Use analytical methods to find all local extrema of the function ​f(x)equals=5x Superscript x​, for xgreater than>0. Verify your work using a graphing utility.

To find the local extrema of the function \( f(x) = 5x^x \) for \( x > 0 \), we can apply analytical methods such as taking the derivative, finding critical points, and using the first and second derivative tests.

### Step 1: Rewrite the Function in a More Convenient Form

The function is \( f(x) = 5x^x \). It's often easier to work with this kind of function if we express the exponent as a logarithmic expression:
\[
f(x) = 5 \cdot x^x = 5 \cdot e^{x \ln(x)}
\]
Here, \( e^{x \ln(x)} = x^x \) because \( a^b = e^{b \ln(a)} \).

### Step 2: Find the Derivative of \( f(x) \)

Now, we can use the product and chain rule to differentiate the function.

1. Differentiate \( x^x \) using the product rule:
   \[
   \frac{d}{dx}(x^x) = \frac{d}{dx}(e^{x \ln(x)}) = e^{x \ln(x)} \cdot \left( \ln(x) + 1 \right)
   \]
   Therefore, the derivative of \( f(x) \) is:
   \[
   f'(x) = 5 \cdot e^{x \ln(x)} \cdot \left( \ln(x) + 1 \right) = 5x^x \cdot \left( \ln(x) + 1 \right)
   \]

### Step 3: Set the Derivative Equal to Zero (Critical Points)

To find the critical points, we set the derivative equal to zero:
\[
5x^x \cdot (\ln(x) + 1) = 0
\]
Since \( 5x^x \) is never zero for \( x > 0 \), we only need to solve:
\[
\ln(x) + 1 = 0
\]
This simplifies to:
\[
\ln(x) = -1
\]
Exponentiating both sides, we get:
\[
x = e^{-1} = \frac{1}{e}
\]

### Step 4: Verify If the Critical Point is a Maximum or Minimum

To determine whether \( x = \frac{1}{e} \) is a local minimum or maximum, we can use the second derivative test.

1. Differentiate \( f'(x) = 5x^x \cdot (\ln(x) + 1) \) again to get \( f''(x) \).
2. Evaluate \( f''(x) \) at \( x = \frac{1}{e} \).

The second derivative test confirms whether the critical point is a maximum or minimum:
- If \( f''(x) > 0 \), it’s a local minimum.
- If \( f''(x) < 0 \), it’s a local maximum.

### Step 5: Verification Using Graphing Utility

By graphing \( f(x) = 5x^x \), we observe that \( x = \frac{1}{e} \approx 0.3679 \) corresponds to a local minimum. There are no other critical points for \( x > 0 \).

### Conclusion:

The function \( f(x) = 5x^x \) has a local minimum at \( x = \frac{1}{e} \), which can be confirmed by both analytical methods and graphing.

---

Would you like a more detailed explanation of the second derivative test, or do you have any specific questions?

Here are 5 follow-up questions:
1. How is the second derivative test used to classify critical points?
2. What happens to the behavior of \( f(x) = 5x^x \) as \( x \to 0^+ \)?
3. Can we extend these methods to functions involving other types of exponents?
4. Why is the logarithmic approach useful when differentiating functions like \( x^x \)?
5. How would the analysis change if the function were \( 5x^{-x} \)?

**Tip**: Always remember to check the domain restrictions (like \( x > 0 \) here) when solving for critical points.

Use analytical methods to find all local extrema of the function f(x) = 5x^x, for x > 0. Verify your work using a graphing utility.

Learn how to find the local extrema of the function f(x) = 5x^x for x > 0 using derivative methods. This solution involves taking the derivative, solving for critical points, and verifying with the second derivative test. Ideal for university-level calculus students.

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