Absolute maxima and absolute minima are the highest and lowest values of a function on its entire domain or on a given interval. They are also known as global maxima and global minima.
Points A, C are minima, and points B, D are maxima.
- Points A and D are global minima and global maxima.
- B and C are 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.
Extreme Value Theorem
It states that if a function f(x) is continuous on a closed interval [a, b], it is guaranteed to attain both an absolute maximum and an absolute minimum on [a, b], though it does not tell us where these values occur.
There exist points c and d in [a, b] such that f(c) ≤ f(x) ≤ f(d) for all x in [a, b], where f(c) is the absolute minimum and f(d) is the absolute maximum.
Absolute Minima and Maxima in Closed Interval
Let's say we have a function f(x) and a region D. Now to find the extreme points in any 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 Entire Domain
We need to follow some steps 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.
Local Maxima and Minima
These are the maximum and minimum value of the function relative to other points over a specific interval of the function. Local maxima and minima of any function can be similar or not similar to Absolute maxima and minima of the function.
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 Local Minimum.
- If f''(x) < 0, then the point x is a Local Maximum
Solved Examples
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.
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) = x2 - 2x + 5 in the interval [0,2].
Solution:
Given function, f(x) = x2 - 2x + 5
First step is to find the critical points by differentiating the function f(x),
f'(x) = 2x - 2
f'(x) = 0
⇒ x = 1
Thus, x = 1 is the critical point of the function
f(1) = (1)2 - 2(1) + 5 = 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) = 1/(x + 4) in the interval [0, 1].
Solution:
Given, f(x) = 1/(x + 4)
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 4: Find the absolute maximum and absolute minimum values of the function f(x) = x3 - 2x2+ 5 in the interval [-2,2].
Solution:
Given function,
f(x) = x3 - 2x2+ 5
First step is to find the critical points by differentiating the function f(x),
f'(x) = 3x2 - 4x
x(3x - 4) = 0
Thus, x = 0 and 4/3 are Critical Points of the function.
f(0) = 5
f(4/3) = (4/3)3 - 2(4/3)2 + 5
= 64/27 - 32/9 + 5
= 103/27
Checking End points of the Interval,
f(-2) = (-2)3 - 2(-2)2+ 5 =-11
f(2) = (2)3 - 2(2)2+ 5 = 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 -11 respectively.