Answer :
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].
