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

2. Solution Steps

1. Domain:

For $f(x) = \ln|x^2-1|$ to be defined, we need $|x^2-1|>0$. This means $x^2-1\neq 0$, so $x^2 \neq 1$, which implies $x \neq \pm 1$. Therefore, the domain is $D_f = (-\infty, -1) \cup (-1, 1) \cup (1, \infty)$.

2. Intercepts:

$x$-intercept: We need $f(x) = 0$, so $\ln|x^2-1| = 0$. This means $|x^2-1| = e^0 = 1$, so $x^2-1 = 1$ or $x^2-1 = -1$.

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 $x = -\sqrt{2}$, $x = 0$, and $x = \sqrt{2}$.

$y$-intercept: We evaluate $f(0) = \ln|0^2-1| = \ln|-1| = \ln(1) = 0$. So, the $y$-intercept is $y = 0$.

3. Limits and Asymptotes:

We need to find $\lim_{x \to c} f(x)$, where $c$ is an accumulation point of $D_f$ which is not in $D_f$. So, $c = \pm 1$.

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

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

4. Limits at Infinity:

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

5. Derivatives:

Since $|x^2-1| = \begin{cases} x^2-1, & x<-1 \text{ or } x>1 \\ 1-x^2, & -1<x<1 \end{cases}$,

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

$f'(x) = \frac{2x}{x^2-1}$ for $x<-1$ or $x>1$ and $f'(x) = \frac{-2x}{1-x^2} = \frac{2x}{x^2-1}$ for $-1<x<1$. Therefore, $f'(x) = \frac{2x}{x^2-1}$ for $x \in D_f$.

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

6. Critical Numbers:

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

7. Intervals of Increase and Decrease:

We analyze the sign of $f'(x) = \frac{2x}{x^2-1}$. The critical points are $x=-1, 0, 1$. We have four intervals to check:

$(-\infty, -1)$: $f'(-2) = \frac{-4}{3} < 0$, so $f$ is decreasing.

$(-1, 0)$: $f'(-0.5) = \frac{-1}{-0.75} > 0$, so $f$ is increasing.

$(0, 1)$: $f'(0.5) = \frac{1}{-0.75} < 0$, so $f$ is decreasing.

$(1, \infty)$: $f'(2) = \frac{4}{3} > 0$, so $f$ is increasing.

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

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

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 $x \in D_f$, $f''(x) < 0$ for all $x$ in the domain. Thus, the function is concave down on $(-\infty, -1)$, $(-1, 1)$, and $(1, \infty)$.

9. Sketch:

(Not possible to show a sketch in text format)

The function has vertical asymptotes at $x = -1$ and $x = 1$. It has $x$-intercepts at $-\sqrt{2}$, $0$, and $\sqrt{2}$. The $y$-intercept is $0$. The function is decreasing on $(-\infty, -1)$ and $(0, 1)$ and increasing on $(-1, 0)$ and $(1, \infty)$. It is concave down everywhere.

3. Final Answer

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

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

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. Yes, $x=0$ is a critical number because $f'(0) = 0$ and $0$ is in the domain of $f(x)$.

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

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

9. See solution steps. (Not possible to draw here).