The Candidates
Test
Learn the definitive method for finding absolute (global) extrema of a continuous function on a closed interval.
AP Calculus AB
What Are Absolute Extrema?
The absolute maximum of a function \( f \) on an interval is the largest output value \( f \) achieves anywhere on that interval. The absolute minimum is the smallest output value.
Absolute Max
\( f(c) \geq f(x) \) for every \( x \) in the interval
Absolute Min
\( f(c) \leq f(x) \) for every \( x \) in the interval
Why Closed Intervals Matter
The Extreme Value Theorem guarantees: if \( f \) is continuous on a closed interval \( [a,b] \), then \( f \) must attain both an absolute maximum and an absolute minimum somewhere on \( [a,b] \).
What Is a Critical Point?
A number \( c \) in the domain of \( f \) is a critical point if either:
\( f'(c) = 0 \)
The derivative equals zero: the tangent line is horizontal.
\( f'(c) \) DNE
The derivative does not exist: a cusp, corner, or vertical tangent.
The Candidates Test: Formal Statement
Why the Test Works
1
Extreme Value Theorem guarantees that absolute extrema exist on \( [a, b] \) when \( f \) is continuous.
2
Fermat's Theorem says any interior extreme must be a critical point: \( f'(c)=0 \) or \( f'(c) \) DNE.
3
The only other possibility is an endpoint. So the set {critical points} ∪ {endpoints} contains every possible location.
4
Evaluating \( f \) at each candidate and comparing values identifies the absolute extrema with certainty.
Local vs. Absolute Extrema
| Local Extrema | Absolute Extrema | |
|---|---|---|
| Scope | Nearby points only | Entire interval \( [a, b] \) |
| Location | Interior critical points | Critical points or endpoints |
| Guaranteed? | Not always | Yes (EVT on closed interval) |
| Method | First/Second Derivative Test | Candidates Test |
How to Apply the Candidates Test
1
Verify continuity of \( f \) on \( [a, b] \). If \( f \) is a polynomial, rational (no division by zero), trig, or exponential function on the interval, you are good.
2
Find \( f'(x) \) and solve \( f'(x) = 0 \). Also identify where \( f'(x) \) does not exist. Keep only solutions in \( (a, b) \).
3
Evaluate \( f \) at each critical point and at both endpoints \( a \) and \( b \).
4
Compare all values. The largest is the absolute maximum; the smallest is the absolute minimum.
Example 1: Polynomial
Find the absolute extrema of \( f(x) = x^3 - 3x + 1 \) on \( [-2, 2] \).
1
Differentiate: \( f'(x) = 3x^2 - 3 \).
2
Solve \( f'(x) = 0 \): \( 3x^2 - 3 = 0 \implies x^2 = 1 \implies x = -1, \; 1 \). Both lie in \( (-2, 2) \).
3
Evaluate: \( f(-2) = -1 \), \( f(-1) = 3 \), \( f(1) = -1 \), \( f(2) = 3 \).
4
Compare: Abs max = 3 at \( x = -1 \) and \( x = 2 \). Abs min = -1 at \( x = -2 \) and \( x = 1 \).
Example 1: Visualized
Example 2: Trigonometric
Find the absolute extrema of \( f(x) = 2\sin x + \cos 2x \) on \( [0, 2\pi] \).
1
Differentiate: \( f'(x) = 2\cos x - 2\sin 2x = 2\cos x - 4\sin x\cos x = 2\cos x(1 - 2\sin x) \).
2
Solve \( f'(x) = 0 \): \( \cos x = 0 \implies x = \frac{\pi}{2},\;\frac{3\pi}{2} \); \( \sin x = \frac{1}{2} \implies x = \frac{\pi}{6},\;\frac{5\pi}{6} \).
3
Evaluate: \( f(0) = 1 \), \( f(\frac{\pi}{6}) = \frac{3}{2} \), \( f(\frac{\pi}{2}) = 1 \), \( f(\frac{5\pi}{6}) = \frac{3}{2} \), \( f(\frac{3\pi}{2}) = -3 \), \( f(2\pi) = 1 \).
4
Compare: Abs max = \( \frac{3}{2} \) at \( x = \frac{\pi}{6} \) and \( \frac{5\pi}{6} \). Abs min = \( -3 \) at \( x = \frac{3\pi}{2} \).
Example 3: Non-Differentiable Point
Find the absolute extrema of \( f(x) = x^{2/3} \) on \( [-1, 8] \).
1
Differentiate: \( f'(x) = \frac{2}{3}x^{-1/3} = \frac{2}{3\sqrt[3]{x}} \).
2
Critical points: \( f'(x) \neq 0 \) for any \( x \), but \( f'(0) \) does not exist (division by zero). So \( x = 0 \) is a critical point in \( (-1, 8) \).
3
Evaluate: \( f(-1) = (-1)^{2/3} = 1 \), \( f(0) = 0 \), \( f(8) = 8^{2/3} = 4 \).
4
Compare: Abs max = 4 at \( x = 8 \). Abs min = 0 at \( x = 0 \) (the cusp).
When It Does Not Apply
The Candidates Test requires two conditions. If either fails, the method cannot be used directly:
Open Interval
On \( (a, b) \), there are no endpoints to evaluate: absolute extrema may not exist at all.
Discontinuous \( f \)
If \( f \) has a discontinuity on \( [a, b] \), the Extreme Value Theorem fails: no guarantee extrema exist.
Common Mistakes
✕ Forgetting Endpoints
Students find critical points but skip evaluating \( f(a) \) and \( f(b) \). Endpoints can be the absolute extrema.
✕ Ignoring DNE Points
A point where \( f'(x) \) does not exist is a critical point too: cusps and corners must be checked.
✕ Using Open Intervals
Applying the Candidates Test on \( (a, b) \) without endpoints is invalid: the theorem requires \( [a, b] \).
Summary Checklist
| Step | Action | Key Detail |
|---|---|---|
| 1 | Verify conditions | Continuous on \( [a, b] \) |
| 2 | Compute \( f'(x) \) | Use power, product, chain rules |
| 3 | Solve \( f'(x) = 0 \) and find DNE | Keep only \( x \in (a, b) \) |
| 4 | Evaluate \( f \) | At critical points and at \( a \), \( b \) |
| 5 | Compare | Largest = abs max, smallest = abs min |
Question 1 of 5
What is the absolute maximum of \( f(x) = x^2 - 4x + 3 \) on \( [0, 4] \)?
\( f'(x) = 2x - 4 = 0 \implies x = 2 \). Evaluating: \( f(0) = 3 \), \( f(2) = -1 \), \( f(4) = 3 \). The largest value is 3, occurring at both endpoints.
Question 2 of 5
The Candidates Test can be applied when \( f \) is:
The Candidates Test requires \( f \) to be continuous on a closed interval \( [a, b] \). Differentiability is not required everywhere: the test accounts for points where \( f' \) does not exist.
Question 3 of 5
For \( g(x) = x^3 - 12x \) on \( [-3, 3] \), what is the absolute minimum value?
\( g'(x) = 3x^2 - 12 = 0 \implies x = \pm 2 \). Evaluating: \( g(-3) = -27 + 36 = 9 \), \( g(-2) = -8 + 24 = 16 \), \( g(2) = 8 - 24 = -16 \), \( g(3) = 27 - 36 = -9 \). The smallest value is \( -16 \) at \( x = 2 \).
Question 4 of 5
Which of the following is not a candidate for absolute extrema on \( [a, b] \)?
A point where the second derivative is zero (inflection point) but the first derivative is nonzero is not a critical point and not a candidate for absolute extrema.
Question 5 of 5
Find the absolute maximum of \( h(x) = \sqrt{x} \) on \( [0, 9] \).
\( h'(x) = \frac{1}{2\sqrt{x}} \), which is never zero for \( x > 0 \) and DNE at \( x = 0 \) (an endpoint). So there are no interior critical points. Evaluate: \( h(0) = 0 \), \( h(9) = 3 \). The absolute maximum is 3.