9. a) i) Define step function ii) The function is defined as \[ f(x)=\left\{\begin{array}{rr}-4 & \text { if } x<-2 \\ 2 \text { if }-2 \leq x \leq 3 \\ 5 & \text { if } x \geq 3\end{array}\right. \] Draw the graph of \( f(x) \) and state its domain and range, find \( f(-100) \), \( f\left(\frac{1}{2}\right) \). and \( f(203) \). b) Verify that \( x+4 \) is not a factor of the polynomial function \[ f(x)=x^{3}-9 x^{2}+10 x-24 . \] e) Sketch the graph of the function \( f(x) \) and hence state its domain and range. \( f(x)=\left\{\begin{array}{r}x^{2}+1 \text { for } x>1 \\ |x| \text { for }-2 \leq x \leq 1 \\ x+2 \text { for } x 2\end{array}\right. \)

Certainly! Let's tackle each part of Problem 9 step by step. --- ### **9. a) i) Define Step Function** **Definition:** A **step function** is a piecewise constant function that increases or decreases abruptly from one constant value to another. Graphically, it resembles a series of steps, hence the name. Each "step" corresponds to a specific interval of the independent variable where the function holds a constant value. **Mathematically:** A step function can be defined as: \[ f(x) = \begin{cases} c_1 & \text{if } x \in I_1, \\ c_2 & \text{if } x \in I_2, \\ \vdots & \vdots \\ c_n & \text{if } x \in I_n, \end{cases} \] where \( c_1, c_2, \ldots, c_n \) are constants and \( I_1, I_2, \ldots, I_n \) are mutually exclusive intervals covering the domain of \( f(x) \). --- ### **9. a) ii) Analyzing the Given Function** Given the function: \[ f(x) = \begin{cases} -4 & \text{if } x < -2, \\ 2 & \text{if } -2 \leq x \leq 3, \\ 5 & \text{if } x \geq 3. \end{cases} \] #### **1. Graph of \( f(x) \)** To graph \( f(x) \), plot the constant values over their respective intervals: - **For \( x < -2 \):** \( f(x) = -4 \) - **For \( -2 \leq x \leq 3 \):** \( f(x) = 2 \) - **For \( x \geq 3 \):** \( f(x) = 5 \) **Graph:**  *Note: The graph consists of horizontal lines at \( y = -4 \), \( y = 2 \), and \( y = 5 \) over their respective intervals. Use open or closed circles to indicate inclusivity at interval boundaries.* #### **2. Domain and Range** - **Domain:** All real numbers \( \mathbb{R} \), since \( f(x) \) is defined for every real \( x \). \[ \text{Domain} = (-\infty, \infty) \] - **Range:** The set of output values, which are the constants defined. \[ \text{Range} = \{-4, 2, 5\} \] #### **3. Evaluating \( f(-100) \), \( f\left(\frac{1}{2}\right) \), and \( f(203) \)** - **\( f(-100) \):** Since \( -100 < -2 \), \[ f(-100) = -4 \] - **\( f\left(\frac{1}{2}\right) \):** Since \( -2 \leq \frac{1}{2} \leq 3 \), \[ f\left(\frac{1}{2}\right) = 2 \] - **\( f(203) \):** Since \( 203 \geq 3 \), \[ f(203) = 5 \] --- ### **9. b) Verifying That \( x + 4 \) Is Not a Factor of \( f(x) = x^{3} - 9x^{2} + 10x - 24 \)** To verify if \( x + 4 \) is a factor of \( f(x) \), we can use the **Factor Theorem**. The Factor Theorem states that \( (x - c) \) is a factor of a polynomial \( f(x) \) if and only if \( f(c) = 0 \). Here, \( x + 4 = 0 \) implies \( x = -4 \). We'll evaluate \( f(-4) \): \[ f(-4) = (-4)^3 - 9(-4)^2 + 10(-4) - 24 \] \[ = -64 - 9(16) - 40 - 24 \] \[ = -64 - 144 - 40 - 24 \] \[ = -272 \neq 0 \] **Conclusion:** Since \( f(-4) \neq 0 \), \( x + 4 \) is **not** a factor of \( f(x) \). --- ### **9. e) Sketching the Graph of \( f(x) \) and Determining Its Domain and Range** Given the function: \[ f(x) = \begin{cases} x^{2} + 1 & \text{for } x > 1, \\ |x| & \text{for } -2 \leq x \leq 1, \\ x + 2 & \text{for } x < -2. \end{cases} \] > **Note:** There seems to be a slight typo in the function definition provided. The last piece likely applies to \( x < -2 \) instead of "x 2". We'll proceed under this assumption. #### **1. Graph of \( f(x) \)** Let's analyze each piece: - **For \( x < -2 \):** \( f(x) = x + 2 \) This is a straight line with a slope of 1 and y-intercept at \( (0, 2) \). - **For \( -2 \leq x \leq 1 \):** \( f(x) = |x| \) This is the absolute value function, V-shaped with a vertex at \( (0, 0) \). - **For \( x > 1 \):** \( f(x) = x^{2} + 1 \) This is a parabola opening upwards with vertex at \( (0, 1) \), but only for \( x > 1 \). **Graph:**  *Description:* - **\( x < -2 \):** Line increasing from left, passes through \( (-2, 0) \). - **\( -2 \leq x \leq 1 \):** Absolute value graph from \( (-2, 2) \) to \( (0, 0) \) to \( (1, 1) \). - **\( x > 1 \):** Parabola starting at \( (1, 2) \) and increasing. #### **2. Domain and Range** - **Domain:** All real numbers \( \mathbb{R} \), as each piece covers a part of the real line without gaps. \[ \text{Domain} = (-\infty, \infty) \] - **Range:** Let's determine based on each piece: - **\( x < -2 \):** \( f(x) = x + 2 \). As \( x \) approaches \( -\infty \), \( f(x) \) approaches \( -\infty \). At \( x = -2 \), \( f(-2) = -2 + 2 = 0 \). - **\( -2 \leq x \leq 1 \):** \( f(x) = |x| \). The outputs range from \( 0 \) to \( 2 \). - **\( x > 1 \):** \( f(x) = x^{2} + 1 \). As \( x \) increases, \( f(x) \) increases without bound. At \( x = 1 \), \( f(1) = 1 + 1 = 2 \). Combining these: \[ \text{Range} = (-\infty, 2] \cup [2, \infty) = (-\infty, \infty) \] However, upon closer inspection: - The function approaches \( -\infty \) as \( x \to -\infty \). - The minimum value of \( f(x) \) is \( 0 \) at \( x = 0 \). Therefore, the corrected range is: \[ \text{Range} = [0, \infty) \] --- **Summary:** - **9.a.i)** Defined the step function. - **9.a.ii)** Graphed the step function, determined its domain \((-∞, ∞)\) and range \(\{-4, 2, 5\}\), and evaluated \( f(-100) = -4 \), \( f\left(\frac{1}{2}\right) = 2 \), and \( f(203) = 5 \). - **9.b)** Verified using the Factor Theorem that \( x + 4 \) is not a factor of \( f(x) = x^{3} - 9x^{2} + 10x - 24 \). - **9.e)** Sketched the piecewise function \( f(x) \), determined its domain as all real numbers \((-∞, ∞)\), and its range as \([0, ∞)\). Feel free to reach out if you have any further questions or need additional clarifications!