Find the absolute extrema for the given function on the interval [1.2,3.1]
. Write your answer in the form (x,f(x))
. Round your answers to two decimal places.

f(x)=8x^2/ln(x)

To find the absolute extrema of the function \( f(x) = \frac{8x^2}{\ln(x)} \) on the interval \([1.2, 3.1]\), we need to follow these steps:

1. **Find the critical points** by setting the derivative \( f'(x) \) to zero.
2. **Evaluate the function** at the critical points and the endpoints of the interval.
3. **Compare these values** to determine the absolute extrema.

### Step 1: Find the derivative of \( f(x) \)

Given:
\[ f(x) = \frac{8x^2}{\ln(x)} \]

Using the quotient rule for differentiation:
\[ f'(x) = \frac{(8 \cdot 2x \cdot \ln(x)) - (8x^2 \cdot \frac{1}{x})}{(\ln(x))^2} \]
\[ f'(x) = \frac{16x \ln(x) - 8x}{(\ln(x))^2} \]
\[ f'(x) = \frac{8x(2 \ln(x) - 1)}{(\ln(x))^2} \]

### Step 2: Set the derivative equal to zero and solve for \( x \)

\[ \frac{8x(2 \ln(x) - 1)}{(\ln(x))^2} = 0 \]

The numerator must be zero for the fraction to be zero:
\[ 8x(2 \ln(x) - 1) = 0 \]

Since \( 8x \neq 0 \):
\[ 2 \ln(x) - 1 = 0 \]
\[ 2 \ln(x) = 1 \]
\[ \ln(x) = \frac{1}{2} \]
\[ x = e^{\frac{1}{2}} \]
\[ x = \sqrt{e} \approx 1.65 \]

### Step 3: Evaluate the function at the critical points and endpoints

Endpoints:
- \( x = 1.2 \)
- \( x = 3.1 \)

Critical point:
- \( x \approx 1.65 \)

Calculate the function values:

1. \( f(1.2) \):
\[ f(1.2) = \frac{8 \cdot (1.2)^2}{\ln(1.2)} = \frac{8 \cdot 1.44}{\ln(1.2)} \]
\[ f(1.2) \approx \frac{11.52}{0.182} \approx 63.30 \]

2. \( f(3.1) \):
\[ f(3.1) = \frac{8 \cdot (3.1)^2}{\ln(3.1)} = \frac{8 \cdot 9.61}{\ln(3.1)} \]
\[ f(3.1) \approx \frac{76.88}{1.131} \approx 68.01 \]

3. \( f(\sqrt{e}) \):
\[ f(\sqrt{e}) = \frac{8 \cdot (\sqrt{e})^2}{\ln(\sqrt{e})} = \frac{8 \cdot e}{\frac{1}{2}} \]
\[ f(\sqrt{e}) = \frac{8e}{0.5} = 16e \]
\[ 16e \approx 16 \cdot 2.718 \approx 43.49 \]

### Step 4: Determine the absolute extrema

From the values calculated:
- \( f(1.2) \approx 63.30 \)
- \( f(3.1) \approx 68.01 \)
- \( f(\sqrt{e}) \approx 43.49 \)

The absolute minimum is \( (1.65, 43.49) \).

The absolute maximum is \( (3.1, 68.01) \).

### Conclusion

The absolute extrema of \( f(x) = \frac{8x^2}{\ln(x)} \) on the interval \([1.2, 3.1]\) are:
- Minimum: \( (1.65, 43.49) \)
- Maximum: \( (3.1, 68.01) \)

Do you want details or have any questions?

1. What is the process to find critical points for any function?
2. How do you apply the quotient rule in differentiation?
3. Why is setting the derivative equal to zero important for finding extrema?
4. Can you explain why we need to check endpoints for absolute extrema?
5. How do logarithmic properties affect the derivative calculation?
6. What are the properties of the exponential function used here?
7. Why is rounding important in final answers for extrema?
8. What is the significance of evaluating the function at critical points and endpoints?

**Tip:** Always double-check your calculations for derivative and function values to ensure accuracy, especially when working with logarithms and exponentials.

Learn how to find the absolute extrema of the function f(x) = 8x^2 / ln(x) on the interval [1.2, 3.1] with a detailed step-by-step solution. This problem involves calculus concepts including derivatives and extrema. Suitable for undergraduate level students.

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