Finding Absolute Extreme Values Of F(x) = 5(2x)^x On [0.1, 1]
In calculus, finding the absolute extreme values of a function on a given interval is a fundamental problem with wide-ranging applications. These extreme values, also known as global maxima and minima, represent the highest and lowest points the function reaches within the specified interval. Determining these values is crucial in various fields, including optimization problems in engineering, economics, and physics. This article delves into the process of finding the absolute extreme values of the function f(x) = 5(2x)^x on the interval [0.1, 1]. We will explore the necessary calculus techniques, including finding critical points and evaluating the function at endpoints, to accurately identify the absolute maximum and minimum values.
Before diving into the specifics of our function, let's define what absolute extreme values are. The absolute maximum of a function f(x) on an interval [a, b] is the largest value f(x) attains on that interval. Conversely, the absolute minimum is the smallest value f(x) attains on [a, b]. These extreme values can occur at critical points within the interval or at the endpoints of the interval (a and b). A critical point is a point in the domain of the function where the derivative is either zero or undefined. Finding these critical points is a crucial step in determining absolute extreme values. Understanding the behavior of a function, particularly its increasing and decreasing intervals, is also essential in this process. The first derivative test helps us identify intervals where the function is increasing (positive derivative) or decreasing (negative derivative). This information, combined with the critical points and endpoint evaluations, allows us to pinpoint the absolute extreme values with confidence.
To determine the absolute extreme values of f(x) = 5(2x)^x on the interval [0.1, 1], our first step is to find the derivative f'(x). This will allow us to identify the critical points of the function. Let's rewrite the function using exponential properties to make differentiation easier. We can express f(x) as f(x) = 5e^(x ln(2x)). Now, we can apply the chain rule to find the derivative:
f'(x) = 5 * d/dx [e^(x ln(2x))]
Using the chain rule:
f'(x) = 5 * e^(x ln(2x)) * d/dx [x ln(2x)]
Now, we need to find the derivative of x ln(2x) using the product rule:
d/dx [x ln(2x)] = x * d/dx [ln(2x)] + ln(2x) * d/dx [x]
d/dx [ln(2x)] = (1/(2x)) * 2 = 1/x
d/dx [x] = 1
So,
d/dx [x ln(2x)] = x * (1/x) + ln(2x) * 1 = 1 + ln(2x)
Substituting this back into the expression for f'(x):
f'(x) = 5 * e^(x ln(2x)) * (1 + ln(2x))
Since e^(x ln(2x)) = (2x)^x, we can rewrite f'(x) as:
f'(x) = 5(2x)^x (1 + ln(2x))
This derivative is crucial for finding critical points and understanding the function's behavior.
Critical points are essential in determining absolute extreme values. As mentioned earlier, critical points occur where the derivative f'(x) is either equal to zero or undefined. From the previous section, we found the derivative of our function f(x) = 5(2x)^x to be:
f'(x) = 5(2x)^x (1 + ln(2x))
To find where f'(x) = 0, we set the expression equal to zero:
5(2x)^x (1 + ln(2x)) = 0
Since 5(2x)^x is always positive for x in the interval [0.1, 1], we only need to consider the second factor:
1 + ln(2x) = 0
ln(2x) = -1
To solve for x, we exponentiate both sides:
e^(ln(2x)) = e^(-1)
2x = e^(-1)
x = e^(-1) / 2
x = 1 / (2e) ≈ 0.1839
This value, approximately 0.1839, lies within our interval [0.1, 1], so it is a critical point. Now, we need to check where f'(x) is undefined. The term (2x)^x is defined for all x in [0.1, 1]. The natural logarithm ln(2x) is defined for 2x > 0, which means x > 0. Thus, f'(x) is defined for all x in our interval. Therefore, x = 1 / (2e) is the only critical point within the interval [0.1, 1].
With the critical points identified, the next step in finding the absolute extreme values is to evaluate the function f(x) = 5(2x)^x at these critical points and the endpoints of the interval. Our interval is [0.1, 1], and we found one critical point at x = 1 / (2e) ≈ 0.1839. Therefore, we need to evaluate f(x) at x = 0.1, x = 1 / (2e), and x = 1.
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Evaluating f(0.1):
f(0.1) = 5(2 * 0.1)^(0.1) = 5(0.2)^(0.1) ≈ 5(0.8513) ≈ 4.2565
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Evaluating f(1 / (2e)) ≈ f(0.1839):
f(1 / (2e)) = 5(2 * (1 / (2e)))^(1 / (2e)) = 5(1 / e)^(1 / (2e)) = 5(e(-1))(1 / (2e)) = 5e^(-1 / (2e)) ≈ 5e^(-0.1839) ≈ 5(0.8329) ≈ 4.1645
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Evaluating f(1):
f(1) = 5(2 * 1)^1 = 5(2)^1 = 10
By evaluating the function at these points, we have obtained the following values:
- f(0.1) ≈ 4.2565
- f(1 / (2e)) ≈ 4.1645
- f(1) = 10
These values will help us determine the absolute maximum and minimum of f(x) on the interval [0.1, 1].
After evaluating the function f(x) = 5(2x)^x at the critical point and endpoints within the interval [0.1, 1], we can now determine the absolute extreme values. From our calculations in the previous section, we have:
- f(0.1) ≈ 4.2565
- f(1 / (2e)) ≈ 4.1645
- f(1) = 10
Comparing these values, we can see that the smallest value is f(1 / (2e)) ≈ 4.1645, which occurs at the critical point x = 1 / (2e) ≈ 0.1839. Therefore, the absolute minimum of f(x) on the interval [0.1, 1] is approximately 4.1645.
The largest value is f(1) = 10, which occurs at the endpoint x = 1. Thus, the absolute maximum of f(x) on the interval [0.1, 1] is 10.
In summary:
- Absolute Minimum: Approximately 4.1645 at x = 1 / (2e) ≈ 0.1839
- Absolute Maximum: 10 at x = 1
In this article, we successfully determined the absolute extreme values of the function f(x) = 5(2x)^x on the interval [0.1, 1]. We began by understanding the concept of absolute extreme values and the importance of critical points and endpoints. We then calculated the derivative of the function, f'(x) = 5(2x)^x (1 + ln(2x)), and identified the critical point at x = 1 / (2e) ≈ 0.1839. By evaluating the function at this critical point and the endpoints of the interval, we found that the absolute minimum is approximately 4.1645 at x ≈ 0.1839, and the absolute maximum is 10 at x = 1. This process demonstrates a systematic approach to finding absolute extreme values, which is a crucial skill in calculus and its applications. Understanding how to find these values allows us to solve optimization problems, analyze the behavior of functions, and make informed decisions in various fields. The techniques discussed here can be applied to a wide range of functions and intervals, making them a valuable tool for anyone studying or working with calculus.
