Get the answers you need at Westonci.ca, where our expert community is always ready to help with accurate information. Join our platform to get reliable answers to your questions from a knowledgeable community of experts. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
To find the absolute extrema of the function [tex]\( g(x) = \frac{x}{\ln(x)} \)[/tex] on the interval [tex]\([2, 3]\)[/tex], follow these steps:
1. Identify the Given Function and Interval:
The function is [tex]\( g(x) = \frac{x}{\ln(x)} \)[/tex], and we are tasked with finding the absolute minimum on the interval [tex]\([2, 3]\)[/tex].
2. Evaluate the Function at the Endpoints:
- At [tex]\( x = 2 \)[/tex]:
[tex]\[ g(2) = \frac{2}{\ln(2)} \][/tex]
- At [tex]\( x = 3 \)[/tex]:
[tex]\[ g(3) = \frac{3}{\ln(3)} \][/tex]
3. Calculate the Derivative of the Function:
To find the critical points, we must calculate the derivative of [tex]\( g(x) \)[/tex] and set it to zero.
[tex]\[ g'(x) = \frac{d}{dx} \left( \frac{x}{\ln(x)} \right) \][/tex]
Using the quotient rule:
[tex]\[ g'(x) = \frac{\ln(x) \cdot 1 - x \cdot \frac{1}{x}}{(\ln(x))^2} = \frac{\ln(x) - 1}{(\ln(x))^2} \][/tex]
Setting [tex]\( g'(x) = 0 \)[/tex]:
[tex]\[ \frac{\ln(x) - 1}{(\ln(x))^2} = 0 \][/tex]
Solving [tex]\( \ln(x) - 1 = 0 \)[/tex] gives:
[tex]\[ \ln(x) = 1 \implies x = e \][/tex]
Since [tex]\( e \approx 2.72 \)[/tex], [tex]\( x = e \)[/tex] is within the interval [tex]\([2, 3]\)[/tex].
4. Evaluate the Function at the Critical Point:
- At [tex]\( x = 2.72 \)[/tex]:
[tex]\[ g(2.72) = \frac{2.72}{\ln(2.72)} \][/tex]
5. Compare Function Values:
To determine the absolute minimum, we compare the values of the function at the endpoints and the critical point:
- [tex]\( g(2) \approx 2.89 \)[/tex]
- [tex]\( g(3) \approx 2.73 \)[/tex]
- [tex]\( g(2.72) \approx 2.72 \)[/tex]
6. Determine the Minimum:
The minimum value among them is:
[tex]\[ g(2.72) \approx 2.72 \][/tex]
Therefore, the absolute minimum of the function on the interval [tex]\([2, 3]\)[/tex] is approximately [tex]\( 2.72 \)[/tex], occurring at [tex]\( x = 2.72 \)[/tex].
1. Identify the Given Function and Interval:
The function is [tex]\( g(x) = \frac{x}{\ln(x)} \)[/tex], and we are tasked with finding the absolute minimum on the interval [tex]\([2, 3]\)[/tex].
2. Evaluate the Function at the Endpoints:
- At [tex]\( x = 2 \)[/tex]:
[tex]\[ g(2) = \frac{2}{\ln(2)} \][/tex]
- At [tex]\( x = 3 \)[/tex]:
[tex]\[ g(3) = \frac{3}{\ln(3)} \][/tex]
3. Calculate the Derivative of the Function:
To find the critical points, we must calculate the derivative of [tex]\( g(x) \)[/tex] and set it to zero.
[tex]\[ g'(x) = \frac{d}{dx} \left( \frac{x}{\ln(x)} \right) \][/tex]
Using the quotient rule:
[tex]\[ g'(x) = \frac{\ln(x) \cdot 1 - x \cdot \frac{1}{x}}{(\ln(x))^2} = \frac{\ln(x) - 1}{(\ln(x))^2} \][/tex]
Setting [tex]\( g'(x) = 0 \)[/tex]:
[tex]\[ \frac{\ln(x) - 1}{(\ln(x))^2} = 0 \][/tex]
Solving [tex]\( \ln(x) - 1 = 0 \)[/tex] gives:
[tex]\[ \ln(x) = 1 \implies x = e \][/tex]
Since [tex]\( e \approx 2.72 \)[/tex], [tex]\( x = e \)[/tex] is within the interval [tex]\([2, 3]\)[/tex].
4. Evaluate the Function at the Critical Point:
- At [tex]\( x = 2.72 \)[/tex]:
[tex]\[ g(2.72) = \frac{2.72}{\ln(2.72)} \][/tex]
5. Compare Function Values:
To determine the absolute minimum, we compare the values of the function at the endpoints and the critical point:
- [tex]\( g(2) \approx 2.89 \)[/tex]
- [tex]\( g(3) \approx 2.73 \)[/tex]
- [tex]\( g(2.72) \approx 2.72 \)[/tex]
6. Determine the Minimum:
The minimum value among them is:
[tex]\[ g(2.72) \approx 2.72 \][/tex]
Therefore, the absolute minimum of the function on the interval [tex]\([2, 3]\)[/tex] is approximately [tex]\( 2.72 \)[/tex], occurring at [tex]\( x = 2.72 \)[/tex].
We hope this was helpful. Please come back whenever you need more information or answers to your queries. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.