We are given the function $f(x) = \ln|x^2 - 1|$. We need to find the domain, intercepts, limits, first and second derivatives, critical numbers, intervals of increase and decrease, concavity, and sketch the graph of the function.

1. Problem Description

We are given the function $f(x) = \ln|x^2 - 1|$. We need to find the domain, intercepts, limits, first and second derivatives, critical numbers, intervals of increase and decrease, concavity, and sketch the graph of the function.

2. Solution Steps

1. Domain of $f(x)$:

The natural logarithm is defined for positive values. Therefore, we need $|x^2 - 1| > 0$. This is equivalent to $x^2 - 1 \neq 0$, which means $x^2 \neq 1$, so $x \neq \pm 1$.

Thus, the domain of $f(x)$ is $D_f = (-\infty, -1) \cup (-1, 1) \cup (1, \infty)$.

2. Intercepts:

x-intercept: We set $f(x) = 0$.

$\ln|x^2 - 1| = 0$

$|x^2 - 1| = e^0 = 1$

$x^2 - 1 = 1$ or $x^2 - 1 = -1$

$x^2 = 2$ or $x^2 = 0$

$x = \pm \sqrt{2}$ or $x = 0$.

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

y-intercept: We set $x = 0$.

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

So the y-intercept is $y = 0$.

3. Limits as $x \to c$, where $c$ is an accumulation point of $D_f$ which is not in $D_f$:

The points 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 $x \to -1$, $x^2 \to 1$, so $|x^2 - 1| \to 0$.

Since $\lim_{u \to 0^+} \ln u = -\infty$, we have $\lim_{x \to 1} \ln|x^2 - 1| = -\infty$ and $\lim_{x \to -1} \ln|x^2 - 1| = -\infty$.

Thus, $x = 1$ and $x = -1$ are vertical asymptotes.

4. Limits as $x \to \pm \infty$:

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

As $x \to \pm \infty$, $x^2 \to \infty$, so $|x^2 - 1| \to \infty$.

Since $\lim_{u \to \infty} \ln u = \infty$, we have $\lim_{x \to \pm \infty} \ln|x^2 - 1| = \infty$.

5. Rewriting $f(x)$ without absolute values:

$f(x) = \ln|x^2 - 1|$. Since $x \neq \pm 1$, we have two cases:

Case 1: $x^2 > 1$, i.e., $x < -1$ or $x > 1$. Then $|x^2 - 1| = x^2 - 1$, so $f(x) = \ln(x^2 - 1)$.

Case 2: $x^2 < 1$, i.e., $-1 < x < 1$. Then $|x^2 - 1| = -(x^2 - 1) = 1 - x^2$, so $f(x) = \ln(1 - x^2)$.

Now we find the first and second derivatives.

$f'(x) = \frac{1}{x^2 - 1} \cdot 2x = \frac{2x}{x^2 - 1}$ for $x < -1$ or $x > 1$.

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

Thus, $f'(x) = \frac{2x}{x^2 - 1}$ for $x \neq \pm 1$.

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

6. Critical numbers:

Critical numbers occur where $f'(x) = 0$ or $f'(x)$ is undefined.

$f'(x) = \frac{2x}{x^2 - 1} = 0$ when $2x = 0$, so $x = 0$. Since $x = 0$ is in the domain, it is a critical number.

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

Thus, $x = 0$ is the only critical number.

7. Intervals of increase and decrease:

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

- If $x < -1$, then $2x < 0$ and $x^2 - 1 > 0$, so $f'(x) < 0$. Thus, $f(x)$ is decreasing on $(-\infty, -1)$.

- If $-1 < x < 0$, then $2x < 0$ and $x^2 - 1 < 0$, so $f'(x) > 0$. Thus, $f(x)$ is increasing on $(-1, 0)$.

- If $0 < x < 1$, then $2x > 0$ and $x^2 - 1 < 0$, so $f'(x) < 0$. Thus, $f(x)$ is decreasing on $(0, 1)$.

- If $x > 1$, then $2x > 0$ and $x^2 - 1 > 0$, so $f'(x) > 0$. Thus, $f(x)$ is increasing on $(1, \infty)$.

8. Concavity:

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

Since $x^2 + 1 > 0$ and $(x^2 - 1)^2 > 0$ for all $x$ in the domain, we have $f''(x) < 0$ for all $x$ in the domain.

Thus, $f(x)$ is concave down on $(-\infty, -1)$, $(-1, 1)$, and $(1, \infty)$.

9. Sketch the graph:

The graph has x-intercepts at $-\sqrt{2}, 0, \sqrt{2}$.

The y-intercept is

0. Vertical asymptotes at $x = -1$ and $x = 1$.

The function is concave down everywhere in its domain.

Decreasing on $(-\infty, -1)$ and $(0, 1)$.

Increasing on $(-1, 0)$ and $(1, \infty)$.

Since $f'(0) = 0$ and the function changes from increasing to decreasing at $x=0$, there is a local maximum at $(0,0)$.

The limit as $x$ approaches $\pm \infty$ is $\infty$.

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: $x = 0$

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

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

9. (Description of graph provided above)