5.5 - Using the Candidates Test to Determine Absolute (Global) Extrema

Candidates for Absolute Extrema

To find the absolute max and absolute min of a function on a closed interval, you only have to check two kinds of points:

Critical points (where $f'(x) = 0$ or $f'(x)$ doesn’t exist)

Endpoints of the interval

5.5 - Using the Candidates Test to Determine Absolute (Global) Extrema

The absolute extrema have to occur at one of these candidates. Nothing else needs to be checked.

Why this works: if

f

is continuous on

[a, b]

, the Extreme Value Theorem (5.2) guarantees the absolute max and min exist. An absolute extremum inside the interval is also a relative extremum, so it has to be at a critical point. The only other place it can be is an endpoint.

The Procedure

Find all critical points by solving $f'(x) = 0$ and finding where $f'(x)$ doesn’t exist

Identify the endpoints of the interval $[a, b]$

Evaluate $f(x)$ at every critical point and endpoint

Compare all the values. Largest is the absolute max, smallest is the absolute min.

Example 1

Find the absolute max and min of

f(x) = x^3 - 6x^2 + 9x + 2

[0, 5]

Step 1: Critical points.

f'(x) = 3x^2 - 12x + 9 = 3(x - 1)(x - 3)

3(x - 1)(x - 3) = 0 \quad \Rightarrow \quad x = 1 \text{ or } x = 3

Both are inside

[0, 5]

Step 2: Endpoints.

x = 0

and

x = 5

Step 3: Evaluate at all candidates.

$x$	$f(x)$	Calculation
$0$	$2$	$0 - 0 + 0 + 2 = 2$
$1$	$6$	$1 - 6 + 9 + 2 = 6$
$3$	$2$	$27 - 54 + 27 + 2 = 2$
$5$	$22$	$125 - 150 + 45 + 2 = 22$

Step 4: Compare.

Absolute maximum value:

22

, at

x = 5

Absolute minimum value:

2

, at

x = 0

and

x = 3

Using the Calculator Table Feature

For ugly expressions, use the table function:

Enter the function into $Y_1$

Press 2ND then TBLSET

Set “Indpnt: Ask”

Press 2ND then TABLE

Enter each candidate value and record the output

This is really handy for fractions, radicals, or big numbers.

Example 2: Using the Calculator

Find the absolute extrema of

g(x) = x^4 - 8x^2 + 5

[-1, 3]

Critical points:

g'(x) = 4x^3 - 16x = 4x(x-2)(x+2)

Critical points:

x = -2, 0, 2

. But only

x = 0

and

x = 2

are inside

[-1, 3]

Candidates:

x = -1, 0, 2, 3

$x$	$g(x)$
$-1$	$-2$
$0$	$5$
$2$	$-11$
$3$	$14$

Absolute maximum value:

14

Absolute minimum value:

-11

Finding Absolute Extrema from the Graph of $f'(x)$

When you’re only given the graph of the derivative, you have to reason about how

f(x)

behaves.

Strategy

Where

f'(x) > 0

f

is increasing. Where

f'(x) < 0

f

is decreasing.

To compare candidates, think about the area between

f'(x)

and the x-axis:

Positive area (green) is how much the function goes up

Negative area (yellow) is how much it goes down

Working Through the Graph

Interval

[-3, 4]

. Candidates: endpoints

x = -3

and

x = 4

, plus critical points

x = 1

and

x = 3

For the absolute maximum:

At $x = 1$ : the function increased a lot from $x = -3$ to $x = 1$ (big green area)

At $x = 4$ : it increased a bit from $x = 3$ to $x = 4$ (small green area), but it dropped a lot from $x = 1$ to $x = 3$ (big yellow area)

Since the increase over

[-3, 1]

outweighs the net change over

[1, 4]

, the absolute maximum is at $x = 1$ .

For the absolute minimum:

At $x = -3$ : this is the starting point, before any of the increase

At $x = 3$ : after the big drop from $x = 1$

From

x = -3

x = 3

, the function first goes up by the big green area, then down by the smaller yellow area. It ends up higher than where it started, so

f(3) > f(-3)

. So the absolute minimum is at

x = -3

Visual Area Comparison

Large positive area means a significant increase

Large negative area means a significant decrease

Compare the net change to figure out which candidate ends up higher or lower

Key Takeaways

Candidates for absolute extrema are critical points and endpoints, that’s it

You must evaluate $f(x)$ at every candidate

Absolute max is the largest value among the candidates, absolute min is the smallest

You don’t need to classify critical points as relative max/min, just evaluate them all

Show your work by listing every candidate and its function value

From a graph of $f'(x)$ , compare areas to figure out net increase and decrease

The calculator table feature speeds up evaluating messy expressions

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