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Absolute Minima and Maxima

  • Last Updated : 26 May, 2021

Sometimes we come across functions that have hills and valleys. Polynomial functions usually have more than one hill and valley point. We know that these points are critical points of the function and can be classified into maxima or minima. Hill points are called maxima and the valley points are called minimas. Since there are multiple maxima and minimas, it becomes essential for us to identify the points where the function takes up the minimum value and the maximum value are called global maxima and global minima.

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. 

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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. 

  1. If f”(x) > 0, then the point x is a maxima.
  2. 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 guarantees both the maximum and minima for a function under certain conditions. This theorem does not tell us where the extreme points will exist, this theorem tells us that they 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 a 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. 

Step 1: Find the critical points of the function in the interval D, 

f'(x) = 0 

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

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

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

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

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

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



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

Let’s see some sample problems 

Sample Problems

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

Solution: 

The 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) is the minimum and f(2) is the maximum value of the function. 

Question 2: Find the absolute maximum and absolute minimum values of the function f(x) = x2 – 2x + 5 in the interval [0,2]. 

Solution: 

The first step is to find the critical points by differentiating the function f(x), 

f'(x) = 2x – 2

f'(x) = 0 

⇒ 2x – 2 = 0 

⇒ x = 1 

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

f(1) = (1)2 – 2(1) + 5

⇒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, we can conclude that, 

x = 1 is the minima and x = 2,0 is the maxima. 

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

Question 3: Find the absolute maximum and absolute minimum values of the function f(x) = x3 – 2x2+ 5 in the interval [-2,2]. 

Solution: 

The first step is to find the critical points by differentiating the function f(x), 

f'(x) = 3x2 – 4x

f'(x) = x(3x – 4)

⇒ x(3x – 4) = 0 

⇒ x = 0 and \frac{4}{3}       

This, x = 0 and \frac{4}{3} are the critical points of the function. 

f(0) = 5

⇒f(1) = x^3 - 2x^2+ 5       

⇒f(1) = (\frac{4}{3})^3 - 2(\frac{4}{3})^2+ 5

⇒f(1) = \frac{64 - 96}{27} + 5

⇒f(1) = \frac{-32}{27} + 5



⇒f(1) = \frac{-32 + 135}{27}

⇒f(1) = \frac{103}{27}

Checking the end points of the interval, 

f(-2) = (-2)3 – 2(-2)2+ 5

⇒ f(-2) = -8 -2(4) + 5

⇒f(-2) = -16 + 5

⇒f(-2) = -11 

f(2) = (2)3 – 2(2)2+ 5

⇒ f(2) = 5

Out of all these values, we can conclude that, 

x = -2 is the minima and x = 0, 2 is the maxima. 

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

Question 4: Find the absolute maximum and absolute minimum values of the function f(x) = \frac{1}{x + 4} in the interval [0,1]. 

Solution:

The first step is to find the critical points by differentiating the function f(x), 

f'(x) = \frac{-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) = \frac{1}{4}

f(1) = \frac{1}{5}

Out of all these values, we can conclude that, 



x = 1 is the minima and x = 0 is the maxima. 

Thus, absolute maximum and absolute minimum values of the function are \frac{1}{4} and \frac{1}{5} respectively. 

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

Solution: 

The first step is to find the critical points by differentiating the function f(x), 

f'(x) = 2ex 

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) = 2e1 – 2

⇒f(1) = 2e – 2

f(0) = 2(1) – 2

⇒f(0) = 0

Out of all these values, we can conclude that, 

x = 0 is the minima and x = 1 is the maxima. 

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

Question 6: Find the absolute maximum and absolute minimum values of the function f(x) = x2 – x in the interval [0,1]. 

Solution: 

The first step is to find the critical points by differentiating the function f(x), 

f'(x) = 2x – 2

f'(x) = 0 

⇒ 2x – 1 = 0 

⇒ x = \frac{1}{2}

This, x = \frac{1}{2} is the critical point of the function. 

f(\frac{1}{2}) = (\frac{1}{2})^2 - \frac{1}{2}

⇒f(\frac{1}{2}) = \frac{1}{4} - \frac{1}{2}

⇒f(\frac{1}{2}) = \frac{-1}{4}

Checking the end points of the interval, 

f(0) = 0 

f(1) = 0

Out of all these values, we can conclude that, 

x = \frac{1}{2} is the minima and x = 1,0 is the maxima. 



Thus, absolute maximum and absolute minimum values of the function are 0 and \frac{-1}{4} respectively. 

Question 7: Find the absolute maxima and the minima for the function f(x) = x4 + 2x2.

Solution: 

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

f(x) = x4 – 2x2

⇒ f'(x) = 4x3  – 4x 

f'(x) = 0 

⇒ 4x3 – 4x = 0  

⇒ 4x(x2 – 1) = 0 

The critical points are x = 0, 1, -1. 

Now since it’s a polynomial function there are no discontinuities. Let’s check for the asymptomatic values of the function. 

When x ⇢ ∞, f(x) ⇢ ∞ similarly, 

 x ⇢ -∞, f(x) ⇢ ∞

So, there is no maximum value for the function. For minimum value, let’s check on the critical points. 

x = 0,-1 and 1 

f(0) = 0 

f(1) = 3 

f(-1) = 3 

Thus, minimum value is f(0) = 0. 

Question 8: Find the absolute minimum and maximum value of the function f(x) = \frac{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. 

f(x) = \frac{1}{x + 3}

f'(x) = \frac{-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 ⇢ -∞ or x ⇢ ∞.  f(x) ⇢ 0. And as x ⇢ -3, f (x) ⇢ ∞

Thus, there is no maximum value and the minimum value exists when x ⇢ ∞ or -∞ and f(x) ⇢ 0. 




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