Given: \( f(x)=\frac{-1}{x-1} \) 5.1 Write down the domain of \( f \) 5.2 Write down the agmptotes of \( f \). 5.3 Sketch the graph of \( f \). clear! showing all intercents with the axes and any aymptoles. 54 For ulvich values of \( x \) uill \( x f^{\prime}(x) \geq 0 \) ?

Let's tackle each part of the problem step by step. ### **5.1 Domain of \( f(x) = \frac{-1}{x - 1} \)** The domain of a function consists of all real numbers \( x \) for which the function is defined. For \( f(x) = \frac{-1}{x - 1} \): - The denominator \( x - 1 \) cannot be zero because division by zero is undefined. \[ x - 1 \neq 0 \implies x \neq 1 \] **Domain:** All real numbers except \( x = 1 \). \[ \boxed{\text{Domain: } \mathbb{R} \setminus \{1\}} \] --- ### **5.2 Asymptotes of \( f(x) = \frac{-1}{x - 1} \)** Asymptotes are lines that the graph of the function approaches but never touches. 1. **Vertical Asymptote:** This occurs where the denominator is zero. \[ x - 1 = 0 \implies x = 1 \] **Vertical Asymptote:** \( x = 1 \) 2. **Horizontal Asymptote:** Analyze the behavior of \( f(x) \) as \( x \) approaches infinity or negative infinity. \[ \lim_{x \to \pm\infty} \frac{-1}{x - 1} = 0 \] **Horizontal Asymptote:** \( y = 0 \) (the x-axis) \[ \boxed{\text{Vertical Asymptote: } x = 1 \quad \text{and} \quad \text{Horizontal Asymptote: } y = 0} \] --- ### **5.3 Graph of \( f(x) = \frac{-1}{x - 1} \)** **Intercepts:** 1. **Y-intercept:** Set \( x = 0 \): \[ f(0) = \frac{-1}{0 - 1} = 1 \] **Y-intercept:** \( (0, 1) \) 2. **X-intercept:** Set \( f(x) = 0 \): \[ \frac{-1}{x - 1} = 0 \] This equation has no solution because a non-zero constant divided by any finite number cannot be zero. **X-intercept:** None **Asymptotes:** - **Vertical:** \( x = 1 \) - **Horizontal:** \( y = 0 \) **Sketch:** The graph is a hyperbola with two branches: - **Left of the vertical asymptote (\( x < 1 \))**: - Approaches \( x = 1 \) from the left and \( y = 0 \) from below. - Passes through the y-intercept at \( (0, 1) \). - **Right of the vertical asymptote (\( x > 1 \))**: - Approaches \( x = 1 \) from the right and \( y = 0 \) from above. \[ \boxed{ \begin{array}{c} \text{Graph Description:} \\ \text{A hyperbola with a vertical asymptote at } x = 1 \text{ and a horizontal asymptote at } y = 0. \\ \text{Y-intercept at } (0, 1). \\ \text{No x-intercept.} \end{array} } \] --- ### **5.4 Values of \( x \) for Which \( x f'(x) \geq 0 \)** First, find the derivative \( f'(x) \): \[ f(x) = \frac{-1}{x - 1} = - (x - 1)^{-1} \] \[ f'(x) = -(-1)(x - 1)^{-2} = \frac{1}{(x - 1)^2} \] Now, consider the inequality: \[ x f'(x) \geq 0 \implies x \cdot \frac{1}{(x - 1)^2} \geq 0 \] Since \( (x - 1)^2 \) is always positive except at \( x = 1 \) (where the function is undefined), the sign of \( x f'(x) \) depends solely on the sign of \( x \). \[ \frac{x}{(x - 1)^2} \geq 0 \] - **When \( x > 0 \):** The expression is positive. - **When \( x = 0 \):** The expression is zero. - **When \( x < 0 \):** The expression is negative. - **At \( x = 1 \):** Undefined. **Solution:** All real numbers \( x \) such that \( x \geq 0 \) except \( x = 1 \). \[ \boxed{x \in [0, 1) \cup (1, \infty)} \]