Let's solve each part in ** steps** as requested: --- ## a. \( f(x) = -\frac{1}{x+2} \) ### **Step 1: Analyze the Function and Its Derivative** - The function is a **rational function**, undefined at \( x = -2 \). - Derivative: \[ f'(x) = -\frac{d}{dx}\left(\frac{1}{x+2}\right) = -\left(-\frac{1}{(x+2)^2}\right) = \frac{1}{(x+2)^2} \] - \( f'(x) > \) for all \( x \neq -2 \). #### **Explanation:** The derivative is always **positive** except where undefined, so the function is **increasing** everywhere it is defined. --- ### **Step 2: Intervals of Increase/Decrease** - **Increasing:** \( (-\infty, -2) \cup (-2, \infty) \) - **Decreasing:** Nowhere #### **Explanation:** There are **no intervals where the function is decreasing** since the derivative is always positive. --- **Final Answer for (a):** - **Increasing:** \( (-\infty, -2) \cup (-2, \infty) \) - **Decreasing:** None --- ## b. \( g(x) = \frac{2}{-x+3} + 4 \) ### **Step 1: Analyze the Function and Its Derivative** - The function is undefined at \( x = 3 \). - Derivative: \[ g'(x) = \frac{d}{dx}\left(\frac{2}{-x+3}\right) = 2 \cdot \frac{d}{dx}(-x+3)^{-1} = 2(-1)(-x+3)^{-2} = \frac{2}{(-x+3)^2} \] #### **Explanation:** The derivative is always **positive** except at \( x = 3 \) (undefined). --- ### **Step 2: Intervals of Increase/Decrease** - **Increasing:** \( (-\infty, 3) \cup (3, \infty) \) - **Decreasing:** None #### **Explanation:** There are **no intervals where \( g(x) \) is decreasing** because the derivative is always positive. --- **Final Answer for (b):** - **Increasing:** \( (-\infty, 3) \cup (3, \infty) \) - **Decreasing:** None --- ## c. \( h(x) = 1 - \frac{1}{1-x} \) ### **Step 1: Analyze the Function and Its Derivative** - The function is undefined at \( x = 1 \). - Derivative: \[ h'(x) = -\frac{d}{dx}\left(\frac{1}{1-x}\right) = -(-1)(1-x)^{-2} = \frac{1}{(1-x)^2} \] #### **Explanation:** The derivative is **always positive** except at \( x = 1 \) (undefined). --- ### **Step 2: Intervals of Increase/Decrease** - **Increasing:** \( (-\infty, 1) \cup (1, \infty) \) - **Decreasing:** None #### **Explanation:** There are **no intervals where \( h(x) \) is decreasing** because the derivative is always positive. --- **Final Answer for (c):** - **Increasing:** \( (-\infty, 1) \cup (1, \infty) \) - **Decreasing:** None --- ## **Summary Table** | Function | Increasing Intervals | Decreasing Intervals | |--------------------|----------------------------------------|---------------------| | \( f(x) \) | \( (-\infty, -2) \cup (-2, \infty) \) | None | | \( g(x) \) | \( (-\infty, 3) \cup (3, \infty) \) | None | | \( h(x) \) | \( (-\infty, 1) \cup (1, \infty) \) | None |

# Function Analysis and Slope Behavior This document provides a structured analysis of three functions, detailing their increasing and decreasing intervals based on their derivatives. ## a. \( f(x) = -\frac{1}{x+2} \) ### **Step 1: Analyze the Function and Its Derivative** - The function \( f(x) \) is a rational function and is undefined at \( x = -2 \). - Derivative: \[ f'(x) = -\frac{d}{dx}\left(\frac{1}{x+2}\right) = -\left(-\frac{1}{(x+2)^2}\right) = \frac{1}{(x+2)^2} \] - Since \( f'(x) > 0 \) for all \( x \neq -2 \), the function is always increasing where defined. #### **Explanation:** The positive derivative indicates that the function increases in the intervals where it is defined. --- ### **Step 2: Intervals of Increase/Decrease** - **Increasing:** \( (-\infty, -2) \cup (-2, \infty) \) - **Decreasing:** None #### **Explanation:** There are no intervals of decrease since the derivative is always positive. --- **Final Answer for (a):** - **Increasing:** \( (-\infty, -2) \cup (-2, \infty) \) - **Decreasing:** None --- ## b. \( g(x) = \frac{2}{-x+3} + 4 \) ### **Step 1: Analyze the Function and Its Derivative** - The function \( g(x) \) is undefined at \( x = 3 \). - Derivative: \[ g'(x) = \frac{d}{dx}\left(\frac{2}{-x+3}\right) = 2 \cdot \frac{d}{dx}(-x+3)^{-1} = 2(-1)(-x+3)^{-2} = \frac{2}{(-x+3)^2} \] - The derivative is positive for all \( x \neq 3 \). #### **Explanation:** The positive derivative indicates that the function is increasing where defined. --- ### **Step 2: Intervals of Increase/Decrease** - **Increasing:** \( (-\infty, 3) \cup (3, \infty) \) - **Decreasing:** None #### **Explanation:** The function does not decrease as the derivative remains positive. --- **Final Answer for (b):** - **Increasing:** \( (-\infty, 3) \cup (3, \infty) \) - **Decreasing:** None --- ## c. \( h(x) = 1 - \frac{1}{1-x} \) ### **Step 1: Analyze the Function and Its Derivative** - The function \( h(x) \) is undefined at \( x = 1 \). - Derivative: \[ h'(x) = -\frac{d}{dx}\left(\frac{1}{1-x}\right) = -(-1)(1-x)^{-2} = \frac{1}{(1-x)^2} \] - The derivative is positive for all \( x \neq 1 \). #### **Explanation:** The positive derivative indicates the function is increasing where it is defined. --- ### **Step 2: Intervals of Increase/Decrease** - **Increasing:** \( (-\infty, 1) \cup (1, \infty) \) - **Decreasing:** None #### **Explanation:** The function does not decrease, as the derivative remains positive. --- **Final Answer for (c):** - **Increasing:** \( (-\infty, 1) \cup (1, \infty) \) - **Decreasing:** None --- ## **Summary Table** | Function | Increasing Intervals | Decreasing Intervals | |--------------------|----------------------------------------|---------------------| | \( f(x) \) | \( (-\infty, -2) \cup (-2, \infty) \) | None | | \( g(x) \) | \( (-\infty, 3) \cup (3, \infty) \) | None | | \( h(x) \) | \( (-\infty, 1) \cup (1, \infty) \) | None |