absolute extrema might not exist, even if you find critical points.
• If $f$ is continuous on $[a,b]$ but $f'(x)$ is undefined at some interior point, that point is still a valid candidate.
• If a derivative-undefined point is also where $f$ is undefined (like a vertical asymptote), it is **not** a candidate because $f(c)$ doesn’t exist.

Examples & Applications

Example 1: Standard polynomial on a closed interval

Find absolute extrema of $f(x)=x^3-3x^2+1$ on $[0,3]$.

1. Derivative: $f'(x)=3x^2-6x=3x(x-2)$
2. Critical numbers: $f'(x)=0 \Rightarrow x=0,2$
   • Only interior critical numbers matter for this step: $x=2$ is in $(0,3)$. (Endpoint $x=0$ will be included anyway.)
3. Candidates: $x=0,2,3$
4. Evaluate:
   • $f(0)=1$
   • $f(2)=8-12+1=-3$
   • $f(3)=27-27+1=1$
5. Compare values:
   • Absolute minimum value is $-3$ at $x=2$.
   • Absolute maximum value is $1$ at $x=0$ and $x=3$.

AP-style conclusion: “The absolute minimum value is $-3$ at $x=2$. The absolute maximum value is $1$ at $x=0$ and $x=3$.”

Example 2: Derivative undefined but function defined (corner/cusp)

Find absolute extrema of $f(x)=|x-1|$ on $[-1,3]$.

1. $f'(x)$ does not exist at $x=1$ (cusp), but $f(1)=0$ exists.
2. Candidates: endpoints and cusp point: $x=-1,1,3$
3. Evaluate:
   • $f(-1)=|-2|=2$
   • $f(1)=|0|=0$
   • $f(3)=|2|=2$
4. Conclusions:
   • Absolute minimum value is $0$ at $x=1$.
   • Absolute maximum value is $2$ at $x=-1$ and $x=3$.

Key takeaway: Extrema can happen where $f'(x)$ DNE (corners/cusps), so you must include them.

Example 3: Rational function—watch the domain and interval

Find absolute extrema of $f(x)=\dfrac{x}{x^2+1}$ on $[-2,2]$.

1. Domain: $x^2+1>0$ always, so $f$ is continuous on $[-2,2]$.
2. Derivative (quotient rule):
   $f'(x)=\dfrac{(x^2+1)\cdot 1 - x\cdot (2x)}{(x^2+1)^2}=\dfrac{1-x^2}{(x^2+1)^2}$
3. Critical numbers: $f'(x)=0 \Rightarrow 1-x^2=0 \Rightarrow x=\pm 1$
4. Candidates: $x=-2,-1,1,2$
5. Evaluate:
   • $f(-2)=\dfrac{-2}{5}=-0.4$
   • $f(-1)=\dfrac{-1}{2}=-0.5$
   • $f(1)=\dfrac{1}{2}=0.5$
   • $f(2)=\dfrac{2}{5}=0.4$
6. Conclusions:
   • Absolute minimum value is $-\dfrac{1}{2}$ at $x=-1$.
   • Absolute maximum value is $\dfrac{1}{2}$ at $x=1$.

Key takeaway: When continuous on $[a,b]$, evaluating candidates is enough—no need for sign charts unless asked.

Example 4: Discontinuous on a closed interval (no guarantee!)

Consider $f(x)=\dfrac{1}{x-1}$ on $[0,2]$. “Find the absolute extrema.”

• Problem: $f$ is **not defined** at $x=1$, so it’s not continuous on $[0,2]$.
• As $x \to 1^-$, $f(x) \to -\infty$, and as $x \to 1^+$, $f(x) \to +\infty$.
• Even though you can compute endpoint values:
  • $f(0)=-1$
  • $f(2)=1$
• There is no absolute minimum (it goes to $-\infty$) and **no absolute maximum** (it goes to $+\infty$).

Exam insight: If discontinuity/vertical asymptote occurs inside the interval, be ready to say “no absolute max/min” and justify with limiting behavior.

Common Mistakes & Traps

1. Forgetting endpoints

   • What happens: You solve $f'(x)=0$ and pick the biggest/smallest from those points only.
   • Why wrong: Absolute extrema on $[a,b]$ often occur at $a$ or $b$.
   • Fix: Always include $x=a$ and $x=b$ as candidates for a closed interval.

2. Including points not in the interval

   • What happens: You find critical numbers, then evaluate all of them even if they’re outside $[a,b]$.
   • Why wrong: Extrema are only being asked on that interval.
   • Fix: After solving, filter: keep only $c$ with $c \in (a,b)$.

3. Treating “derivative undefined” as automatic disqualification

   • What happens: You ignore cusps/corners/vertical tangents.
   • Why wrong: Absolute extrema can occur where $f'(c)$ DNE as long as $f(c)$ exists.
   • Fix: Check points where $f'(x)$ DNE and test them as candidates.

4. Calling every critical point an extremum without comparing values

   • What happens: You find $f'(c)=0$ and declare “max” or “min.”
   • Why wrong: Critical points can be neither (example: inflection point with horizontal tangent).
   • Fix: For absolute extrema, you must evaluate $f$ at all candidates and compare.

5. Missing domain restrictions (especially rational, root, log functions)

   • What happens: You include candidates where $f$ isn’t defined.
   • Why wrong: If $f(c)$ doesn’t exist, it can’t be an extremum value.
   • Fix: Domain-check first; only keep candidates where $f(c)$ is defined.

6. Assuming absolute extrema exist on an open interval $(a,b)$

   • What happens: You use the same endpoint-including method.
   • Why wrong: Endpoints aren’t included; extrema might not occur (e.g., maximum approached but never reached).
   • Fix: For open intervals, you usually need critical points and boundary-limit analysis.

7. Rounding too early and choosing the wrong “largest” value

   • What happens: You approximate values and comparisons get sloppy.
   • Why wrong: Two candidate values can be close.
   • Fix: Keep exact values (fractions/radicals) until the final comparison when possible.

8. Not stating both location and value

   • What happens: You write “absolute max at $x=2$” but don’t give the max value.
   • Why wrong: AP often wants the point (where) and the value.
   • Fix: Write: “Absolute maximum value is $f(2)=\dots$ at $x=2$.”

Memory Aids & Quick Tricks

Quick efficiency tip: If the problem only asks for the absolute maximum value, you still must check all candidates—an interior critical point could beat an endpoint.

Quick Review Checklist

• [ ] Interval is identified (usually $[a,b]$) and domain makes sense on it.
• [ ] If needed, you note EVT: $f$ continuous on $[a,b]$ ⇒ absolute max/min exist.
• [ ] Compute $f'(x)$.
• [ ] Solve $f'(x)=0$ for interior points in $(a,b)$.
• [ ] Find where $f'(x)$ DNE (but $f(x)$ exists) in $(a,b)$.
• [ ] Candidates = endpoints + interior critical numbers.
• [ ] Evaluate $f$ at every candidate.
• [ ] Largest value ⇒ absolute max; smallest value ⇒ absolute min.
• [ ] State both the value and where it occurs.

You’ve got this—if you always check endpoints and compare candidate values carefully, these problems become automatic.