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Lecture 19: The Extreme Value Theorem

Goal: Be able to compute the maximum and minimum of a differentiable function on a closed and bounded domain.

Definition:

Let \( f \) be a function with domain \( D \):

We say \( f \) has an absolute (or global) maximum at \( c \) if \( f(x) \le f(c) \) for all \( x \) in \( D \).
Likewise, \( f \) has an absolute minimum at \( c \) if \( f(c) \le f(x) \) for all \( x \) in \( D \).

Sometimes functions have both an absolute maximum and minimum, like the function in Figure 1:

Sine-like curve on axes with red dashed lines marking Abs. Max at top and Abs. Min at bottom — Figure 1

A hand-drawn graph of a sine-like wave on horizontal and vertical axes. The horizontal axis is marked with 0 at the origin and tick labels \( \frac{\pi}{2} \), \( \pi \), \( \frac{3\pi}{2} \), \( 2\pi \), and \( \frac{5\pi}{2} \). The curve rises from the origin to a peak between 0 and \( \pi \), falls through the axis near \( \pi \) to a deep trough near \( \frac{3\pi}{2} \), then rises again crossing near \( 2\pi \) and ends at a high point near \( \frac{5\pi}{2} \). A red horizontal dashed line runs across the top touching the first peak and the endpoint at \( \frac{5\pi}{2} \), which is circled in red and labeled Abs. Max. A red horizontal dashed line runs across the bottom touching the trough, which is circled in red and labeled Abs. Min. A red dotted vertical segment connects the Abs. Max point down to the horizontal axis.

But, they can have one and not the other.

Curve with blue-labeled relative minimum and relative maximum, and a red-labeled absolute minimum — Figure 2

A hand-drawn graph on horizontal and vertical axes showing a curve that comes down steeply from an upward arrow on the left, dips to a small local low point circled in blue, rises to a local high point circled in blue just left of the vertical axis, then falls steeply to a deep low point below the horizontal axis circled in red, and finally rises steeply to an upward arrow on the right. A blue arrow points up to the small low point with the blue label: Relative Minimum that is not an absolute min, with the word not underlined. A blue arrow points to the local high point with the blue label: Relative Maximum that is not an absolute max, with the word not underlined. The deep low point is labeled in red: Abs. Min, with short red dashes drawn on either side of it.

They can also have neither:

Increasing S-shaped curve through the origin with arrows at both ends, no maximum or minimum — Figure 3

A hand-drawn graph on horizontal and vertical axes showing a continuously increasing S-shaped (cubic-like) curve that passes through the origin. The curve descends to a downward arrow at the lower left and ascends to an upward arrow at the upper right, flattening briefly near the origin, illustrating a function with neither an absolute maximum nor an absolute minimum.

Question: Is there a way to guarantee that a function has both an absolute maximum and minimum?

Answer: Yes! ... if we make a few assumptions.

Theorem (Extreme Value Theorem)

Let \( f \) be a function defined on a closed interval \( [a,b] \)

If \( f \) is continuous on \( [a,b] \), then \( f \) has both an absolute maximum and minimum.
If \( f \) is differentiable on \( (a,b) \), then there are only 3 places the absolute maximum and minimum can occur:
1. \( x = a \)
2. \( x = b \)
3. At a critical number on \( (a,b) \)

NOTE (Why (1)?)

The main idea is the endpoints \( (a, f(a)) \) and \( (b, f(b)) \) are fixed. So, if there wasn't a maximum, the graph would skyrocket to infinity, but needs to return to \( (b, f(b)) \). That can not be done without breaking continuity.

Blue sketch of a curve from point a rising to infinity with dotted line returning to point b — Figure 4

A small hand-drawn blue sketch above a horizontal interval. The interval is drawn as a long bracketed segment with its left endpoint labeled a and its right endpoint labeled b. From the point above a, a curve rises steeply and shoots upward toward an infinity symbol drawn at the top. A dotted line then descends from the top down to a point above b, illustrating that a graph which skyrockets to infinity cannot return to the fixed endpoint without breaking continuity.

In other words, we can find the absolute max and min if we have a continuous function on a closed interval. But, if our function is differentiable, there is a procedure to find it. We will demonstrate this in the examples.

Example 1

Find the maximum and minimum of \( f(x) = x^3 - 3x^2 + 1 \) on \( \left[-\tfrac{1}{2}, 4\right] \)

Step 1 Find the critical numbers

\[ \begin{aligned} f'(x) &= 3x^2 - 6x \\ &= 3x(x-2) \overset{\text{set}}{=} 0 \end{aligned} \]\[ \Rightarrow x = 0, 2 \]

Step 2 Compute the value of \( f \) at the endpoints and at each critical number

Values of \( f(x) \) at the endpoints and critical numbers
\( x \)	\( -\tfrac{1}{2} \)	\( 0 \)	\( 2 \)	\( 4 \)
\( f(x) \)	\( \tfrac{1}{8} \)	\( 1 \)	\( -3 \)	\( 17 \)

Step 3 State Conclusion

Conclusion: \( f \) has a minimum value of \( -3 \) at \( x = 2 \) and a maximum value of \( 17 \) at \( x = 4 \)

Red sketch of a cubic curve with marked points at (4,17) and (2,-3) on an x-axis labeled -1, -1/2, 2, 4 — Figure 5

A freehand red graph of a cubic curve drawn on a horizontal x-axis with a vertical y-axis. Tick marks on the x-axis are labeled \( -1 \), \( -\tfrac{1}{2} \), \( 2 \), and \( 4 \). A dot marks the left endpoint of the curve near \( x = -\tfrac{1}{2} \), slightly above the axis. From there the curve rises to a local maximum just right of the y-axis, then falls steeply, crossing the x-axis between the y-axis and \( x = 2 \), reaching a marked low point labeled \( (2,-3) \), then rises steeply back up, crossing the axis near \( x = 4 \), ending at a marked dot labeled \( (4,17) \) at the top right.

Example 2

Find the maximum and minimum for \( f(x) = x^4 - 2x^2 + 3 \) on \( [0,2] \)

Step 1 Critical numbers in \( [0,2] \)

\[ \begin{aligned} f'(x) &= 4x^3 - 4x = 4x(x^2-1) \\ &= 4x(x-1)(x+1) \overset{\text{set}}{=} 0 \end{aligned} \]\[ \Rightarrow x = 0, 1 \]

NOTE:

\( x = -1 \) is also a critical number for \( f \). However, it is not in the interval \( [0,2] \), so it is not taken into consideration.

Step 2

Values of \( f(x) \) at the endpoints and critical numbers
\( x \)	\( 0 \)	\( 1 \)	\( 2 \)
\( f(x) \)	\( 3 \)	\( 2 \)	\( 11 \)

Conclusion: \( f \) has a minimum value of \( 2 \) at \( x = 1 \) while it has a maximum value of \( 11 \) at \( x = 2 \).

Example 3

Find the maximum and minimum of \( f(x) = \sin x + \cos x \) on \( \left[0, \frac{\pi}{3}\right] \)

Step 1

\[ \begin{gathered} f'(x) = \cos x - \sin x \overset{\text{set}}{=} 0 \\[0.5em] \cos x = \sin x \\[0.5em] \tan x = 1 \end{gathered} \]

The only solution on \( \left[0, \frac{\pi}{3}\right] \) is \( x = \frac{\pi}{4} \)

Unit circle with rays drawn at angles pi over 3 and pi over 4 in the first quadrant — Figure 6

A hand-drawn unit circle centered at the origin of a pair of coordinate axes. In the first quadrant, a blue ray is drawn from the origin labeled \( \frac{\pi}{3} \) with a blue arc marking the angle from the positive x-axis, and a red ray is drawn at a slightly smaller angle labeled \( \frac{\pi}{4} \) with a red arc marking its angle. The arc of the circle between the positive x-axis and the \( \frac{\pi}{3} \) ray is shaded in blue, indicating the interval \( \left[0, \frac{\pi}{3}\right] \). A red line also extends from the origin into the third quadrant, continuing the \( \frac{\pi}{4} \) ray through the circle.

Step 2

Values of \( f(x) \) at the endpoints and critical number
\( x \)	\( 0 \)	\( \pi/4 \)	\( \frac{\pi}{3} \)
\( f(x) \)	\( 1 \)	\( \sqrt{2} \approx 1.414 \)	\( \frac{\sqrt{3}}{2} + \frac{1}{2} \approx 1.366 \)

Conclusion: \( f \) has a minimum of \( 1 \) at \( x = 0 \) and a maximum of \( \sqrt{2} \) at \( x = \frac{\pi}{4} \)

Example 4 (HW 19, Q 8)

Find the absolute extrema of \( f(x) = 5x^2 e^x + 8 \) on \( [-4, 2] \)

Step 1

\[ \begin{aligned} f'(x) &= 10xe^x + 5x^2 e^x \\ &= 5x e^x (2 + x) \overset{\text{set}}{=} 0 \end{aligned} \] \[ \Rightarrow x = 0, -2 \]

Step 2

Values of \( f(x) \) at the endpoints and critical numbers
\( x \)	\( -4 \)	\( -2 \)	\( 0 \)	\( 2 \)
\( f(x) \)	\( \frac{80}{e^4} + 8 \approx 9.465 \)	\( \frac{20}{e^2} + 8 \approx 10.707 \)	\( 8 \)	\( 20e^2 + 8 \approx 155.78 \)

Conclusion: \( f \) has an absolute min at \( (0, 8) \) and an absolute max at \( (2, 20e^2 + 8) \).

When \( f \) Has Only One Critical Number

Fact: Regardless of the interval, if \( f \) has only one critical number \( c \). Then, if \( f \) has a relative extremum at \( c \), it is an absolute extremum.

NOTE (Why?):

Since it is a relative extremum, if it was not absolute, then to get more extreme the graph would have to "bend" meaning \( f \) would have more than one critical number.

Sketch of a curve rising to a local max then falling, with a horizontal axis marked at c — Figure 7

A small freehand sketch. A blue curve starts at the upper left, dips down to a blue dot at a local low point, then rises in red to a red dot at a local high point marked with a red X above it, and then falls steeply in red to a blue dot near the horizontal axis on the right. Below the curve a blue horizontal line serves as an axis with a single tick mark labeled c, positioned beneath the red peak region. The sketch illustrates that a graph needing to become more extreme would have to bend, producing more than one critical number.

Example 5

Consider \( f(x) = x^2 + 2x \) on \( (-2, 0) \)

\[ \begin{aligned} f'(x) &= 2x + 2 \\ &= 2(x+1) \overset{\text{set}}{=} 0 \end{aligned} \]

Ie, \( x = -1 \) is the only critical number. From a sign chart

Sign chart for \( f' \)
	\( (-2, -1) \)	\( (-1, 0) \)
Test Point	\( -\frac{3}{2} \)	\( -\frac{1}{2} \)
Sign of \( f' \)	\( - \)	\( + \)

\( f \) has a relative minimum by the 1^st Derivative Test. By the previous fact, \( f \) has an absolute minimum on \( (-2, 0) \) at \( x = -1 \).

NOTE:

The work above tells us nothing about an absolute max. In fact, it does not exist. We can see this by looking at the graph of \( f \) on \( (-2, 0) \).

Graph of f on the open interval from -2 to 0, a downward arc with open circles at both ends — Figure 8

A hand-drawn sketch showing a horizontal x-axis labeled x, with tick marks at \( -2 \), \( -1 \), and \( 0 \). Open circles sit on the axis at \( x = -2 \) and \( x = 0 \), indicating the endpoints are excluded. A smooth downward curve starts at the open circle at \( x = -2 \), dips below the axis reaching its lowest point near \( x = -1 \), and rises back up to the open circle at \( x = 0 \). Dashed vertical lines are drawn downward from the open circles at \( x = -2 \) and \( x = 0 \), with a small circle marked beside the dashed line near \( x = 0 \), illustrating that the function values approach but never attain a maximum on the open interval.

Example 6

Consider \( y = \frac{1}{4x^2 + 3} \) on \( [-1, 1) \)

\[ \begin{aligned} y' &= -\frac{1}{(4x^2+3)^2}(8x) \overset{\text{set}}{=} 0 \\ -8x &= 0 \\ x &= 0 \end{aligned} \]

So \( x = 0 \) is the only critical number. From the sign chart:

Sign chart for \( y' \)
	\( [-1, 0) \)	\( (0, 1) \)
Test Point	\( -\frac{1}{2} \)	\( \frac{1}{2} \)
Sign of \( y' \)	\( + \)	\( - \)

\( y \) has a relative (hence absolute) max at \( 0 \) by the 1^st Derivative Test. We can not say anything about a minimum without extra work. Looking at the graph, the function has an absolute minimum at \((-1, \frac{1}{7})\).

Bell-shaped curve on the interval from -1 to 1 with a maximum at x = 0 — Figure 9

A hand-drawn sketch of a bell-shaped curve above a horizontal axis. The axis is marked with tick labels at \( -1 \), \( 0 \), and \( 1 \). The curve starts at a small open circle above \( x = -1 \), rises to a rounded peak directly above \( x = 0 \), then falls to a small open circle above \( x = 1 \). A dashed vertical line runs from the peak of the curve down to the axis at \( x = 0 \), indicating the location of the absolute maximum.