Continuous Image of a Closed Interval
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In this tutorial, we will focus on the Closed Interval Method which provides a way of finding the absolute maximum and absolute minimum values of a continuous function on a closed interval. We will first explain extreme values. Then, we will work through various examples showing how we can apply the Closed Interval Method.
What are Maximum and Minimum Values?
Some of the most significant application of differential calculus are optimization problems in which we are asked to find the best way of doing something. For example:
- What is the shape of a can that minimizes manufacturing costs?
- What is the maximum concentration of an antibiotic taken by a patient in the first 12 hours?
- or What is the maximum acceleration of a space shuttle?
These problems can be solved by finding the maximum or minimum values of a function; the maximum and minimum are also called extreme values.
There are two types of maximum/ minimum values:
- absolute/ global maximum and absolute/ global minimum values
- local maximum and local minimum values
Let c be a number in the domain D of a function f . Then f(c) is the
- absolute/ global maximum value of f on D if f(c)>= f(x) for all x in D .
- absolute/ global minimum value of f on D if f(c)<= f(x) for all x in D .
The number f(c) is a
- local maximum value of f if if f(c)>= f(x) when x is near c .
- local minimum value of f if if f(c)<= f(x) when x is near c .
Some function have extreme values, while others do not (please see below).
The Extreme Value Theorem
Before we refer to the Closed Interval Method, we need to address the Extreme Value Theorem. The Extreme Value Theorem specifies the conditions under which a function is guaranteed to have extreme values.
The Extreme Value Theorem
If f is continuous on a closed interval [a, b] , then f attains an absolute maximum value f(c) and an absolute minimum value f(d) at some numbers c and d in [a, b] .
Functions continuous on a closed interval always attain extreme values.
Visually, we can depict the Extreme Value Theorem below.
Please note that if any of the required conditions of the Extreme Value Theorem are not met, then the function f may not have these extreme values ; in other words, if f is not continuous on [a, b] or if f is continuous on (a, b) , then f need not have these extreme values (see below).
Notice that the Extreme Value Theorem tells us that a continuous function on a closed interval has an absolute maximum and an absolute minimum, but it does not specify how to find these values. This is why we need to use the Closed Interval Method.
The Closed Interval Method
This is a method we can apply to find the absolute maximum and absolute minimum of a continuous function on a closed interval; this is also also called the Closed Interval Test or the Closed Interval Theorem.
The Closed Interval Method
To find the absolute maximum and absolute minimum of a continuous function f on a closed interval [a, b] , do the following:
1. Find the values of f at the critical numbers of f in (a, b) .
2. Find the values of f at the endpoints a, b of the interval.
3. The largest of the values from Steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value.
Examples
Let's now apply the Closed Interval Method to find the extreme values.
Example 1: Applying the Closed Interval Method
Find the absolute maximum and minimum values of the function f(x)= x^3- 3x^2+ 1 where -\frac{1}{2}<= x<= 4 .
We know that f is a polynomial of degree 3 and, therefore, it is continuous on its specified domain -\frac{1}{2}<= x<= 4 (see our Continuity on an Interval tutorial for more information). Therefore, using the Extreme Value Theorem, we know that f has an absolute maximum and absolute minimum on -\frac{1}{2}<= x<= 4 .
Using Step 1 of the Closed Interval Method, we can now differentiate f(x) to get the critical numbers:
f^{\prime}(x)= 3x^2- 6x= 3x(x- 2)
The critical numbers can be found wherever f^{\prime}(x) does not exist or f^{\prime}(x)= 0 . In this case, f^{\prime}(x) is defined everywhere on -\frac{1}{2}<= x<= 4 and
f^{\prime}(x)= 0 \Longleftrightarrow 3x(x- 2)= 0 \Longleftrightarrow x= 0, x= 2
are the critical numbers.
Notice that 0 and 2 are in the interval -\frac{1}{2}<= x<= 4 and f(0)= 1 and f(2)= 2^3-3 \cdot 2^2+1= 8- 12+ 1=-3 .
Applying Step 2 of the Closed Interval Method, we can now evaluate f at the endpoints, -\frac{1}{2} and 4 . Therefore,
f(-\frac{1}{2})= (-\frac{1}{2})^3- 3 \cdot (-\frac{1}{2})^2+ 1= -\frac{1}{8} -\frac{3}{4}+ 1= \frac{1}{8}
f(4)= 4^3- 3 \cdot 4^2+ 1= 64- 48+ 1= 17
Using Step 3, we can compare the values obtained in Steps 1 and 2. We conclude that
f(4)= 17 is the absolute maximum
f(2)= -3 is the absolute minimum
Example 2: Applying the Closed Interval Method
Determine the absolute maximum and minimum values of the function f(x)= x+ \frac{1}{x} where x \in [0.2, 4] .
We know that f is a polynomial function which is continuous on [0.2, 4] (see our Continuity on an Interval tutorial for more information). Therefore, the Extreme Value Theorem states that f has an absolute maximum and absolute minimum on [0.2, 4] .
We can also apply the Closed Interval Test to find the critical numbers.
f^{\prime}(x)= 1- \frac{1}{x^2}
f^{\prime}(x) does not exist when x= 0 .
As a result, f^{\prime}(x)= 0 when
1- \frac{1}{x^2}= 0 \Longleftrightarrow x^2= 1 \Longleftrightarrow x= \pm 1
But, x= 0 \notin [0.2, 4] . Therefore, x= \pm 1 are the critical numbers.
f(-1)= -2 and f(1)= 2 .
We can now also calculate f at the endpoints, so that:
f(0.2)= 0.2+ \frac{1}{0.2}= 5.2
f(4)= 4+ \frac{1}{4}= 4.25
So, we find that
f(0.2)= 5.2 is the absolute maximum and
f(-1)= -2 is the absolute minimum
Example 3 : Applying the Closed Interval Method
Find the absolute maximum and minimum values of the function f(x)= 2\cos x+ \sin2x where x \in [0, \frac{\pi}{2}] .
This is a trigonometric function which is continuous on [0, \frac{\pi}{2}] . Therefore, we know that it has an absolute maximum and absolute minimum based on the Extreme Value Theorem.
We find
f^{\prime}(x)= -2 \sin x+ 2 \cos 2x
f^{\prime}(x) is defined everywhere.
If f^{\prime}(x)= 0 , then
-2(\sin x- \cos 2x)= 0 \Longleftrightarrow \sin x= \cos 2x \Longleftrightarrow x= \frac{\pi}{6} is the critical number.
Also,
f(\frac{\pi}{6})= 2 \cos \frac{\pi}{6}+ \sin \frac{\pi}{3}= 2 \frac{\sqrt{3}}{2}+ \frac{1}{2}= \frac{3\sqrt{3}}{2}
We also find that:
f(0)= 2 \cos 0+ \sin 2(0)= 2
f(\frac{\pi}{2})= 2 \cos \frac{\pi}{2}+ \sin 2(\frac{\pi}{2})= 0
Therefore,
f(\frac{\pi}{6})=\frac{3\sqrt{3}}{2} is the absolute maximum.
f(\frac{\pi}{2})=0 is the absolute minimum.
Practice Problems
Now it's your turn to practice the ideas explained in this tutorial. Find the absolute maximum and absolute minimum of:
- f(x)= 12+ 4x- x^2, x \in [0, 5]
- f(x)= x^{-2} \ln x, x \in [\frac{1}{2}, 4]
- Determine the absolute maximum and minimum of f(x)= \ln (x^2+x+1), x \in [-1, 1]
Practice Solutions
- f(-1)= 8 (absolute maximum) and f(2)= -19 (absolute minimum).
- f(e^{\frac{1}{2}})= \frac{1}{2e} (absolute maximum) and f(\frac{1}{2})= -4 \ln 2 (absolute minimum).
- The answer is: f(1)= \ln 3 (absolute maximum) and f(-\frac{1}{2})= \ln \frac{3}{4} (absolute minimum).
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Source: https://mathleverage.com/closed-interval-method/
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