How to Determine Extrema using Candidates Test

What You Need to Know

Why this matters

The Candidates Test is your go-to, reliable method for finding absolute (global) maxima/minima of a function on a closed interval. On AP Calc, it shows up constantly in FRQs and MC because it’s systematic and hard to mess up if you follow the checklist.

Core idea (the theorem-level statement)

If $f$ is **continuous** on a closed interval $[a,b]$, then by the **Extreme Value Theorem (EVT)**, $f$ attains an **absolute maximum** and **absolute minimum** somewhere in $[a,b]$.

The Candidates Test tells you where to look:

• Endpoints: $x=a$ and $x=b$
• Critical numbers in $(a,b)$: where $f'(x)=0$ or $f'(x)$ **does not exist** (but $f$ does)

Then you evaluate $f$ at all candidates; the largest value is the absolute max and the smallest value is the absolute min.

Key definitions (you need these precise)

• Absolute maximum on $[a,b]$: a point $x=c$ in $[a,b]$ such that $f(c) \ge f(x)$ for all $x \in [a,b]$.
• Absolute minimum on $[a,b]$: a point $x=c$ in $[a,b]$ such that $f(c) \le f(x)$ for all $x \in [a,b]$.
• Critical number: a value $c$ in the **domain** of $f$ where $f'(c)=0$ or $f'(c)$ is undefined.

Big warning: EVT (and thus the guarantee that absolute extrema exist) requires continuity on a closed interval. If the interval is open or the function is discontinuous, absolute extrema may not exist.

When you use the Candidates Test

Use it when the problem asks for:

• Absolute max/min on a closed interval $[a,b]$
• “Find the maximum/minimum value of $f$ on $[a,b]$”
• Optimization restricted to a domain like $0 \le x \le 5$

It’s also a great fallback when you’re unsure whether a “peak” is local or absolute.

Step-by-Step Breakdown

The Candidates Test algorithm (absolute extrema on $[a,b]$)

1. Check the setup

   • Make sure you’re working on a closed interval $[a,b]$.
   • Ensure $f$ is **continuous** on $[a,b]$ (common in AP problems; still check for piecewise, radicals, denominators).

2. Compute the derivative

   • Find $f'(x)$ (or use a given derivative/graph/table).

3. Find interior critical numbers

   • Solve $f'(x)=0$ for $x \in (a,b)$.
   • Find where $f'(x)$ **does not exist** (corners, cusps, vertical tangents, discontinuities in $f'$) and keep those $x$ values if $f$ is defined there and they lie in $(a,b)$.

4. List all candidates

   • Include endpoints: $a$ and $b$.
   • Include all critical numbers in $(a,b)$.

5. Evaluate $f$ at each candidate

   • Compute values $f(a), f(b), f(c_1), f(c_2), \dots$.

6. Compare values

   • Biggest function value $\rightarrow$ absolute maximum value.
   • Smallest function value $\rightarrow$ absolute minimum value.

7. State your answer clearly

   • Give both the x-location(s) and the value(s).
   • If multiple points tie, say so.

Decision point: If the question is about absolute extrema, you must compare function values (not just derivative signs).

Mini worked walkthrough (annotated)

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

1. Closed interval $[-2,2]$ and polynomial $\Rightarrow$ continuous.
2. $f'(x)=3x^2-3=3(x^2-1)$.
3. Critical numbers: $f'(x)=0 \Rightarrow x=\pm 1$ (both in $(-2,2)$).
4. Candidates: $x=-2,-1,1,2$.
5. Evaluate:
   • $f(-2)=-8+6=-2$
   • $f(-1)=-1+3=2$
   • $f(1)=1-3=-2$
   • $f(2)=8-6=2$
6. Compare: max value $2$ at $x=-1,2$; min value $-2$ at $x=-2,1$.

Key Formulas, Rules & Facts

Candidates Test essentials

What counts as “$f'(x)$ does not exist” (common sources)

• Corner in piecewise/absolute value: left and right derivatives differ.
• Cusp: slopes blow up with opposite signs.
• Vertical tangent: derivative infinite/undefined but function may be continuous.
• Derivative discontinuity: often piecewise-defined derivatives.

Reminder: A point where $f$ is not defined is not a critical number (because critical numbers must be in the domain).

Local vs absolute (don’t mix them)

• Local extrema: compare nearby values; can happen at critical numbers.
• Absolute extrema on $[a,b]$: compare against every point in interval.
• A local max might not be the absolute max if an endpoint is higher.

Graph/table version of Candidates Test

If you’re given a graph of $f'$ or a table:

• Candidates still come from:
  • endpoints
  • where $f'(x)=0$ (x-intercepts of $f'$)
  • where $f'$ is undefined
• To compare absolute extrema you still need function values:
  • from a table of $f$
  • by integrating: $f(x)=f(x_0)+\int_{x_0}^{x} f'(t)\,dt$ if $f'$ is given and an initial value is provided

Examples & Applications

Example 1: Absolute extrema on a closed interval (classic)

Find absolute extrema of $f(x)=x^2-4x+1$ on $[0,5]$.

• $f'(x)=2x-4$
• Critical number: $2x-4=0 \Rightarrow x=2$ (in interval)
• Candidates: $x=0,2,5$
• Evaluate:
  • $f(0)=1$
  • $f(2)=4-8+1=-3$
  • $f(5)=25-20+1=6$
• Absolute min value $-3$ at $x=2$; absolute max value $6$ at $x=5$.

Exam angle: Quadratic has a vertex, but the endpoint can still win for max.

Example 2: Derivative undefined at a corner (must include!)

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

• $f$ is continuous on $[-1,3]$.
• $f'$ is undefined at $x=1$ (corner), so $x=1$ is a critical number.
• Candidates: $x=-1,1,3$
• Evaluate:
  • $f(-1)=|-2|+2=4$
  • $f(1)=|0|+2=2$
  • $f(3)=|2|+2=4$
• Absolute min value $2$ at $x=1$; absolute max value $4$ at $x=-1,3$.

Exam angle: If you only solve $f'(x)=0$ you miss the corner completely.

Example 3: Restricted domain + interior critical point

Find absolute extrema of $f(x)=\sqrt{x}(4-x)$ on $[0,4]$.

• Endpoints matter because of the closed interval and because radicals often peak inside.
• Differentiate (product rule + power rule):
  • $f(x)=4x^{1/2}-x^{3/2}$
  • $f'(x)=2x^{-1/2}-\frac{3}{2}x^{1/2}$
• Solve $f'(x)=0$ for $x \in (0,4)$:
  • $2x^{-1/2}=\frac{3}{2}x^{1/2}$
  • Multiply by $2x^{1/2}$: $4=3x$
  • $x=\frac{4}{3}$
• Candidates: $x=0,\frac{4}{3},4$
• Evaluate:
  • $f(0)=0$
  • $f(4)=0$
  • $f\left(\frac{4}{3}\right)=\sqrt{\frac{4}{3}}\left(4-\frac{4}{3}\right)=\frac{2}{\sqrt{3}}\cdot\frac{8}{3}=\frac{16}{3\sqrt{3}}$
• Absolute max is $\frac{16}{3\sqrt{3}}$ at $x=\frac{4}{3}$; absolute min is $0$ at $x=0$ and $x=4$.

Exam angle: Don’t ignore endpoints just because they “look boring.”

Example 4: Given $f'$ and an initial value (BC-flavored)

Suppose $f'(x)=x(x-2)$ and $f(0)=1$. Find where $f$ has absolute extrema on $[0,3]$.

• Critical numbers from $f'(x)=0$: $x=0,2$, plus endpoints $0,3$.
• Candidates: $x=0,2,3$.
• Need function values. Use accumulation:
  • $f(x)=f(0)+\int_{0}^{x} f'(t)\,dt =1+\int_{0}^{x} t(t-2)\,dt$
  • Integrand: $t(t-2)=t^2-2t$
  • Antiderivative: $\frac{t^3}{3}-t^2$
• Evaluate:
  • $f(0)=1$
  • $f(2)=1+\left[\frac{t^3}{3}-t^2\right]_{0}^{2}=1+\left(\frac{8}{3}-4\right)=1-\frac{4}{3}=-\frac{1}{3}$
  • $f(3)=1+\left[\frac{t^3}{3}-t^2\right]_{0}^{3}=1+(9-9)=1$
• Absolute min value $-\frac{1}{3}$ at $x=2$; absolute max value $1$ at $x=0$ and $x=3$.

Exam angle: When $f'$ is given, you often must integrate to compare candidate values.

Common Mistakes & Traps

1. Forgetting endpoints

   • Wrong: Only checking where $f'(x)=0$.
   • Why wrong: Absolute extrema on $[a,b]$ can occur at $a$ or $b$.
   • Fix: Always write “Candidates: endpoints + critical numbers.”

2. Using points where $f$ is not defined

   • Wrong: Treating a discontinuity (like a vertical asymptote) as a candidate.
   • Why wrong: Critical numbers must be in the domain of $f$.
   • Fix: Check domain first; if $f(c)$ doesn’t exist, $c$ cannot be an extremum.

3. Not checking where $f'(x)$ does not exist

   • Wrong: Solving $f'(x)=0$ and stopping.
   • Why wrong: Corners/cusps/vertical tangents can produce extrema with undefined derivative.
   • Fix: After differentiating, ask: “Where is $f'$ undefined inside $(a,b)$?”

4. Finding critical numbers outside the interval and still using them

   • Wrong: Solving $f'(x)=0$ and evaluating all solutions.
   • Why wrong: The test is for extrema on $[a,b]$; only candidates in $[a,b]$ matter.
   • Fix: Filter critical numbers to $(a,b)$ (and include endpoints separately).

5. Mixing up x-values and y-values in the final answer

   • Wrong: Saying “absolute max is $x=5$” when they asked for the maximum value.
   • Why wrong: Problems sometimes ask for the value $f(x)$, sometimes the point, sometimes both.
   • Fix: State both: “Absolute max value is $f(5)=6$ at $x=5$.”

6. Assuming a critical point guarantees an absolute extremum

   • Wrong: Seeing $f'(c)=0$ and declaring max/min.
   • Why wrong: $f'(c)=0$ is only a candidate; it could be a local min/max or neither.
   • Fix: Compare $f$ at all candidates (or use first/second derivative tests for local behavior, then still compare values for absolute).

7. Using derivative sign (increasing/decreasing) but never computing values

   • Wrong: Concluding absolute max where $f'$ changes from $+$ to $-$.
   • Why wrong: That gives a local max; an endpoint might be larger.
   • Fix: For absolute extrema on $[a,b]$, you must compare actual $f$-values.

8. Applying EVT when continuity fails

   • Wrong: Assuming absolute max/min exist on $[a,b]$ without checking continuity.
   • Why wrong: Discontinuities can prevent extrema from existing (even on a closed interval).
   • Fix: Quick continuity scan: holes, jumps, asymptotes, piecewise mismatches.

Memory Aids & Quick Tricks

Quick Review Checklist

• [ ] Are you on a closed interval $[a,b]$?
• [ ] Is $f$ **continuous** on $[a,b]$ (EVT applicable)?
• [ ] Did you compute or use $f'(x)$ correctly?
• [ ] Did you find all critical numbers in $(a,b)$ where $f'(x)=0$?
• [ ] Did you include points in $(a,b)$ where $f'(x)$ **does not exist** (but $f$ does)?
• [ ] Did you include both endpoints $a$ and $b$?
• [ ] Did you evaluate $f$ at every candidate?
• [ ] Did you choose the largest value for absolute max and smallest for absolute min?
• [ ] Did you report answers with clear x-location(s) and function value(s)?

You’ve got this—run E.C.C. carefully and the problem basically grades itself.