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To find the absolute extrema of the function 

\[
f(x) = \ln(x + 72) + \frac{1}{x}
\]

on the closed interval \([-71, -1]\), we need to:

1. **Check the values of \( f(x) \) at the endpoints**: \( x = -71 \) and \( x = -1 \).
2. **Find the critical points** within the interval by setting \( f'(x) = 0 \).
3. **Evaluate \( f(x) \) at the critical points** and compare these values with the values at the endpoints to determine the absolute maximum and minimum.

### Step 1: Evaluate \( f(x) \) at the Endpoints
- **At \( x = -71 \)**:
  \[
  f(-71) = \ln(-71 + 72) + \frac{1}{-71} = \ln(1) - \frac{1}{71} = 0 - \frac{1}{71} = -\frac{1}{71}
  \]

- **At \( x = -1 \)**:
  \[
  f(-1) = \ln(-1 + 72) + \frac{1}{-1} = \ln(71) - 1
  \]

### Step 2: Find the Derivative \( f'(x) \) and Solve \( f'(x) = 0 \)
The derivative of \( f(x) \) is:

\[
f'(x) = \frac{1}{x + 72} - \frac{1}{x^2}
\]

Setting \( f'(x) = 0 \) gives:

\[
\frac{1}{x + 72} = \frac{1}{x^2}
\]

This can be rearranged to:

\[
x^2 = x + 72
\]

or 

\[
x^2 - x - 72 = 0
\]

This is a quadratic equation that can be solved using the quadratic formula:

\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-72)}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 + 288}}{2} = \frac{1 \pm \sqrt{289}}{2} = \frac{1 \pm 17}{2}
\]

So,

\[
x = \frac{18}{2} = 9 \quad \text{or} \quad x = \frac{-16}{2} = -8
\]

Only \( x = -8 \) lies within the interval \([-71, -1]\).

### Step 3: Evaluate \( f(x) \) at \( x = -8 \)
- **At \( x = -8 \)**:
  \[
  f(-8) = \ln(-8 + 72) + \frac{1}{-8} = \ln(64) - \frac{1}{8} = \ln(64) - \frac{1}{8}
  \]
  Since \( \ln(64) = 4 \), we have:
  \[
  f(-8) = 4 - \frac{1}{8} = \frac{32 - 1}{8} = \frac{31}{8}
  \]

### Step 4: Compare Values
- \( f(-71) = -\frac{1}{71} \)
- \( f(-1) = \ln(71) - 1 \)
- \( f(-8) = \frac{31}{8} \)

Thus:
- The **absolute maximum** occurs at \( x = -8 \) with \( f(-8) = \frac{31}{8} \).
- The **absolute minimum** occurs at \( x = -71 \) with \( f(-71) = -\frac{1}{71} \).

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Would you like further clarification on any step? 

### Related Questions:
1. How do you find critical points for a function with logarithmic and rational terms?
2. Why is it necessary to check endpoint values when finding absolute extrema on closed intervals?
3. How do you interpret the result of the derivative set to zero when it includes fractions?
4. What are the implications if no critical points fall within the interval?
5. Can absolute extrema occur at points where the function is not differentiable?

### Tip:
For functions involving logarithmic terms, make sure the argument of the logarithm remains positive in the domain.

Determine where the absolute extrema of f(x) = ln(x + 72) + 1/x occur on [-71, -1]. Give exact answers.

Learn how to determine the absolute extrema of the function f(x) = ln(x + 72) + 1/x on the interval [-71, -1]. This solution involves finding critical points, evaluating endpoints, and applying the Extreme Value Theorem for closed intervals. Suitable for advanced high school calculus students.

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