# Absolute Minima and Maxima

Absolute Maxima and Minima are the maximum and minimum values of the function defined on a fixed interval. A function in general can have high values or low values as we move along the function. The maximum value of the function in any interval is called the maxima and the minimum value of the function is called the minima. These maxima and minima if defined on the whole functions are called the Absolute Maxima and Absolute Minima of the function.

In this article, we will learn about Absolute Maxima and Mimima, How to calculate absolute maxima and minima, their examples, and others in detail.

Table of Content

## What are Absolute Maxima and Minima?

Absolute maxima and minima are the maximum and minimum values of the function on the entire given range. Absolute Maxima and Minima are also called the global maxima and minima of the function it is the maximum and the minimum value that the function can achieve in its entire domain. Suppose we are given a function f(x) = sin x, defined on the interval R then we know that, **-1 â‰¤ sin x â‰¤ 1**

Thus,

- Maximum value of f(x) is 1
- Minimum value of f(x) is -1

Thus, absolute maxima and minima of f(x) = sin x defined over R is 1 and -1.

## Critical Points and Extrema Value Theorem

Let’s say we have a function f(x), critical points are the points where the derivative of the function becomes zero. These points can either be maxima or minima. A critical point is minima or maxima is determined by the second derivative test. Since there can be more than one point where the derivative of the function is zero, more than minima or maxima is possible. The figure below shows a function that has multiple critical points.

Notice that points A, C are minimas, and points B, D are maximas. B and C are called local maxima and local minima respectively. This means that these points are maximum and minimum in their locality but not necessarily on a global level. Points A and D are called global minima and global maxima.

Let’s say we have a function f(x) which is twice differentiable. Its critical points are given by the f'(x) = 0. Second Derivative Test allows us to check whether the calculated critical point is minima or maxima.

- If f”(x) > 0, then the point x is a maxima.
- If f”(x) < 0, then the point x is a minima.

Now, this test tells us which point is a minimum or a maximum, but it still fails to give us information about the global maxima and global minima. Extrema value Theorem comes to our rescue.

**Extrema Value Theorem**

**Extrema Value Theorem**

Extrema value theorem guarantees both the maxima and minima for a function under certain conditions. This theorem does not tell us where the extreme points will exist, this theorem tells us only that extreme value will exist. The theorem states that,

If a function f(x) is continuous on a closed interval [a, b], then f(x) has both at least one maximum and minimum value on [a, b].

**Absolute Minima and Maxima in Closed Interval**

**Absolute Minima and Maxima in Closed Interval**

Now to find the extreme points in any interval, we need to follow some basic steps. Let’s say we have a function f(x) and a region D. We want to find the extreme value of the function in this interval.

### How to Find Absolute Maxima and Minima in Closed Interval?

Find the critical points of the function in the interval D,Step 1:f'(x) = 0

Find the value of the function at the extreme points of interval D.Step 2:

The largest value and smallest value found in the above two steps are the absolute maximum and absolute minimum of the function.Step 3:

**Absolute Minima and Maxima in Entire Domain**

**Absolute Minima and Maxima in Entire Domain**

Absolute minimum and maximum values of the function in the entire domain are the highest and lowest value of the function wherever it is defined. A function can have both maximum and minimum values, either one of them or neither of them. For example, a straight line extends up to infinity in both directions so it neither has a maximum value nor minimum value.

### How to Find Absolute Maxima and Minima in Entire Domian?

We need to follow some steps similar to the previous case to find out the absolute maxima and minima for the entire domain.

Find the critical points of the function wherever it is defined.Step 1:

Find the value of the function at these extreme points.Step 2:

Check for the value of the function when x tends to infinity and negative infinity. Also, check for the points of discontinuity.Step 3:

Maximum and minimum of all these values give us the absolute maximum and absolute minimum for the function in its entire domain.Step 4:

## What are Local Maxima and Minima?

Local Maxima and Local Minima are the maximum and minimum value of the function relative to other points over a specific interval of the function. They are generally calculated in the same way we calculate Absolute maxima and minima. Local maxima and minima of any function can be similar or not similar to Absolute maxima and minima of the function.

Suppose we have a function f(x) = cos x defined on [-Ï€, Ï€] then is maximum value is 1 and its minimum value is -1 this is the local maxima and minima of the function. Now the function f(x) defined on R also has the maximum and minimum value of the function to be 1 and -1 this is absolute maxima and minima of the f(x). Here, we can see that local maxima and minima of the function are similar.

**Read More,**

## Solved Examples of Absolute Maxima and Minima

Let’s see some examples on Absolute Maxima and Minima to better understand the concept of Absolute Maxima and Minima

**Example 1: Find the absolute maximum and absolute minimum values of the function f(x) = 5x + 2 in the interval [0,2]. **

**Solution:**

Given function,

- f(x) = 5x + 2
First step is to find the critical points by differentiating the function f(x),

f'(x) = 5

This equation has no roots, therefore there are no critical points which means no maxima or minima. This function is continuously increasing. Thus, the maxima and minima will occur at the end points of the interval.

- f(0) = 2
- f(2) = 12
Thus, f(0) = 2 is the minimum and f(2) = 12 is the maximum value of the function.

**Example 2: Find the absolute maximum and absolute minimum values of the function f(x) = x**^{2}** – 2x + 5 in the interval [0,2]. **

**Solution:**

Given function,

- f(x) = x
^{2}– 2x + 5First step is to find the critical points by differentiating the function f(x),

f'(x) = 2x – 2

f'(x) = 0

â‡’ 2x – 2 = 0

â‡’ x = 1

Thus, x = 1 is the critical point of the function

f(1) = (1)

^{2}– 2(1) + 5â‡’f(1) = 1 – 2 + 5

- f(1) = 4
Checking the End Points of the Interval,

- f(0) = 5
- f(2) = 5
Out of all these values,

Minimum value is at x = 1, f(1) = 4

Maximum value is at x = 0 and 2, f(0) = f(2) = 5

Thus, absolute maximum and absolute minimum values of the function are 5 at x = 0 and 2, and 4 at x= 1 respectively.

**Example 3: Find the absolute maximum and absolute minimum values of the function f(x) = x**^{3}** – 2x**^{2}**+ 5 in the interval [-2,2]. **

**Solution:**

Given function,

- f(x) = x
^{3}– 2x^{2}+ 5First step is to find the critical points by differentiating the function f(x),

f'(x) = 3x

^{2}– 4xf'(x) = x(3x – 4)

â‡’ x(3x – 4) = 0

â‡’ x = 0 and 4/3

Thus, x = 0 and 4/3 are Critical Points of the function.

f(0) = 5

â‡’f(4/3) = (4/3)

^{3}– 2(4/3)^{2}+ 5â‡’f(4/3) = 64/27 – 32/9 + 5

â‡’f(4/3) = (64 – 96)/27 + 5 = -32/27 + 5/1

â‡’f(4/3) = (-32 + 135)/27 = 103/27

Checking End points of the Interval,

f(-2) =

(-2)^{3}– 2(-2)^{2}+ 5â‡’ f(-2) = -8 -2(4) + 5

â‡’f(-2) = -16 + 5

â‡’f(-2) = -11

f(2) = (2)

^{3}– 2(2)^{2}+ 5â‡’ f(2) = 5

Out of all these values,

Minimum value is at x = -2, f(-2) = 11

Maximum value is at x = 0 and 2, f(0) = f(2) = 5

Thus, absolute maximum and absolute minimum values of the function are 5 and 4 respectively.

**Example 4: Find the absolute maximum and absolute minimum values of the function f(x) = 1/(x + 4) in the interval [0,1]. **

**Solution:**

Given,

- f(x) = 1/(x + 4)
First step is to find the critical points by differentiating the function f(x),

f'(x) = -1/(x + 4)

^{2}This equation will not be zero for any value of x in the interval. So, it is monotonically increasing or decreasing in the interval. Checking at the boundary points.

f(0) = 1/4

f(1) = 1/5

Out of all these values,

Maximum Value is at x = 0, f(0) = 1/4

Minimum Value is at x = 1, f(1) = 1/5

Thus, absolute maximum and absolute minimum values of the function are 1/4 and 1/5 respectively.

**Example 5: Find the absolute maximum and absolute minimum values of the function f(x) = 2e**^{x}** – 2 in the interval [0,1]. **

**Solution:**

Given function,

- f(x) = 2e
^{x}– 2First step is to find the critical points by differentiating the function f(x),

f'(x) = 2e

^{x}This equation will not be zero for any value of x in the interval. So, it is monotonically increasing or decreasing in the interval. Checking at the boundary points.

f(1) = 2e

^{1}– 2â‡’f(1) = 2e – 2

f(0) = 2(1) – 2

â‡’f(0) = 0

Out of all these values,

Maximum Value is at x = 1, f(1) = 2e – 2

Minimum Value is at x = 0, f(0) = 0

Thus, absolute maximum and absolute minimum values of the function are 2e – 2 and 0 respectively.

**Example 6: Find the absolute maximum and absolute minimum values of the function f(x) = x**^{2}** – x in the interval [0,1]. **

**Solution:**

Given function,

- f(x) = x
^{2}– xFirst step is to find the critical points by differentiating the function f(x),

f'(x) = 2x – 2

f'(x) = 0

â‡’ 2x – 1 = 0

â‡’ x = 1/2

This, x = 1/2 is the critical point of the function.

f(1/2) = (1/2)

^{2}– (1/2)â‡’f(1/2) = 1/4 – 1/2

â‡’f(1/2) = -1/4

Checking the end points of the interval,

f(0) = 0

f(1) = 0

Out of all these values,

Maximum Value is at x = 0, and 1, f(0) = f(1) = 0

Minimum Value is at x = 1/2, f(1/2) = -1/4

Thus, absolute maximum and absolute minimum values of the function are 0 and -1/4 respectively.

** Example 7: Find the absolute minimum and maximum value of the function f(x) =** 1/(x + 3).

**Solution:**

Since no interval is given, we need to calculate the minimum and the maximum value of the function on its domain which is R – {-3}. First let’s check for the critical points.

Given function,

f(x) = 1/(x + 3)

f'(x) = -1/(x + 3)

^{2}There is no point in the function’s domain where f'(x) = 0

So now we need to check for the values of the function when x tends to infinity.

As,

- x â‡¢ -âˆž â‡’ f(x) â‡¢ 0
- x â‡¢ âˆž â‡’ f(x) â‡¢ 0
- x â‡¢ -3 â‡’ f (x) â‡¢ âˆž
Thus, there is no maximum value and the minimum value exists when x â‡¢ âˆž or -âˆž and f(x) â‡¢ 0

## Practice Questions on Absolute Maxima and Minima

**Q1: Find the absolute extreme points of a function f(x) = 2x**^{3}** – 3x**^{2}** + 5 in the interval [-2, 2]**

**Q2: Find the absolute maxima and minima of function f(x) = 2sin x + cox in the interval [-1, 1]**

**Q3: Find the absolute extreme points of function f(x) = log x in the interval [-1, 1]**

## FAQs on Absolute Maxima and Minima

### 1. What is Absolute Maxima of the Function?

The maximum value of the function over its entire domain is called the Absolute maxima of the function. Suppose we are given a function f(x) = 2x

^{2}then its domain is R and its maximum value is âˆž at x –> âˆž. Thus, its Absolute maxima is âˆž.

### 2. What is Absolute Minima of the Function?

The minimum value of the function over its entire domain is called the Absolute minima of the function. Suppose we are given a function f(x) = 2x

^{2}then its domain is R and its minimum value is 0 at x = 0. Thus, its Absolute minima is 0.

### 3. What are the Conditions for Absolute Maxima and Minima?

The absolute maxima and minima of a function {say f(x)} occur at critical points and endpoints within a given interval.

### 4. How to Find the Absolute Maxima and Minima?

To find the absolute maxima and minima of the given function follow the following steps.

Differentiate the given function f(x) to f'(x)Step 1:

Equate the f'(x) to zero to get the critical points of the function.Step 2:

Find the value of the function f(x) at all critical points and at the end points of the given interval.Step 3:

Comapare all the values of f(x) to find the required value of Absolute Maxima and Absolute Minima.Step 4: