Application of Derivatives · Lesson 6 of 7

Absolute Extrema on Closed Intervals

“Find greatest and least values by comparing critical inputs and endpoints, while checking existence and domain conditions.”

Learning Objectives

• Distinguish absolute comparisons from local derivative tests. • State the closed-interval existence theorem with its continuity condition. • Include all relevant interior critical points and both endpoints. • Handle fractional powers and undefined derivatives correctly. • Justify a global conclusion separately when the domain is not a closed interval.

Comparing the Whole Interval

A local test examines a small neighbourhood. An absolute maximum or minimum compares every output in the stated domain. On a closed interval, an endpoint can win that comparison even though no horizontal tangent or interior turning point occurs there.

Consider $f (x) = x + 2$ on $(0, 1)$ . It approaches 2 and 3 without attaining either as an extremum. On $[0, 1]$ , both endpoint inputs are allowed: the absolute minimum is $f (0) = 2$ and the absolute maximum is $f (1) = 3$ . Adding endpoints changes attainment, even though the function remains increasing.

Definition

Absolute extrema

An absolute maximum is the greatest attained output on the entire stated domain. An absolute minimum is the least attained output. “Global maximum,” “greatest value,” and “absolute maximum” mean the same comparison; similarly for their minimum counterparts.

We use “local extremum” at an interior point with a two-sided neighbourhood, as in the earlier definitions. Endpoint absolute extrema are therefore reported as endpoint extrema without forcing them into that interior local classification.

An endpoint can beat every local peak— For this curve on [−2, 3], the right endpoint is higher than the local peak. The left endpoint ties the local valley for the least value.

Why a Finite Candidate List Works

Continuity on a closed, bounded interval guarantees that the highest and lowest values are actually reached. At an interior point where the function is differentiable, an attained absolute extremum is also local and must have derivative zero. Other possibilities are nondifferentiable interior points and the two endpoints.

Definition

Existence of absolute extrema

Every function continuous on a closed, bounded interval $[a, b]$ attains an absolute maximum and an absolute minimum at least once in that interval. The theorem guarantees existence, not that either extremum is unique or interior.

The working rule is to collect all interior critical points, add the endpoints, evaluate the original function at every candidate, and compare the outputs. In standard examples there are finitely many candidates. If a function has an entire interval of critical points, handle its constant values rather than pretending the list must be finite.

Step	What to do	Why it is necessary
Check the setting	Verify domain and continuity on $[a, b]$	The existence theorem has hypotheses
Interior candidates	Find zero or undefined derivatives at allowed interior inputs	Both kinds can give extrema
Boundary candidates	Add $a$ and $b$	Extrema can occur at endpoints
Evaluate	Compute $f$ at every candidate	Derivative values are not extreme function values
Compare	Choose largest and smallest outputs, retaining all ties	Several inputs may attain the same value

An undefined function value is different from an undefined derivative

If $f (c)$ exists but $f^{'} (c)$ does not, $c$ can be a candidate. If $f (c)$ itself is undefined, it is not part of the graph. A function with a missing or discontinuous point does not automatically satisfy the closed-interval existence theorem.

Example — Endpoint values beat the local candidates

Problem
Find absolute extrema of $f (x) = 2 x^{3} - 15 x^{2} + 36 x + 1$ on $[1, 5]$ .

1.The polynomial is continuous on the interval. Differentiate: $f^{'} (x) = 6 x^{2} - 30 x + 36 = 6 (x - 2) (x - 3)$ .
2.Interior candidates are 2 and 3. Add endpoints 1 and 5.
3.Evaluate: $f (1) = 24$ , $f (2) = 29$ , $f (3) = 28$ , and $f (5) = 56$ .
4.Comparing these outputs gives absolute minimum 24 at $x = 1$ and absolute maximum 56 at $x = 5$ .
5.The interior values include a local peak and valley, but neither wins the overall comparison.

Candidate input	Why included	Function value
1	Left endpoint	24
2	Zero derivative	29
3	Zero derivative	28
5	Right endpoint	56

Including a Point Where the Derivative Does Not Exist

Fractional powers are a useful reminder that continuity and differentiability are different properties. A cube root is defined at negative inputs and continuous at zero, but its derivative can fail there. Check the original function before discarding a point because a derivative formula has a zero denominator.

Example — A fractional-power function

Problem
Find absolute extrema of $f (x) = 12 x^{4/3} - 6 x^{1/3}$ on $[- 1, 1]$ . Use real cube roots.

1.The function is continuous on $[- 1, 1]$ . For $x \neq = 0$ , $f^{'} (x) = 16 x^{1/3} - 2/ x^{2/3} = 2 (8 x - 1) / x^{2/3}$ .
2.The derivative is zero at $x = 1/8$ and undefined at 0, while the original function is defined at 0. Both are interior candidates.
3.Add the endpoints. The values are $f (- 1) = 18$ , $f (0) = 0$ , $f (1/8) = 12 (1/16) - 6 (1/2) = - 9/4$ , and $f (1) = 6$ .
4.The absolute maximum is 18 at $x = - 1$ ; the absolute minimum is $- 9/4$ at $x = 1/8$ .
5.The point 0 did not win, but it still had to be considered. Also, $(- 1)^{1/3} = - 1$ while $(- 1)^{4/3} = 1$ under the real cube-root interpretation.

When the Domain Has No Endpoints

The closed-interval rule cannot simply be copied onto the whole real line. An unbounded domain has no endpoint values to add, and the existence theorem does not apply automatically. Instead, show directly how the objective behaves over that domain.

For a squared-distance objective with one critical point, proving that it decreases everywhere before that point and increases everywhere afterwards establishes a global minimum. Merely discovering one critical point is not enough: a function may have no extrema at all or may behave differently far away.

Example — Nearest distance to a parabolic path

Problem
A moving object follows $y = x^{2} + 7$ . Find its least distance from a fixed observation point $(3, 7)$ .

1.A point on the path is $(x, x^{2} + 7)$ . Its squared distance is $G (x) = (x - 3)^{2} + x^{4}$ , defined for every real $x$ .
2.Differentiate: $G^{'} (x) = 2 (x - 3) + 4 x^{3} = 2 (x - 1) (2 x^{2} + 2 x + 3)$ .
3.The quadratic factor is $2 (x + 1/2)^{2} + 5/2 > 0$ . Hence the derivative is negative for $x < 1$ and positive for $x > 1$ .
4.Thus $G$ has its global minimum at $x = 1$ , where $G (1) = 4 + 1 = 5$ . The least distance is $5$ , attained at path point $(1, 8)$ .
5.This sign argument covers the entire real line. It supplies the global justification that the closed-interval theorem cannot supply here.

Use every condition in an application

For a profit model such as $41 - 72 x - 18 x^{2}$ , the unconstrained real-input maximum is at $x = - 2$ . If $x$ counts produced items, a domain such as $x \geq 0$ excludes that input. On that domain the same function decreases and is greatest at 0. A physical interpretation must supply or respect the feasible domain.

An absolute-extrema answer should state the greatest or least output and all inputs attaining it. A table helps avoid losing ties. If continuity fails, or an interval is open, discuss existence rather than assuming a candidate list guarantees a winner.

Check Your Understanding

Use these questions to check both your calculations and your interpretation of the ideas. Read the conditions carefully before choosing an answer.

Quick check

For a continuous function on $[a, b]$ , which candidate set is required?

Quick check

What must be compared to find absolute extrema?

Quick check

A continuous function has an undefined derivative at an interior point. What should be done?

Quick check

Where does $f (x) = x$ attain its absolute minimum on $[0, 1]$ ?

Quick check

What are the absolute maximum and minimum values of $2 x^{3} - 15 x^{2} + 36 x + 1$ on $[1, 5]$ ?

Quick check

The least squared distance is 5. What is the least distance?

Practice Problems

Find the absolute extrema of $x^{3}$ on $[- 2, 2]$ , and explain why the stationary point at 0 does not give either answer.
Find absolute extrema of $sin x + cos x$ on $[0, π]$ .
Find absolute extrema of $3 x^{4} - 8 x^{3} + 12 x^{2} - 48 x + 25$ on $[0, 3]$ .
Find all inputs in $[0, 2 π]$ where $sin 2 x$ attains its maximum value.
Find maxima of $2 x^{3} - 24 x + 107$ on $[1, 3]$ and $[- 3, - 1]$ . Explain why the different intervals give different answers.
If $x^{4} - 62 x^{2} + a x + 9$ has its absolute maximum on $[0, 2]$ at $x = 1$ , determine $a$ and verify the global claim.
Find absolute extrema of $x + sin 2 x$ on $[0, 2 π]$ .
Find the least value of $(1 - x + x^{2}) / (1 + x + x^{2})$ over the real line, justifying the domain and global conclusion.
Find the maximum of $[x (x - 1) + 1]^{1/3}$ on $[0, 1]$ and list all attaining inputs.
Find the nearest points on $x^{2} = 2 y$ to $(0, 5)$ ; remember that there may be two equally near points.

Key Takeaways

• Continuity on a closed, bounded interval guarantees attained absolute extrema. • Interior zero derivatives, interior undefined derivatives, and endpoints all belong in the search. • Compare original function values, retaining every input tied for an extreme value. • Local extrema need not be absolute extrema. • For open or unbounded domains, give a separate existence and global comparison argument.

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Building and Solving Optimization Models

Next · Lesson 7

Chapter Summary and Practice

Comparing the Whole Interval

Why a Finite Candidate List Works

Including a Point Where the Derivative Does Not Exist

When the Domain Has No Endpoints

Check Your Understanding

Quiz

Practice Problems

Key Takeaways