Sure, let's walk through how you can find the critical points and the absolute extrema of the function [tex]\( f(x) = x \ln(x) - 2 \)[/tex] on the interval [tex]\([0.2, 5]\)[/tex].
### Step 1: Determine the Critical Points
1. Find the derivative of the function:
The first step is to find the first derivative of the function [tex]\( f(x) \)[/tex]. Using the product rule and the fact that the derivative of [tex]\(\ln(x)\)[/tex] is [tex]\(\frac{1}{x}\)[/tex], we get:
[tex]\[
f'(x) = \frac{d}{dx} (x \ln(x) - 2) = \ln(x) + 1
\][/tex]
2. Set the derivative equal to zero to find critical points:
We solve the equation [tex]\( f'(x) = 0 \)[/tex]:
[tex]\[
\ln(x) + 1 = 0
\][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[
\ln(x) = -1 \implies x = e^{-1} \implies x = \frac{1}{e} \approx 0.367879441171442
\][/tex]
3. Check if the critical point is in the interval:
Here, [tex]\( x = 0.367879441171442 \)[/tex] is within the interval [tex]\([0.2, 5]\)[/tex].
### Step 2: Evaluate the Function at Critical Points and Endpoints
1. Evaluate [tex]\( f(x) \)[/tex] at the critical point:
[tex]\[
f\left( \frac{1}{e} \right) = \frac{1}{e} \ln\left( \frac{1}{e} \right) - 2 = \frac{1}{e} (-1) - 2 = -\frac{1}{e} - 2 \approx -2.36787944117144
\][/tex]
2. Evaluate [tex]\( f(x) \)[/tex] at the endpoints of the interval:
Evaluate at [tex]\( x = 0.2 \)[/tex]:
[tex]\[
f(0.2) = 0.2 \ln(0.2) - 2 \approx -2.32188758248682
\][/tex]
Evaluate at [tex]\( x = 5 \)[/tex]:
[tex]\[
f(5) = 5 \ln(5) - 2 \approx 6.04718956217050
\][/tex]
### Step 3: Identify the Absolute Extrema
Using the evaluations, compile a list of the function values at the critical points and the endpoints:
- [tex]\( f(0.367879441171442) \approx -2.36787944117144 \)[/tex]
- [tex]\( f(0.2) \approx -2.32188758248682 \)[/tex]
- [tex]\( f(5) \approx 6.04718956217050 \)[/tex]
### Step 4: Determine the Absolute Minimum and Maximum
- The absolute minimum value occurs at [tex]\( x = 0.367879441171442 \)[/tex] with [tex]\( f(x) \approx -2.36787944117144 \)[/tex].
- The absolute maximum value occurs at [tex]\( x = 5 \)[/tex] with [tex]\( f(x) \approx 6.04718956217050 \)[/tex].
### Summary
[tex]\( f(x) = x \ln(x) - 2 \)[/tex] has:
- An absolute minimum of approximately [tex]\(-2.36787944117144\)[/tex] at [tex]\( x \approx 0.367879441171442 \)[/tex].
- An absolute maximum of approximately [tex]\(6.04718956217050\)[/tex] at [tex]\( x = 5 \)[/tex].
