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.