Let's analyze the function $f(x) = \frac{\log(x^2 + 1)}{x}$ and answer each part based on the problem requirements.

Step-by-Step Analysis

1. Continuity and Limit as $x \to 0$:

   To determine the continuity of $f(x)$ at $x = 0$, we need to examine the limit: $\lim_{x \to 0} \frac{\log(x^2 + 1)}{x}$ Using L'Hôpital's Rule because it results in a $\frac{0}{0}$ indeterminate form: $\lim_{x \to 0} \frac{\log(x^2 + 1)}{x} = \lim_{x \to 0} \frac{\frac{d}{dx}(\log(x^2 + 1))}{\frac{d}{dx}(x)} = \lim_{x \to 0} \frac{\frac{2x}{x^2 + 1}}{1} = \lim_{x \to 0} \frac{2x}{x^2 + 1} = 0$ Therefore, $\lim_{x \to 0} f(x) = 0$, and since $f(x)$ is undefined at $x = 0$, it has a removable discontinuity at $x = 0$.

2. Symmetry:

   Since $f(x) = \frac{\log(x^2 + 1)}{x}$, we note that $f(-x) = -f(x)$. This implies that $f(x)$ is odd.

3. Where $f(x)$ is Positive and Negative:

   • For $f(x) > 0$: The function $f(x)$ is positive when $x > 0$, because $\log(x^2 + 1)$ is always positive and dividing by a positive $x$ results in a positive value. So, $f(x) > 0$ for $x \in (0, \infty)$.

   • For $f(x) < 0$: The function $f(x)$ is negative when $x < 0$, as $\log(x^2 + 1)$ remains positive but dividing by a negative $x$ results in a negative value. Thus, $f(x) < 0$ for $x \in (-\infty, 0)$.

Answers Based on the Image

1. Limit as $x \to 0$:

   • Answer: $0$

2. Type of Discontinuity:

   • Answer: Removable

3. Symmetry:

   • Answer: Odd

4. Intervals:

   • $f(x) > 0$: Interval $(0, \infty)$
   • $f(x) < 0$: Interval $(-\infty, 0)$

Would you like a more detailed explanation, or do you have any questions?

Related Questions:

1. How does the behavior of $f(x)$ near $x = 0$ change if we modify the function slightly?
2. Can we define $f(x)$ at $x = 0$ to make it continuous everywhere?
3. How would $f(x)$ change if we replaced $x^2 + 1$ with $x^2 - 1$ in the logarithmic expression?
4. What would the graph of $f(x) = \frac{\log(x^2 + 1)}{x}$ look like near $x = 0$?
5. How can we check if $f(x)$ has any maximum or minimum points?

Tip:

Remember that applying L'Hôpital's Rule is useful for limits resulting in indeterminate forms, especially when dealing with logarithmic functions divided by variables approaching zero.