To solve for where the function $f(x) = x \log(x)$ is increasing and decreasing on the interval $(0, \infty)$, let's analyze the first derivative.

1. Compute the first derivative $f'(x)$: $f(x) = x \log(x)$ Using the product rule, where $u = x$ and $v = \log(x)$, we get: $f'(x) = u'v + uv' = 1 \cdot \log(x) + x \cdot \frac{1}{x} = \log(x) + 1$

2. Set $f'(x) = 0$ to find critical points: $\log(x) + 1 = 0$ $\log(x) = -1$ Solving for $x$, we get: $x = e^{-1} = \frac{1}{e}$

   So, $x = \frac{1}{e}$ is the critical point.

3. Determine intervals of increase and decrease:

   • For $x < \frac{1}{e}$, choose a test point (e.g., $x = 0.1$): $f'(0.1) = \log(0.1) + 1 < 0$ This implies $f(x)$ is decreasing on $\left( 0, \frac{1}{e} \right)$.
   • For $x > \frac{1}{e}$, choose a test point (e.g., $x = 1$): $f'(1) = \log(1) + 1 = 1 > 0$ This implies $f(x)$ is increasing on $\left( \frac{1}{e}, \infty \right)$.

Conclusion:

• The function $f(x) = x \log(x)$ is:
  • Decreasing on $\left( 0, \frac{1}{e} \right)$
  • Increasing on $\left( \frac{1}{e}, \infty \right)$

Would you like further details on any of these steps?

Related Questions

1. What is the behavior of $f(x) = x \log(x)$ near $x = 0$?
2. How can we verify the concavity of $f(x) = x \log(x)$ using the second derivative?
3. What happens to $f(x) = x \log(x)$ as $x \to \infty$?
4. Can we determine any points of inflection for $f(x) = x \log(x)$?
5. How would the intervals change if the function were $f(x) = x \log(x) + c$, where $c$ is a constant?

Tip

For functions involving logarithmic terms, always check that the argument of the logarithm is positive within the interval of interest.