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# Absolute Maxima and Minima

If you have ever hiked along a mountain ridge, you know all about hills and valleys. These hills and valleys are quite similar to those of maxima and minima of a function.

And, if you have ever had that hiking buddy who tells your group that you have made it to the top, only for you all to have to hike up again, you know these hills were merely local maximum heights you reached.

Where is the maximum height of your hike? Well, just like where you start and stop your hike, a function that is on a closed interval has a beginning and end point. And, just like you can view an elevation map of your hike, you can look at the graph of a function and determine the absolute maxima and minima it has!

## Absolute Maxima and Minima: Definitions

If you studied the maxima and minima article, then you already know that

• the absolute maximum (also called the global maximum) of a function is the largest output value of a function on its entire domain, and

• the absolute minimum (also called the global minimum) of a function is the smallest output value of a function on its entire domain.

But what about a more formal definition?

To formally define absolute maxima and absolute minima, consider the functions

$f(x) = x^{2} + 1$

and

$g(x) = -x^{2} -1,$

each over the interval of $$( - \infty, \infty )$$.

Consider their graphs as well:

The function $$f(x)$$ has an absolute minimum at $$(0, 1)$$ and does not have an absolute maximum. The function $$g(x)$$ has an absolute maximum at $$(0, -1)$$ and does not have an absolute minimum.

For $$f(x)$$:

• As $$x \to \pm \infty, f(x) \to \infty$$.

• This shows you that $$f(x)$$ has no absolute maximum value.

• But, since $$x^{2} \geq 1$$ for all real numbers of $$x$$ and $$x^{2} + 1 = 1$$ when $$x = 0$$, you know that $$f(x)$$ has a smallest value – an absolute minimum – of $$1$$ when $$x = 0$$.

For $$g(x)$$:

• As $$x \to \pm \infty, g(x) \to - \infty$$.

• This shows you that $$g(x)$$ has no absolute minimum value.

• But, since $$-x^{2} \leq -1$$ for all real numbers of $$x$$ and $$-x^{2} - 1 = -1$$ when $$x = 0$$, you know that $$g(x)$$ has a largest value – an absolute maximum – of $$-1$$ when $$x = 0$$.

This leads you to conclude that the formal definition of absolute maxima, absolute minima, and absolute extremum are as follows:

Let a function $$f$$ be defined over an interval $$I$$ with a value $$c$$ that is a subset of $$I$$.

1. You say that $$f$$ has an absolute maximum on $$I$$ at $$c$$ if $$f(c) \ge f(x)$$ for all $$x$$ that is a subset of $$I$$.
2. You say that $$f$$ has an absolute minimum on $$I$$ at $$c$$ if $$f(c) \le f(x)$$ for all $$x$$ that is a subset of $$I$$.
3. Finally, if $$f$$ has either an absolute maximum or an absolute minimum on $$I$$ at $$c$$, you say that $$f$$ has an absolute extremum on $$I$$ at $$c$$.

But, before moving on, there are two issues of note regarding these definitions:

1. The term absolute does not mean absolute value.
• Absolute extrema can be positive, negative, or zero.
2. If a function has an absolute extremum over $$I$$ at $$c$$, then the absolute extremum is $$f(c)$$.
• The real number $$c$$ is a point on the function's domain at which the absolute extremum occurs.

Consider the function

$f(x) = \frac{1}{x^{2} + 1}$

over the interval $$( - \infty, \infty)$$.

The function $$f(x)$$ on the interval of $$(-\infty, \infty)$$ has an absolute maximum of $$1$$ at $$x = 0$$ and no absolute minimum.

Because

$f(0) = 1 \ge \frac{1}{x^{2} + 1} = f(x)$

for all real numbers $$x$$, you say that $$f(x)$$ has an absolute maximum over the interval $$( - \infty, \infty)$$ at $$x = 0$$. The absolute maximum is $$f(0) = 1$$, and it occurs at $$x = 0$$.

## Absolute Maximum and Minimum Graphs

A function may have:

• both an absolute maximum and an absolute minimum,
• either an absolute maximum or an absolute minimum, or
• neither an absolute maximum nor an absolute minimum.

The graphs below show several possibilities regarding how many of which absolute extrema a function could have.

(a) The function $$f(x) = x^{3}$$ on the interval $$(-\infty, \infty)$$ has no absolute maximum and no absolute minimum.

(a) The function $$f(x) = x^{3}$$ on the interval $$(-\infty, \infty)$$ has no absolute maximum and no absolute minimum.

(b) The function $$f(x) = \frac{-1}{x^{2}+1}$$ on the interval $$(-\infty, \infty)$$ has an absolute minimum of $$-1$$ at $$x = 0$$ and no absolute maximum.

(b) The function $$f(x) = \frac{-1}{x^{2}+1}$$ on the interval $$(-\infty, \infty)$$ has an absolute minimum of $$-1$$ at $$x = 0$$ and no absolute maximum.

(c) The function $$f(x) = cos(x)$$ on the interval $$(-\infty, \infty)$$ has an absolute maximum of $$1$$ at $$x = 0, \pm 2 \pi, \pm 4 \pi, \dots$$ and an absolute minimum of $$-1$$ at $$x = \pm \pi, \pm 3 \pi, \dots$$.

(c) The function $$f(x) = cos(x)$$ on the interval $$(-\infty, \infty)$$ has an absolute maximum of $$1$$ at $$x = 0, \pm 2 \pi, \pm 4 \pi, \dots$$ and an absolute minimum of $$-1$$ at $$x = \pm \pi, \pm 3 \pi, \dots$$

(d) The function $f(x) = \begin{cases} 2-x^{2} & 0 \le x < 2 \\ x-3 & 2 \le x \le 4 \end{cases}$ has an absolute maximum of $$2$$ at $$x = 0$$ and no absolute minimum.

(d) The function $$f(x) = \begin{cases} 2-x^{2} & 0 \leq x < 2 \\ x-3 & 2 \leq x \leq 4 \end{cases}$$ has an absolute maximum of $$2$$ at $$x = 0$$ and no absolute minimum.

(e) The function $$f(x) = (x-2)^{2}$$ on the interval $$[1, 4]$$ has an absolute maximum of $$4$$ at $$x = 4$$ and an absolute minimum of $$0$$ at $$x = 2$$.

(e) The function $$f(x) = (x-2)^{2}$$ on the interval $$[1, 4]$$ has an absolute maximum of $$4$$ at $$x = 4$$ and an absolute minimum of $$0$$ at $$x = 2$$.

(f) The function $$f(x) = \frac{x}{2-x}$$ on the interval $$[0, 2)$$ has an absolute minimum of $$0$$ at $$x = 0$$ and no absolute maximum.

(f) The function $$f(x) = \frac{x}{2-x}$$ on the interval $$[0, 2)$$ has an absolute minimum of $$0$$ at $$x = 0$$ and no absolute maximum.

The first three graphs, graphs (a), (b), and (c), show how a function with a domain of $$(-\infty, \infty)$$ can have either:

• no absolute extrema,

• one absolute extrema, or

• both an absolute maximum and an absolute minimum.

The second three graphs, graphs (d), (e), and (f), show how a function on a bounded interval can have either:

• a single absolute extremum, or

• both an absolute maximum and an absolute minimum.

A function can NOT have more than one absolute maximum or more than one absolute minimum! However, as in graph (c) above, the absolute maximum and absolute minimum can occur at more than one value of $$x$$.

### Extreme Value Theorem

The Extreme Value Theorem states that a function that is continuous over a closed and bounded interval has both an absolute maximum and an absolute minimum. The formal statement of the Extreme Value Theorem is below.

If a function $$f$$ is continuous over a closed and bounded interval $$[a, b]$$, then

• there is a point in the interval $$[a, b]$$ where $$f$$ has an absolute maximum and
• there is a point in the interval $$[a, b]$$ where $$f$$ has an absolute minimum.

For the extreme value theorem to apply, the function must be continuous over a closed and bounded interval. If, for example, the interval is open, or the function has even one point of discontinuity, it is possible that the function could not have an absolute maximum or an absolute minimum.

Reconsider the functions shown in the second three graphs above – graphs (d), (e), and (f).

All three of these functions are defined over bounded, but not necessarily closed, intervals.

• Because of this, the function in graph (e) is the only one that has both an absolute maximum and an absolute minimum over its domain.

The extreme value theorem cannot be applied to the functions in graphs (d) and (f) because neither of these functions is continuous over a closed and bounded interval.

• Considering graph (d):
• Although the function is defined over the closed and bounded interval of $$[0, 4]$$, it has a discontinuity at $$x = 2$$.
• Therefore, the function has an absolute maximum over the closed and bounded interval of $$[0, 4]$$, but it does not have an absolute minimum.
• Considering graph (f):
• The function is continuous over the half-open interval of $$[0, 2)$$, but it is not defined at $$x = 2$$.
• Therefore, it is not continuous over a closed and bounded interval.
• The function has an absolute minimum over the interval of $$[0, 2)$$, but it does not have an absolute maximum over the interval of $$[0, 2)$$.
• The graphs (d) and (f) illustrate why a function over a bounded interval may fail to have an absolute maximum and/or an absolute minimum.

## Absolute Maxima and Minima on Closed Bounded Regions

The extreme value theorem says that a continuous function over a closed and bounded interval must have an absolute maximum and an absolute minimum.

• As shown in the graphs above, one or both of these absolute extrema could occur at an endpoint of the function.

• If, however, an absolute extremum does not occur at an endpoint of the function, it must occur at an interior point.

• This means that the absolute extremum is also a local extremum.

• Therefore, by Fermat's Theorem, the point $$c$$ at which the local extremum occurs must be a critical point.

Theorem – The Location of Absolute Extrema

Let a function $$f$$ be continuous over a closed and bounded interval $$I$$.

• The absolute maximum of $$f$$ over $$I$$ must occur either at an endpoint of $$I$$ or at a critical point of $$f$$ in $$I$$.
• The absolute minimum of $$f$$ over $$I$$ must occur either at an endpoint of $$I$$ or at a critical point of $$f$$ in $$I$$.

With this in mind, let's develop the strategy for finding the absolute extrema of a function.

### Strategy: Locating Absolute Maxima and Minima Over a Closed Interval

To locate the absolute extrema of a function, the function must be continuous and defined over a closed and bounded interval $$[a, b]$$.

1. Solve the function at its endpoints, i.e., where $$x = a$$ and $$x = b$$.

2. Find all the critical points of the function that are on the open interval $$(a, b)$$ and solve the function at each critical point.

1. Take the first derivative of the given function.

2. Set $$f'(x) = 0$$ and solve for $$x$$ to find all critical points.

3. Take the second derivative of the given function.

4. Plug in the critical points from step $$2$$ into the second derivative.

• If $$f''(c) < 0$$, then the critical point of $$f(x)$$ is a maximum.

• If $$f''(c) > 0$$, then the critical point of $$f(x)$$ is a minimum.

3. Compare all the values from steps $$1$$ and $$2$$.

• The largest of the values is the absolute maximum of the function.

• The smallest of the values is the absolute minimum of the function.

• All other values are relative/local extrema of the function.

Let's review these steps in an example.

Find the absolute maximum and the absolute minimum of the function

$f(x) = x^{2} + 2$

over the interval $$[-2, 3]$$.

Solution:

So, in this case, $$f(x)$$ is continuous and defined over the closed and bounded interval of $$[-2, 3]$$.

• This means that $$a = -2$$ and $$b = 3$$.
1. Solve the function at its endpoints, i.e., where $$x = a$$ and $$x = b$$.

• For $$x = a = -2$$:$f(a) = f(-2) = (-2)^{2} + 2 = 6$

• For $$x = b = 3$$:$f(b) = f(3) = (3)^{2} + 2 = 11$

2. Find all the critical points of the function that are on the open interval $$(a, b)$$ and solve the function at each critical point.

1. Take the first derivative of the given function.$f'(x) = 2x$

2. Set $$f'(x) = 0$$ and solve for $$x$$ to find all critical points.\begin{align}f'(x) = 0 &= 2x \\0 &= 2x \\x &= 0\end{align}

3. Take the second derivative of the given function.$f''(x) = 2$

4. Plug in the critical points from step $$2$$ into the second derivative.$f''(0) = 2$

• If $$f''(c) < 0$$, then the critical point of $$f(x)$$ is a maximum.

• If $$f''(c) > 0$$, then the critical point of $$f(x)$$ is a minimum.

3. Compare all the values from steps $$1$$ and $$2$$.

• The largest of the values is the absolute maximum of the function.

• The smallest of the values is the absolute minimum of the function.

• All other values are relative/local extrema of the function.

 $$x$$ $$f(x)$$ Conclusion $$-2$$ $$6$$ relative max $$0$$ $$2$$ absolute min $$3$$ $$11$$ absolute max

You can graph the function on its closed and bounded interval to validate your conclusions from the table above:

The function $$f(x)$$ has a relative maximum at $$(-2, 6)$$, an absolute minimum at $$(0, 2)$$, and an absolute maximum at $$(3, 11)$$.

Therefore,

• The absolute maximum of $$f(x)$$ is $$11$$ and it occurs where $$x = 3$$.
• The absolute minimum of $$f(x)$$ is $$2$$ and it occurs where $$x = 0$$.

## Absolute Maximum and Minimum of a Function

Most functions you deal with in calculus do not have an absolute maximum or an absolute minimum value over their entire domain.

However, some functions do have an absolute extremum over their entire domain. For example, the function,

$f(x) = xe^{3x}.$

If you take the derivative of this function, you get,

$f'(x) = e^{3x}(1+3x).$The only critical point of this function is where $$x = -\frac{1}{3}$$.

The function $$f(x) = xe^{3x}$$, over its entire domain, has an absolute minimum at $$x = -\frac{1}{3}$$, but it has no absolute maximum.

Looking at the graph of this function, you can see that, over its entire domain, it definitely has an absolute minimum at $$x = -\frac{1}{3}$$, but it has no absolute maximum.

### Strategy: Locating Absolute Maxima and Minima on a Function's Entire Domain

To locate absolute maxima and minima over a function's entire domain, you follow the same process as you would to find local maxima and minima, as there are no endpoints. The steps are summarized below.

1. Take the first derivative of the given function.

2. Set $$f'(x) = 0$$ and solve for $$x$$ to find all critical points.

3. Take the second derivative of the given function.

4. Plug in the critical points from step $$2$$ into the second derivative.

• If $$f''(c) < 0$$, then the critical point of $$f(x)$$ is a maximum.

• If $$f''(c) > 0$$, then the critical point of $$f(x)$$ is a minimum.

Let's review these steps with an example.

Find any absolute extrema of the function

$f(x) = xe^{x}$.

Solution:

1. Take the first derivative of the given function.$f'(x) = e^{x}(1 + x)$

2. Set $$f'(x) = 0$$ and solve for $$x$$ to find all critical points.\begin{align}f'(x) = 0 &= e^{x}(1 + x) \\0 &= 1 + x \\x &= -1\end{align}

3. Take the second derivative of the given function.$f''(x) = e^{x}(x + 2)$

4. Plug in the critical points from step $$2$$ into the second derivative.\begin{align}f''(-1) &= e^{-1}(-1 + 2) \\&= e^{-1}(1) \\&= e^{-1} = 0.37\end{align}

• If $$f''(c) < 0$$, then the critical point of $$f(x)$$ is a maximum.

• If $$f''(c) > 0$$, then the critical point of $$f(x)$$ is a minimum.

You can graph the function to validate your conclusions:

The function has an absolute minimum at $$(-1, -0.37)$$ and no other relative or absolute extrema.

Therefore, the only absolute extremum of the function is an absolute minimum of $$-0.37$$ that occurs where $$x = -1$$.

## Absolute Maxima and Minima: Examples

Find the absolute maximum and the absolute minimum of the function

$f(x) = -2x^{2} + 3x - 2$

over the interval $$[-1, 3]$$.

State where the absolute extremum occur.

Solution:

1. Solve the function at its endpoints, i.e., where $$x = -1$$ and $$x = 3$$.

• For $$x = -1$$:$f(-1) = -2(-1)^{2} + 3(-1) - 2 = -7$

• For $$x = 3$$:$f(3) = -2(3)^{2} + 3(3) - 2 = -11$

2. Find all the critical points of the function that are on the open interval $$(1, 3)$$ and solve the function at each critical point.

1. Take the first derivative of the given function.$f'(x) = -4x + 3$

2. Set $$f'(x) = 0$$ and solve for $$x$$ to find all critical points.\begin{align}f'(x) = 0 &= -4x + 3 \\4x &= 3 \\x &= \frac{3}{4}\end{align}

3. Take the second derivative of the given function.$f''(x) = -4$

4. Plug in the critical points from step $$2$$ into the second derivative.$f''(\frac{3}{4}) = -4$

• Since $$f''(\frac{3}{4}) < 0$$, then the critical point of $$f(x)$$ is a maximum.

3. Compare all the values from steps $$1$$ and $$2$$.

 $$x$$ $$f(x)$$ Conclusion $$-1$$ $$-7$$ $$\frac{3}{4}$$ $$\frac{7}{8}$$ absolute max $$3$$ $$-11$$ absolute min

The function $$f(x)$$ has an absolute maximum at $$x = \frac{3}{4}$$ and an absolute minimum at $$x = 3$$.

Therefore,

• The absolute maximum of $$f(x)$$ is $$\frac{7}{8}$$ and it occurs where $$x = \frac{3}{4}$$.
• The absolute minimum of $$f(x)$$ is $$-11$$ and it occurs where $$x = 3$$.

## Absolute Maxima and Minima – Key takeaways

• The absolute maximum (also called the global maximum) of a function is the largest output value of a function on its entire domain.
• The absolute minimum (also called the global minimum) of a function is the smallest output value of a function on its entire domain.
• A function may have:
• both an absolute maximum and an absolute minimum,
• either an absolute maximum or an absolute minimum, or
• neither an absolute maximum nor an absolute minimum.
• The Extreme Value Theorem states that a continuous function over a closed and bounded interval must have an absolute maximum and an absolute minimum.
• These absolute extrema must occur at either a critical point or an endpoint of the function.

An absolute maximum of a function is the largest output value of a function on its entire domain.

An absolute minimum of a function is the smallest output value of a function on its entire domain.

To find the absolute extrema of a function, it must be continuous and defined over a closed interval [a, b].

1. Solve the function at its endpoints, i.e., where x = a and x = b.
2. Find all the critical points of the function that are on the open interval (a, b) and solve the function at each critical point.
3. Compare all the values from steps 1 and 2.
• The largest of the values is the absolute maximum of the function.
• The smallest of the values is the absolute minimum of the function.
• All other values are relative/local extrema of the function.

The difference between local maxima and absolute maxima is the same as the difference between local minima and absolute minima.

Local maxima and maxima are the maximum and minimum values of a function in a specific region (or over a certain interval).

Absolute maxima and minima are the maximum and minimum values of the function over its entire domain.

There is no formula for absolute maxima and minima. However, there is a problem-solving strategy to find them.

## Final Absolute Maxima and Minima Quiz

Question

What is the absolute maximum of a function?

The absolute maximum (also called the global maximum) of a function is the largest output value of a function on its entire domain.

Show question

Question

What is the absolute minimum of a function?

The absolute minimum (also called the global minimum) of a function is the smallest output value of a function on its entire domain.

Show question

Question

The formal definition of absolute maxima is

Let a function $$f$$ be defined over an interval $$I$$ with a value $$c$$ that is a subset of $$I$$.

You say that $$f$$ has an absolute maximum on $$I$$ at $$c$$ if $$f(c) \geq f(x)$$ for all $$x$$ that is a subset of $$I$$.

Show question

Question

The formal definition of absolute minima is

Let a function $$f$$ be defined over an interval $$I$$ with a value $$c$$ that is a subset of $$I$$.

You say that $$f$$ has an absolute minimum on $$I$$ at $$c$$ if $$f(c) \leq f(x)$$ for all $$x$$ that is a subset of $$I$$.

Show question

Question

The formal definition of absolute extremum is

Let a function $$f$$ be defined over an interval $$I$$ with a value $$c$$ that is a subset of $$I$$.

If $$f$$ has either an absolute maximum or an absolute minimum on $$I$$ at $$c$$, you say that $$f$$ has an absolute extremum on $$I$$ at $$c$$.

Show question

Question

Absolute extrema can be

positive

Show question

Question

If a function has an absolute extremum over $$I$$ at $$c$$, then

the absolute extremum is $$f(c)$$.

Show question

Question

A function may have:

both an absolute maximum and an absolute minimum

Show question

Question

The absolute maximum and absolute minimum can occur at more than one value of $$x$$.

True

Show question

Question

The extreme value theorem states that

If a function $$f$$ is continuous over a closed and bounded interval $$[a, b]$$, then

• there is a point in the interval $$[a, b]$$ where $$f$$ has an absolute maximum and
• there is a point in the interval $$[a, b]$$ where $$f$$ has an absolute minimum.

Show question

Question

For the extreme value theorem to apply, the function must be

continuous over a closed and bounded interval

Show question

Question

What is the theorem for locating absolute extrema?

Let a function $$f$$ be continuous over a closed and bounded interval $$I$$.

• The absolute maximum of $$f$$ over $$I$$ must occur either at an endpoint of $$I$$ or at a critical point of $$f$$ in $$I$$.
• The absolute minimum of $$f$$ over $$I$$ must occur either at an endpoint of $$I$$ or at a critical point of $$f$$ in $$I$$.

Show question

Question

What are the steps to locate absolute maxima and absolute minima over a closed and bounded interval?

1. Solve the function at its endpoints, i.e., where $$x = a$$ and $$x = b$$.

2. Find all the critical points of the function that are on the open interval $$(a, b)$$ and solve the function at each critical point.

1. Take the first derivative of the given function.

2. Set $$f'(x) = 0$$ and solve for $$x$$ to find all critical points.

3. Take the second derivative of the given function.

4. Plug in the critical points from step $$2$$ into the second derivative.

• If $$f''(c) < 0$$, then the critical point of $$f(x)$$ is a maximum.

• If $$f''(c) > 0$$, then the critical point of $$f(x)$$ is a minimum.

3. Compare all the values from steps $$1$$ and $$2$$.

• The largest of the values is the absolute maximum of the function.

• The smallest of the values is the absolute minimum of the function.

• All other values are relative/local extrema of the function.

Show question

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