5.2 - Extreme Value Theorem, Global Versus Local Extrema, and Critical Points

The Extreme Value Theorem

If a function

f

is continuous on a closed interval

[a, b]

, then

f

has to have both: - At least one absolute maximum - At least one absolute minimum

This seems obvious, but it’s a fundamental guarantee. Every continuous function on a closed interval reaches both a highest and a lowest point.

Special case: for a horizontal line segment, every point is both an absolute maximum and an absolute minimum at the same time, since all the

y

-values are equal.

Both conditions matter. If either one is missing, there’s no guarantee:

Not a closed interval: $f(x) = x^2$ on the open interval $(0, 2)$ has no absolute maximum or minimum. Its values get closer and closer to $0$ and $4$ but never reach them, because the endpoints aren’t included.

Not continuous: $g(x) = \frac{1}{x}$ on $[-1, 1]$ has no absolute maximum or minimum. It isn’t continuous at $x = 0$ (a vertical asymptote), where it shoots up to $\infty$ on one side and down to $-\infty$ on the other.

A function that breaks a condition can still have extrema. The graph under Identifying Extrema on a Graph below has a jump at

x = 1

, and it still has an absolute max and an absolute min. The theorem just doesn’t promise them.

Justification to write: “Since

f

is continuous on the closed interval

[a, b]

, the Extreme Value Theorem guarantees that

f

has an absolute maximum and an absolute minimum on

[a, b]

.”

Terminology: Two Ways to Say the Same Thing

Different textbooks and teachers use different words for the same ideas.

Common Term 1	Common Term 2	Meaning
Global extrema	Absolute extrema	Highest or lowest point on the entire domain
Local extrema	Relative extrema	Highest or lowest point in a neighborhood

Extrema, the plural of extremum, means maximum or minimum points.

These notes use the “absolute” and “relative” terminology.

Absolute vs. Relative Extrema

Absolute Maximum/Minimum

The highest or lowest point on the entire function, or on the given interval.

Absolute maximum:

f(c) \geq f(x)

for all

x

in the domain

Absolute minimum:

f(c) \leq f(x)

for all

x

in the domain

Relative Maximum/Minimum

A point that is higher or lower than all the nearby points.

Relative maximum:

f(c) \geq f(x)

for all

x

in some open interval around

c

Relative minimum:

f(c) \leq f(x)

for all

x

in some open interval around

c

Test for relative extrema: construct a small open interval around the point. If the point is higher (or lower) than every other point in that interval, it’s a relative max (or min).

Identifying Extrema on a Graph

5.2 - Extreme Value Theorem, Global Versus Local Extrema, and Critical Points

Analysis of the extrema:

At $x = -3$ : absolute minimum of 1, since it’s the lowest

y

-value on the whole graph

At $x = 0$ : absolute maximum of 5, since it’s the highest

y

-value on the whole graph

At $x = 1$ : relative minimum of 2. Construct an open interval around

x = 1

, and the point

(1, 2)

is lower than the points to its left and right, so it’s a relative minimum

At $x = 2$ : relative maximum of 4, since it’s higher than the nearby points, a peak

At $x = 3$ : relative minimum of 1, since it’s lower than the nearby points, a valley. It also ties x = -3 for the lowest y-value on the whole graph, so it’s an absolute minimum too

At $x = 4$ : controversial, see the endpoint discussion below

The Endpoint Controversy

The question: can an endpoint be a relative extremum if it isn’t absolute?

Consider the endpoint at

x = 4

in the graph above, where

f(4) = 2

Argument FOR it being a relative maximum: - It’s higher than all the points in its domain to the left - It’s “relatively” higher than the nearby points that actually exist

Argument AGAINST it being a relative maximum: - An open interval around

x = 4

extends to the right, where no points exist - “Relative” should mean relative to points on both sides - You can’t compare to points that don’t exist

Bottom line: mathematicians debate this, and different textbooks take different stances.

For this course: - Absolute extrema at endpoints are always valid - Relative extrema at endpoints: check with your instructor - AP exams avoid asking about this controversial case

Practice: Identifying Extrema

Extrema identification:

Absolute maximum of 3 at $x = -2$ , since it’s the highest

y

-value on the graph

Relative minimum of

-1

x = 0

, since it’s lower than the surrounding points, a valley. It also ties

x = 4

for the lowest

y

-value on the whole graph, so it’s an absolute minimum too

Relative maximum of 1 at $x = 1$ , since it’s higher than the surrounding points, a peak

x = 4

: absolute minimum of

-1

, tied with

x = 0

for the lowest

y

-value on the graph. It’s also a relative minimum, if endpoints count as relative extrema

Critical Points: Definition and Importance

Critical point: a point in the domain of

f

where:

\boxed{f'(c) = 0 \quad \text{OR} \quad f'(c) \text{ does not exist}}

Why critical points matter: they’re candidates for local extrema, meaning maxima or minima.

Important distinction: - Critical point ≠ guaranteed extremum - Critical point = possible extremum

Every relative extremum (away from the endpoints) happens at a critical point, but not every critical point is a relative extremum. You have to test critical points to see if they’re actually extrema.

Example:

f(x) = x^3

has

f'(x) = 3x^2

, so

f'(0) = 0

and

x = 0

is a critical point. But

f'(x) > 0

on both sides of

x = 0

, so

f

keeps increasing right through it. The graph levels off for an instant at the origin without making a peak or a valley, so

x = 0

is a critical point that is not an extremum.

Why Critical Points Matter: The Derivative Connection

At every relative extremum, either

f'(c) = 0

f'(c)

does not exist. The two most common shapes:

At a smooth peak or valley:

f'(c) = 0

, which is a horizontal tangent line

At a sharp corner:

f'(c)

does not exist

Observation: both types of points, where

f' = 0

and where

f'

doesn’t exist, are critical points.

Finding Critical Points: The Process

Step 1: Find $f'(x)$ . Take the derivative of the function.

Step 2: Solve $f'(x) = 0$ . Find all the

x

-values where the derivative equals zero.

Step 3: Find where $f'(x)$ does not exist. Identify the

x

-values where the derivative is undefined.

Step 4: Check the domain. This is critical. Only include

x

-values that are in the domain of the original function

f(x)

. If

f(x)

doesn’t exist at some point, that point can’t be a critical point, even if

f'(x)

is undefined there.

Example 1: Polynomial Function

Find all critical points of

f(x) = x^3 - 9x + 2

Step 1: Find the derivative

f'(x) = 3x^2 - 9

Step 2: Solve $f'(x) = 0$

3x^2 - 9 = 0

3x^2 = 9

x^2 = 3

x = \pm\sqrt{3}

Step 3: Where does $f'(x)$ not exist?

f'(x) = 3x^2 - 9

is a polynomial, so it exists for all real numbers. No issues here.

Step 4: Check the domain

Both

x = \sqrt{3}

and

x = -\sqrt{3}

are in the domain of

f

Critical points:

\boxed{x = \sqrt{3} \text{ and } x = -\sqrt{3}}

These are candidates for relative extrema.

Example 2: Rational Function with Radicals

Find all critical points of

g(x) = \frac{1}{\sqrt{4 - x^2}}

Step 1: Rewrite for easier differentiation

g(x) = (4 - x^2)^{-1/2}

Step 2: Find the derivative using the chain rule

g'(x) = -\frac{1}{2}(4 - x^2)^{-3/2} \cdot (-2x)

g'(x) = x(4 - x^2)^{-3/2}

Rewrite as a fraction:

g'(x) = \frac{x}{(4 - x^2)^{3/2}}

Step 3: Solve $g'(x) = 0$

A fraction equals zero only when the numerator equals zero, so:

x = 0

Step 4: Where does $g'(x)$ not exist?

The derivative is undefined when the denominator equals zero:

(4 - x^2)^{3/2} = 0

4 - x^2 = 0

x^2 = 4

x = \pm 2

Step 5: Check the domain of the original function

g(x) = \frac{1}{\sqrt{4 - x^2}}

For

g(x)

to exist, we need

4 - x^2 > 0

, which means

-2 < x < 2

Critical check:

x = 2

and

x = -2

make the original function undefined, so they are NOT in the domain of

g

Even though

g'(\pm 2)

doesn’t exist, those are not critical points, because the original function doesn’t exist there.

Critical point:

\boxed{x = 0}

Only

x = 0

is a critical point, since

g'(0) = 0

and

g(0)

exists.

Example 3: Trig Function

Find all critical points of

h(x) = 2\sin x - \cos(2x)

[0, 2\pi]

Step 1: Find the derivative

h'(x) = 2\cos x - (-\sin(2x)) \cdot 2

h'(x) = 2\cos x + 2\sin(2x)

Step 2: Solve $h'(x) = 0$

2\cos x + 2\sin(2x) = 0

\cos x + \sin(2x) = 0

Using the identity

\sin(2x) = 2\sin x \cos x

\cos x + 2\sin x \cos x = 0

\cos x(1 + 2\sin x) = 0

Case 1:

\cos x = 0

, so

x = \frac{\pi}{2}, \frac{3\pi}{2}

[0, 2\pi]

Case 2:

1 + 2\sin x = 0

, so

\sin x = -\frac{1}{2}

, giving

x = \frac{7\pi}{6}, \frac{11\pi}{6}

[0, 2\pi]

Step 3: Where does $h'(x)$ not exist?

h'(x)

is continuous everywhere, since it’s a combination of sine and cosine.

Critical points on $[0, 2\pi]$ :

\boxed{x = \frac{\pi}{2}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}}

Summary: Critical Points Checklist

Find $f'(x)$

Solve $f'(x) = 0$

Find where $f'(x)$ does not exist

Verify all the points are in the domain of $f(x)$

Remember: critical points are candidates, not guarantees, for extrema

Key Takeaways

Extreme Value Theorem: continuous functions on closed intervals have an absolute max and an absolute min

Absolute extrema: highest or lowest on the entire domain

Relative extrema: highest or lowest in a neighborhood

Critical points occur where $f'(x) = 0$ or $f'(x)$ does not exist

Critical points are candidates. Not all critical points are extrema

Domain matters. If $f(c)$ doesn’t exist, then $c$ can’t be a critical point

Endpoint controversy: whether endpoints can be relative extrema is debated, but absolute extrema at endpoints are always valid

Every relative extremum occurs at a critical point (absolute extrema can also be at endpoints), which is why finding critical points is essential

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