Let's solve the problem step-by-step!

Given: $f(x) = \frac{\ln(x)}{x^5}.$

Step 1: Find $f'(x)$ to determine the critical number(s).

We need to apply the quotient rule: $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2},$ where $u = \ln(x)$ and $v = x^5$.

• $u' = \frac{1}{x}$ (the derivative of $\ln(x)$).
• $v' = 5x^4$ (the derivative of $x^5$).

Now apply the quotient rule: $f'(x) = \frac{\frac{1}{x} \cdot x^5 - \ln(x) \cdot 5x^4}{(x^5)^2}.$ Simplify the numerator: $\frac{x^5}{x} = x^4, \quad \text{so} \quad x^4 - 5x^4 \ln(x) = x^4(1 - 5 \ln(x)).$ Thus, $f'(x) = \frac{x^4(1 - 5 \ln(x))}{x^{10}} = \frac{1 - 5 \ln(x)}{x^6}.$

Step 2: Set $f'(x) = 0$ to find critical points.

$\frac{1 - 5 \ln(x)}{x^6} = 0 \quad \Rightarrow \quad 1 - 5 \ln(x) = 0.$ Solve for $x$: $5 \ln(x) = 1 \quad \Rightarrow \quad \ln(x) = \frac{1}{5}.$ $x = e^{1/5}.$ So the critical number is: $A = e^{1/5}.$

Step 3: Find $f''(x)$ and evaluate it at $A = e^{1/5}$.

We need to take the derivative of $f'(x) = \frac{1 - 5 \ln(x)}{x^6}$.

Using the product rule and chain rule, the second derivative involves some computation, but we are only interested in the sign of $f''(e^{1/5})$. After some detailed differentiation, we get:

$f''(x) = \frac{-30 + 25 \ln(x)}{x^7}.$ Now evaluate at $A = e^{1/5}$: $\ln(e^{1/5}) = \frac{1}{5}, \quad \text{so} \quad f''(e^{1/5}) = \frac{-30 + 25 \cdot \frac{1}{5}}{(e^{1/5})^7}.$ Simplify: $f''(e^{1/5}) = \frac{-30 + 5}{e^{7/5}} = \frac{-25}{e^{7/5}}.$

Since $f''(e^{1/5}) < 0$, the function has a local maximum at $A = e^{1/5}$.

Final Answer:

• Critical number: $A = e^{1/5}$.
• $f''(A) = \frac{-25}{e^{7/5}}$.
• Conclusion: $f(x)$ has a local maximum at $A = e^{1/5}$.

Would you like more details about any part of this process?

Here are 5 related questions for further exploration:

1. How do we simplify expressions involving natural logarithms?
2. What other methods can we use to confirm local maxima or minima?
3. How does the quotient rule work for more complex derivatives?
4. What is the behavior of $f(x)$ as $x \to \infty$ or $x \to 0^+$?
5. Why is the second derivative test useful for classifying critical points?

Tip: The second derivative test can fail if $f''(A) = 0$, so checking the first derivative's sign change around critical points is a good backup plan!