The problem involves analyzing the function $f(x) = \ln|x^2 - 1|$. We need to find the domain, intercepts, limits at accumulation points not in the domain, limits at infinity, first and second derivatives, critical numbers, intervals of increase and decrease, concavity, and finally sketch the graph.

1. Problem Description

The problem involves analyzing the function $f(x) = \ln|x^2 - 1|$. We need to find the domain, intercepts, limits at accumulation points not in the domain, limits at infinity, first and second derivatives, critical numbers, intervals of increase and decrease, concavity, and finally sketch the graph.

2. Solution Steps

1. Domain:

The domain of $f(x) = \ln|x^2 - 1|$ is all $x$ such that $|x^2 - 1| > 0$, which means $x^2 - 1 \neq 0$. Therefore, $x^2 \neq 1$, so $x \neq \pm 1$. Thus, the domain is $D_f = (-\infty, -1) \cup (-1, 1) \cup (1, \infty)$.

2. Intercepts:

x-intercepts: $f(x) = 0 \implies \ln|x^2 - 1| = 0 \implies |x^2 - 1| = e^0 = 1 \implies x^2 - 1 = \pm 1$.

If $x^2 - 1 = 1$, then $x^2 = 2$, so $x = \pm\sqrt{2}$.

If $x^2 - 1 = -1$, then $x^2 = 0$, so $x = 0$.

The x-intercepts are $x = -\sqrt{2}, 0, \sqrt{2}$.

y-intercept: $f(0) = \ln|0^2 - 1| = \ln|-1| = \ln(1) = 0$.

The y-intercept is $y = 0$.

3. Limits at accumulation points not in the domain:

The accumulation points of $D_f$ that are not in $D_f$ are $x = \pm 1$.

We need to find $\lim_{x \to 1} \ln|x^2 - 1|$ and $\lim_{x \to -1} \ln|x^2 - 1|$.

As $x \to 1$ or $x \to -1$, $|x^2 - 1| \to 0$, so $\ln|x^2 - 1| \to -\infty$.

Therefore, $\lim_{x \to 1} \ln|x^2 - 1| = -\infty$ and $\lim_{x \to -1} \ln|x^2 - 1| = -\infty$.

Vertical asymptotes are $x = -1$ and $x = 1$.

4. Limits at infinity:

$\lim_{x \to \pm \infty} \ln|x^2 - 1| = \lim_{x \to \pm \infty} \ln(x^2 - 1) = \infty$.

5. Derivatives:

We can rewrite $f(x)$ as:

$f(x) = \begin{cases} \ln(x^2 - 1), & x < -1 \text{ or } x > 1 \\ \ln(1 - x^2), & -1 < x < 1 \end{cases}$

$f'(x) = \begin{cases} \frac{2x}{x^2 - 1}, & x < -1 \text{ or } x > 1 \\ \frac{-2x}{1 - x^2} = \frac{2x}{x^2 - 1}, & -1 < x < 1 \end{cases}$

So, $f'(x) = \frac{2x}{x^2 - 1}$ for $x \in (-\infty, -1) \cup (-1, 1) \cup (1, \infty)$.

$f''(x) = \frac{2(x^2 - 1) - 2x(2x)}{(x^2 - 1)^2} = \frac{2x^2 - 2 - 4x^2}{(x^2 - 1)^2} = \frac{-2x^2 - 2}{(x^2 - 1)^2} = \frac{-2(x^2 + 1)}{(x^2 - 1)^2}$.

6. Critical numbers:

$f'(x) = 0 \implies \frac{2x}{x^2 - 1} = 0 \implies 2x = 0 \implies x = 0$.

Also, $f'(x)$ is undefined at $x = \pm 1$, but these are not in the domain.

$x = 0$ is in the domain, so it is a critical number.

7. Intervals of increase and decrease:

We analyze the sign of $f'(x) = \frac{2x}{x^2 - 1}$.

- $(-\infty, -1)$: $x < -1 \implies 2x < 0$ and $x^2 - 1 > 0$, so $f'(x) < 0$. Decreasing.

- $(-1, 0)$: $-1 < x < 0 \implies 2x < 0$ and $x^2 - 1 < 0$, so $f'(x) > 0$. Increasing.

- $(0, 1)$: $0 < x < 1 \implies 2x > 0$ and $x^2 - 1 < 0$, so $f'(x) < 0$. Decreasing.

- $(1, \infty)$: $x > 1 \implies 2x > 0$ and $x^2 - 1 > 0$, so $f'(x) > 0$. Increasing.

8. Concavity:

$f''(x) = \frac{-2(x^2 + 1)}{(x^2 - 1)^2}$.

Since $x^2 + 1 > 0$ and $(x^2 - 1)^2 > 0$ for all $x$ in the domain, $f''(x) < 0$ for all $x$ in the domain. Thus, $f(x)$ is concave down on its entire domain.

9. Sketch the graph: The graph has vertical asymptotes at $x = -1$ and $x = 1$. The x-intercepts are at $x = -\sqrt{2}, 0, \sqrt{2}$ and the y-intercept is at $y=0$. The function is concave down everywhere.

3. Final Answer

1. Domain: $(-\infty, -1) \cup (-1, 1) \cup (1, \infty)$

2. x-intercepts: $x = -\sqrt{2}, 0, \sqrt{2}$; y-intercept: $y = 0$

3. $\lim_{x \to 1} f(x) = -\infty$, $\lim_{x \to -1} f(x) = -\infty$; Vertical asymptotes: $x = -1, x = 1$

4. $\lim_{x \to \pm \infty} f(x) = \infty$

5. $f'(x) = \frac{2x}{x^2 - 1}$, $f''(x) = \frac{-2(x^2 + 1)}{(x^2 - 1)^2}$

6. Critical number: $x = 0$, because $f'(0) = 0$ and $0$ is in the domain.

7. Increasing: $(-1, 0) \cup (1, \infty)$; Decreasing: $(-\infty, -1) \cup (0, 1)$

8. Concavity: Concave down on $(-\infty, -1) \cup (-1, 1) \cup (1, \infty)$

9. (Sketch of the graph - omitted).