1. **Problem Statement:** We will analyze the polynomial function $$f(x) = \frac{x^3 - 4x}{x^2 - 1}$$ to find its domain, range, asymptotes, local behavior, and end behavior. 2. **Domain:** The domain of a function is all real values of $x$ for which the function is defined. Since the function is a rational function, the denominator cannot be zero: $$x^2 - 1 \neq 0$$ $$\Rightarrow (x-1)(x+1) \neq 0$$ $$\Rightarrow x \neq 1, x \neq -1$$ So, the domain is all real numbers except $x=1$ and $x=-1$. 3. **Asymptotes:** - **Vertical asymptotes** occur where the denominator is zero and the numerator is not zero. At $x=1$: $$f(1) = \frac{1^3 - 4(1)}{1^2 - 1} = \frac{1 - 4}{0} = \frac{-3}{0}$$ undefined, so vertical asymptote at $x=1$. At $x=-1$: $$f(-1) = \frac{(-1)^3 - 4(-1)}{(-1)^2 - 1} = \frac{-1 + 4}{0} = \frac{3}{0}$$ undefined, so vertical asymptote at $x=-1$. - **Horizontal or oblique asymptotes:** Since degree numerator (3) > degree denominator (2), there is no horizontal asymptote but possibly an oblique asymptote. Perform polynomial division: $$\frac{x^3 - 4x}{x^2 - 1} = x + \frac{-4x + x}{x^2 - 1} = x + \frac{-3x}{x^2 - 1}$$ As $x \to \pm \infty$, $\frac{-3x}{x^2 - 1} \to 0$, so the oblique asymptote is: $$y = x$$ 4. **Local behavior (critical points and extrema):** Find derivative using quotient rule: $$f(x) = \frac{N(x)}{D(x)} = \frac{x^3 - 4x}{x^2 - 1}$$ $$f'(x) = \frac{N'(x)D(x) - N(x)D'(x)}{(D(x))^2}$$ Calculate derivatives: $$N'(x) = 3x^2 - 4$$ $$D'(x) = 2x$$ So: $$f'(x) = \frac{(3x^2 - 4)(x^2 - 1) - (x^3 - 4x)(2x)}{(x^2 - 1)^2}$$ Simplify numerator: $$= (3x^2 - 4)(x^2 - 1) - 2x(x^3 - 4x)$$ $$= (3x^4 - 3x^2 - 4x^2 + 4) - (2x^4 - 8x^2)$$ $$= (3x^4 - 7x^2 + 4) - (2x^4 - 8x^2)$$ $$= 3x^4 - 7x^2 + 4 - 2x^4 + 8x^2$$ $$= (3x^4 - 2x^4) + (-7x^2 + 8x^2) + 4$$ $$= x^4 + x^2 + 4$$ Since $x^4 + x^2 + 4 > 0$ for all real $x$, the numerator of $f'(x)$ is always positive. The denominator $(x^2 - 1)^2$ is always positive except at $x=\pm 1$ where $f$ is undefined. Therefore, $f'(x) > 0$ for all $x$ in the domain, so $f$ is strictly increasing on each interval of its domain. No local maxima or minima. 5. **End behavior:** As $x \to \pm \infty$, the function behaves like its oblique asymptote: $$f(x) \approx x$$ So, $f(x) \to \infty$ as $x \to \infty$ and $f(x) \to -\infty$ as $x \to -\infty$. 6. **Range:** Since $f$ is continuous and strictly increasing on each interval $(-\infty, -1)$, $(-1, 1)$, and $(1, \infty)$ with vertical asymptotes at $x=\pm 1$, the range is all real numbers. **Final answers:** - Domain: $\{x \in \mathbb{R} : x \neq -1, x \neq 1\}$ - Vertical asymptotes: $x = -1$, $x = 1$ - Oblique asymptote: $y = x$ - No local maxima or minima; function is strictly increasing on each domain interval - End behavior: $f(x) \to \infty$ as $x \to \infty$, $f(x) \to -\infty$ as $x \to -\infty$ - Range: all real numbers $\mathbb{R}$