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Calculus BC · Unit 5 · 5.5
AP Calculus · Analytical Applications of Differentiation
Using the Candidates Test
Build a finite list — critical points plus endpoints — evaluate, and pick the extremes.
key idea$\text{Candidates}=\{\text{critical points}\}\cup\{\text{endpoints}\};\ \text{evaluate }f\text{ at each}$
The procedure
On a closed interval the absolute extremes must occur either at an interior critical point or at an endpoint — there is nowhere else. So you build a finite list and evaluate. No sign chart is needed, and none is expected.
1
Confirm the theorem applies
$f$ continuous on a closed bounded $[a,b]$. If either fails, the Candidates Test is not valid and absolute extrema may not exist at all.
in $f$ and $[a,b]$ out a guarantee
2
Find every critical point in $(a,b)$
Solve $f'=0$ and find where $f'$ is undefined. Discard any that fall outside the open interval — they are not candidates.
out the interior candidates
3
Add the two endpoints
$a$ and $b$ join the list. They are always candidates for an absolute extremum, even though they are not local extrema.
out the complete candidate list
4
Evaluate $f$ at every candidate
Compute the function value at each — not the derivative. This is the step where students substitute into $f'$ by mistake.
out a table of values
5
Select the largest and smallest
The biggest value is the absolute maximum, the smallest the absolute minimum. Report the value and where it occurs.
out the answer, with location
| Do | Do not | Because |
|---|
| evaluate $f$ at each candidate | evaluate $f'$ | you want the height, not the slope |
| include both endpoints | test only critical points | the extreme often sits at an end |
| discard candidates outside $(a,b)$ | keep every root of $f'$ | only points in the interval compete |
| report value and location | report only the number | questions ask for both |
| include where $f'$ is undefined | solve only $f'=0$ | cusps can hold the extreme |
The left column is the whole procedure. Every entry in the middle column is a real and common error.
Interior candidateswhere $f'=0$ or $f'$ is undefined
Boundary candidatesalways both endpoints $a$ and $b$
The comparisonevaluate $f$, never $f'$
Worked examples
EX 1
The standard case
$f(x)=x^{3}-3x$ on $[-2,3]$Find the absolute extrema of $f(x)=x^{3}-3x$ on $[-2,3]$.
1
Check the hypotheses
A polynomial on a closed bounded interval, so absolute extrema exist.
2
Find the critical points
$f'(x)=3x^{2}-3=0$ gives $x=\pm1$; both lie in $(-2,3)$.
3
List the candidates
$x=-2,\ -1,\ 1,\ 3$.
4
Evaluate $f$ at each
$f(-2)=-2$; $f(-1)=2$; $f(1)=-2$; $f(3)=18$.
5
Select
Absolute maximum $18$ at $x=3$; absolute minimum $-2$ at $x=-2$ and $x=1$. The maximum sits at an endpoint, which is why endpoints must be on the list.
EX 2
A critical point outside the interval
discardFind the absolute extrema of $f(x)=x^{2}-8x+3$ on $[0,3]$.
1
Find the critical point
$f'(x)=2x-8=0$ gives $x=4$.
2
Test membership
$4$ is not in $(0,3)$, so it is not a candidate. Discard it.
3
List what remains
Only the endpoints $x=0$ and $x=3$.
4
Evaluate
$f(0)=3$ and $f(3)=9-24+3=-12$.
5
Select
Absolute maximum $3$ at $x=0$; absolute minimum $-12$ at $x=3$. With no interior candidate the function is monotonic, so both extremes are at the ends.
EX 3
A cusp candidate
$f'$ undefinedFind the absolute extrema of $f(x)=x^{2/3}$ on $[-1,8]$.
1
Differentiate
$f'(x)=\dfrac{2}{3x^{1/3}}$, which is never zero.
2
Find where $f'$ is undefined
At $x=0$, and $f(0)=0$ exists, so $x=0$ is a critical point.
3
List the candidates
$x=-1,\ 0,\ 8$.
4
Evaluate
$f(-1)=1$; $f(0)=0$; $f(8)=4$.
5
Select
Absolute maximum $4$ at $x=8$; absolute minimum $0$ at $x=0$ — the cusp. Solving only $f'=0$ would have found no interior candidate and missed the minimum entirely.
EX 4
Applied, with units
optimisation flavourA tank's volume is $V(t)=t^{3}-9t^{2}+24t+5$ litres for $0\le t\le 5$ hours. When is it fullest?
1
Differentiate
$V'(t)=3t^{2}-18t+24=3(t-2)(t-4)$.
2
Find interior critical points
$t=2$ and $t=4$, both inside $(0,5)$.
3
List and evaluate
$V(0)=5$; $V(2)=8-36+48+5=25$; $V(4)=64-144+96+5=21$; $V(5)=125-225+120+5=25$.
4
Select the maximum
The largest value is $25$ litres, attained at both $t=2$ and $t=5$.
5
Answer in context
The tank is fullest at $25$ litres, occurring at $t=2$ hours and again at $t=5$ hours. Reporting only one of the two loses half the answer.
Where it goes wrong
Evaluating $f'$ instead of $f$
The candidates come from $f'$, but the comparison uses $f$. Substituting into the derivative gives slopes, which answer nothing.
Forgetting the endpoints
In Example 1 the maximum is at $x=3$, an endpoint. A critical-points-only search returns the wrong answer.
Keeping out-of-range critical points
A root of $f'$ outside $(a,b)$ is not a candidate. Test membership before adding it to the list.
Missing the points where $f'$ is undefined
Cusps and corners are critical points. Example 3's minimum is at exactly such a point.