Piecewise Graphs Edf45F

1. **Problem Statement:** We need to draw the graphs of two functions and analyze the domain and range of a discrete function. 2. **Function 1:** $f(x) = |x-2|$ - This is an absolute value function shifted right by 2 units. - The graph is a V-shape with vertex at $(2,0)$. 3. **Function 2:** $$f(x) = \begin{cases} 1-x, & x < 0 \\ 1, & x = 0 \\ 1+x, & x > 0 \end{cases}$$ - This is a piecewise linear function with three parts. - For $x<0$, the graph is a line with slope $-1$ and intercept $1$. - At $x=0$, the function value is $1$. - For $x>0$, the graph is a line with slope $1$ and intercept $1$. 4. **Discrete function:** $$A = \{-2, -1, 1, 2\}$$ $$f(x) = \{(x, \frac{1}{x}) : x \in A\}$$ - Domain $D_f = A = \{-2, -1, 1, 2\}$. - Range $R_f = \left\{ \frac{1}{-2}, \frac{1}{-1}, \frac{1}{1}, \frac{1}{2} \right\} = \{-\frac{1}{2}, -1, 1, \frac{1}{2}\}$. 5. **Graph shapes and positions:** - $f(x) = |x-2|$ is a V-shape centered at $x=2$, positioned top-right. - The piecewise function has three linear branches with a point at $(0,1)$, positioned top-right. - The discrete function has points at $(-2,-\frac{1}{2}), (-1,-1), (1,1), (2,\frac{1}{2})$, positioned bottom-left. --- ### Step-by-step for $f(x) = |x-2|$: 1. The absolute value function is defined as $|u| = \begin{cases} u, & u \geq 0 \\ -u, & u < 0 \end{cases}$. 2. Here, $u = x-2$. 3. So, $$f(x) = \begin{cases} x-2, & x \geq 2 \\ -(x-2) = 2 - x, & x < 2 \end{cases}$$ 4. This forms a V-shape with vertex at $(2,0)$. ### Step-by-step for the piecewise function: 1. For $x<0$, $f(x) = 1 - x$ is a line with slope $-1$ and y-intercept $1$. 2. At $x=0$, $f(0) = 1$. 3. For $x>0$, $f(x) = 1 + x$ is a line with slope $1$ and y-intercept $1$. 4. The graph has a sharp corner at $(0,1)$. ### Step-by-step for discrete function $f(x) = \frac{1}{x}$ on $A$: 1. Domain $D_f = A = \{-2, -1, 1, 2\}$. 2. Calculate range: - $f(-2) = -\frac{1}{2}$ - $f(-1) = -1$ - $f(1) = 1$ - $f(2) = \frac{1}{2}$ 3. Range $R_f = \{-1, -\frac{1}{2}, \frac{1}{2}, 1\}$. --- **Final answers:** - $D_f = \{-2, -1, 1, 2\}$ - $R_f = \{-1, -\frac{1}{2}, \frac{1}{2}, 1\}$