Mathematics · Ch 12 — Application of Derivatives

Maximum and Minimum Values of a Function in a Closed Interval

12.4.1

Maximum and Minimum Values of a Function in a Closed Interval

6.4.1 Maximum and Minimum Values of a Function in a Closed Interval

Why the Interval Matters

Consider $f(x) = x + 2$ on the open interval $(0, 1)$ . It is continuous there yet has neither a maximum nor a minimum value (and no local extremum either): the values approach $2$ as $x \to 0^+$ and $3$ as $x \to 1^-$ , but the endpoints are excluded, so the function only gets arbitrarily close without ever reaching those values.

On the closed interval $[0, 1]$ the situation changes completely: $f$ now attains

Maximum value $3 = f(1)$ ,
Minimum value $2 = f(0)$ .

These are the absolute maximum value (global maximum / greatest value) and absolute minimum value (global minimum / least value) of $f$ on $[0, 1]$ .

Important

A continuous function on an open interval may have no absolute maximum or minimum. On a closed interval the endpoints are included, which guarantees both absolute extremes exist.

Local vs. Absolute Extremes — A Graphical View

For a continuous function $f$ on $[a, d]$ (Fig 6.19 of the textbook):

Local minimum at $x = b$ , value $f(b)$
Local maximum at $x = c$ , value $f(c)$
Absolute maximum value $f(a)$ — at the left endpoint
Absolute minimum value $f(d)$ — at the right endpoint

Note

An absolute extreme is the overall highest or lowest value on the entire interval; a local extreme is a peak or trough relative to nearby points only. They may coincide, but need not.

Two Fundamental Theorems (Stated Without Proof)

Theorem 5 — Existence of Absolute Extremes

Let $f$ be a continuous function on $I = [a, b]$ . Then $f$ has an absolute maximum value and an absolute minimum value, each attained at least once in $I$ .

Theorem 6 — Where Absolute Extremes Occur (for Differentiable Functions)

Let $f$ be a differentiable function on a closed interval $I$ , and let $c$ be an interior point of $I$ ( $a < c < b$ ). If $f$ attains its absolute maximum value or its absolute minimum value at $c$ , then $f'(c) = 0$ .

Watch out

Theorem 6 applies only to interior points — at an endpoint the derivative need not be zero. The converse also fails: $f'(c) = 0$ does not guarantee an absolute extreme (it could be a local extreme or a point of inflection).

Working Rule for Finding Absolute Maximum and Minimum Values

Find all critical points in $(a, b)$ — where $f'(x) = 0$ or $f$ is not differentiable.
Include the endpoints $x = a$ and $x = b$ .
Evaluate $f$ at all candidate points from Steps 1 and 2. …

Theorem 5

Theorem 6: Derivative at an Absolute Extremum

Let $f$ be a differentiable function on a closed interval $I$ and let $c$ be any interior point of $I$ . Then:

f'(c) = 0 \quad \text{if } f \text{ attains its absolute maximum value at } c

The same holds for an absolute minimum: $f'(c) = 0$ if $f$ attains its absolute minimum value at $c$ .

Hypotheses explained:

$f$ is differentiable on $I$ — this means $f'(x)$ exists at every point in $I$ .
$I$ is a closed interval — for example $[a, b]$ .
$c$ is an interior point of $I$ — that is, $c$ is strictly between the endpoints, not equal to $a$ or $b$ .
$f$ attains its absolute maximum at $c$ — meaning $f(c) \geq f(x)$ for every $x$ in $I$ .

Watch out

The converse is not true: $f'(c) = 0$ does not guarantee a maximum or minimum at $c$ . For example, $f(x) = x^3$ has $f'(0) = 0$ but no extremum at $x = 0$ . The theorem only tells us that if an absolute extremum occurs at an interior point, then the derivative must be zero there.

›Proof

Proof of Theorem 6(i)

Since $f$ attains its absolute maximum value at $c$ , we have:

$f(c) \geq f(x) \quad \text{for all } x \in I$

Because $c$ is an interior point of $I$ , there exists some $h > 0$ small enough that both $c + h$ and $c - h$ lie inside $I$ .

Step 1: Consider the right-hand difference quotient.

For $h > 0$ , we have $c + h \in I$ , so $f(c) \geq f(c + h)$ . This gives:

$f(c + h) - f(c) \leq 0$

Dividing by the positive number $h$ :

$\frac{f(c + h) - f(c)}{h} \leq 0$

Taking the limit as $h \to 0^+$ (the right-hand derivative):

$f'(c) = \lim_{h \to 0^+} \frac{f(c + h) - f(c)}{h} \leq 0$

Step 2: Consider the left-hand difference quotient.

For $h < 0$ , write $h = -k$ where $k > 0$ . Then $c - k \in I$ , so $f(c) \geq f(c - k)$ . This gives:

$f(c - k) - f(c) \leq 0$

Now $h = -k$ is negative. Dividing by $h$ (a negative number) reverses the inequality:

$\frac{f(c + h) - f(c)}{h} = \frac{f(c - k) - f(c)}{-k} \geq 0$

Taking the limit as $h \to 0^-$ (the left-hand derivative):

$f'(c) = \lim_{h \to 0^-} \frac{f(c + h) - f(c)}{h} \geq 0$

Step 3: Combine the two inequalities.

From Step 1: $f'(c) \leq 0$

From Step 2: $f'(c) \geq 0$

The only number that satisfies both $f'(c) \leq 0$ and $f'(c) \geq 0$ is $f'(c) = 0$ .

…

Theorem 6

Theorem 6: Derivative at an Absolute Extremum

Let $f$ be a differentiable function on a closed interval $I$ and let $c$ be any interior point of $I$ . Then:

f'(c) = 0 \quad \text{if } f \text{ attains its absolute maximum value at } c

The same holds for an absolute minimum: $f'(c) = 0$ if $f$ attains its absolute minimum value at $c$ .

Hypotheses explained:

$f$ is differentiable on $I$ — this means $f'(x)$ exists at every point in $I$ .
$I$ is a closed interval — for example $[a, b]$ .
$c$ is an interior point of $I$ — that is, $c$ is strictly between the endpoints, not equal to $a$ or $b$ .
$f$ attains its absolute maximum at $c$ — meaning $f(c) \geq f(x)$ for every $x$ in $I$ .

Watch out

›Proof

Proof of Theorem 6(i)

Since $f$ attains its absolute maximum value at $c$ , we have:

$f(c) \geq f(x) \quad \text{for all } x \in I$

Because $c$ is an interior point of $I$ , there exists some $h > 0$ small enough that both $c + h$ and $c - h$ lie inside $I$ .

Step 1: Consider the right-hand difference quotient.

For $h > 0$ , we have $c + h \in I$ , so $f(c) \geq f(c + h)$ . This gives:

$f(c + h) - f(c) \leq 0$

Dividing by the positive number $h$ :

$\frac{f(c + h) - f(c)}{h} \leq 0$

Taking the limit as $h \to 0^+$ (the right-hand derivative):

$f'(c) = \lim_{h \to 0^+} \frac{f(c + h) - f(c)}{h} \leq 0$

Step 2: Consider the left-hand difference quotient.

For $h < 0$ , write $h = -k$ where $k > 0$ . Then $c - k \in I$ , so $f(c) \geq f(c - k)$ . This gives:

$f(c - k) - f(c) \leq 0$

Now $h = -k$ is negative. Dividing by $h$ (a negative number) reverses the inequality:

$\frac{f(c + h) - f(c)}{h} = \frac{f(c - k) - f(c)}{-k} \geq 0$

Taking the limit as $h \to 0^-$ (the left-hand derivative):

$f'(c) = \lim_{h \to 0^-} \frac{f(c + h) - f(c)}{h} \geq 0$

Step 3: Combine the two inequalities.

From Step 1: $f'(c) \leq 0$

From Step 2: $f'(c) \geq 0$

The only number that satisfies both $f'(c) \leq 0$ and $f'(c) \geq 0$ is $f'(c) = 0$ .

…

Figure 6.19Graph of a continuous function on a closed interval [a, d] showing local and absolute extrema

Fig. 6.19 — Graph of a continuous function on a closed interval [a, d] showing local and absolute extrema

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Understanding Fig. 6.19: Local and Absolute Extrema on a Closed Interval

The figure plots a continuous function $y = f(x)$ on the $xy$ -plane over the closed interval $[a, d]$ on the $x$ -axis. Four key points are marked on the $x$ -axis: $a$ , $b$ , $c$ , and $d$ , with dashed vertical lines rising from each to meet the curve. The corresponding function values $f(a)$ , $f(b)$ , $f(c)$ , and $f(d)$ are labelled on the $y$ -axis.

The curve itself tells a story: it begins at a high value at $x = a$ , then descends to a low point at $x = b$ , rises to a peak at $x = c$ , and finally falls again to end at $x = d$ . This shape is deliberately chosen to illustrate a critical distinction between two types of extreme values.

What the Curve Reveals

At $x = b$ , the function reaches a local minimum — the lowest value in its immediate neighbourhood. The value $f(b)$ is smaller than values of $f$ at points just to the left and right of $b$ . Similarly, at $x = c$ , the function attains a local maximum — the highest value in its vicinity, with $f(c)$ exceeding nearby function values.

However, when we look at the entire interval $[a, d]$ , the picture changes. The absolute maximum (the greatest value of $f$ on the whole interval) occurs at the left endpoint $x = a$ , giving $f(a)$ . The absolute minimum (the smallest value on the whole interval) occurs at the right endpoint $x = d$ , giving $f(d)$ .

Important

This figure demonstrates a crucial principle: absolute extrema on a closed interval need not coincide with local extrema, and they can occur at the endpoints — even when the interior contains local maxima and minima.

The Key Formulas

The textbook uses this figure to motivate the formal procedure for finding absolute extrema. For a function $f$ continuous on $[a, b]$ , the absolute maximum and minimum are found by:

Finding all critical points of $f$ in $(a, b)$ — points $x = c$ where $f'(c) = 0$ or $f'(c)$ does not exist.
Evaluating $f$ at every critical point and at both endpoints $a$ and $b$ .
Comparing these values: the largest is the absolute maximum, the smallest is the absolute minimum.

\text{Absolute maximum} = \max\{f(a), f(b), f(c_1), f(c_2), \ldots, f(c_n)\}

Why This Matters

The figure makes a subtle but essential point: local extrema are about behaviour in a small neighbourhood, while absolute extrema consider the entire interval. In Fig. 6.19, $f(b)$ is a local minimum but not the absolute minimum — that honour goes to $f(d)$ . Similarly, $f(c)$ is a local maximum but not the absolute maximum — $f(a)$ is larger.

Watch out

A common mistake is to assume that the absolute maximum must occur at a local maximum, or that the absolute minimum must occur at a local minimum. This figure shows both assumptions can be false. Always check the endpoints. …

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