Using the Candidates Test to Determine Absolute (Global) Extrema
Introduced| English | Meaning |
|---|---|
| Candidates Test/ˈkændɪdeɪts test/ | For a continuous function on a closed interval, comparing values at interior critical points and both endpoints to find absolute extrema. |
Finding the true highest and lowest
- On a closed interval, the EVT promises an absolute max and min exist. Where are they?
- They can only be at a critical point or at an endpoint — a short list of suspects.
- The Candidates Test 候选点判别法 just evaluates $f$ at every suspect and compares.
- No sign charts needed — the biggest output wins, the smallest loses.
The three steps
- 1. Find all critical points in $[a,b]$ (where $f'=0$ or is undefined).
- 2. Evaluate $f$ at each critical point and at both endpoints $a$ and $b$.
- 3. Compare the values: the largest is the absolute maximum, the smallest the absolute minimum.
- Report both the value and its location.
Candidates test: compare f at the critical points and both endpoints
Endpoints and critical points
y = ax³ + cx
The absolute max/min on a closed interval hide among the endpoints and the interior critical points — evaluate and compare.
To find absolute extrema on $[a,b]$, evaluate $f$ at which points?
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Critical points plus both endpoints — that is the full candidate list.
Don't forget the endpoints
- On a closed interval, the extremes often sit at the endpoints, not a critical point.
- A monotonic function (always increasing) has its max at the right end and min at the left.
- So you must always include $f(a)$ and $f(b)$ in your candidate list.
- Missing an endpoint is the most common way to get the wrong absolute extremum.
For $f(x)=x^3-3x$ on $[0,2]$, with $f(0)=0$, $f(1)=-2$, $f(2)=2$, what is the absolute maximum value?
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Largest candidate value is $2$ (at the endpoint $x=2$).
Same function and candidates: what is the absolute minimum value?
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Smallest candidate value is $-2$ (at $x=1$).
You can skip the endpoints and still be sure of the absolute extrema.
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Extremes often sit at endpoints; skipping them can give a wrong answer.
Compare, don't classify
- Unlike the First Derivative Test, the Candidates Test doesn't care what kind each point is.
- It simply asks: of all the candidate outputs, which is biggest and which smallest?
- One evaluation per candidate, then a comparison — clean and reliable on a closed interval.
- (It needs a closed interval; on an open interval an absolute extremum may not exist.)
The Candidates Test requires a ____ interval, where the EVT guarantees extrema exist.
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On an open interval an absolute extremum may not exist.
The absolute maximum is the candidate with the largest...
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Compare outputs $f(x)$, not the inputs.
The Candidates Test requires a closed interval $[a,b]$ and must include the endpoints. Evaluating only the critical points misses extremes that live at $a$ or $b$. And compare $f$-values, not $x$-values — the winner is the largest output, wherever it occurs.
Find the absolute extrema of $f(x)=x^3-3x$ on $[0,2]$.
- Critical points in $[0,2]$: $f'=3(x-1)(x+1)=0$ at $x=1$ (only $x=1$ is in range).
- Evaluate candidates: $f(0)=0$, $f(1)=-2$, $f(2)=2$.
- Absolute max $=2$ at $x=2$ (endpoint); absolute min $=-2$ at $x=1$.
The Candidates Test finds absolute extrema on a closed interval: list the critical points and both endpoints, evaluate $f$ at each, and compare. Largest value = absolute maximum, smallest = absolute minimum. Always include the endpoints, and compare outputs, not inputs.