1. Problem: Sketch the graph of the piecewise function $$f(x) = \begin{cases}-2x - 1, & x \leq 2 \\ -x + 4, & x > 2\end{cases}$$ Explanation: For $x \leq 2$, the graph is a line with slope $-2$ and intercept $-1$. At $x=2$, $f(2) = -2(2) - 1 = -5$, so point $(2, -5)$ is on the graph. For $x > 2$, the graph is a line with slope $-1$ and intercept $4$. At $x=2$, $f(2)$ from right side is $-2 + 4 = 2$, creating a jump at $x=2$. 2. Problem: Sketch the graph of the piecewise function $$f(x) = \begin{cases}-4, & x \leq -2 \\ x - 2, & -2 < x < 2 \\ -2x + 4, & x \geq 2 \end{cases}$$ Explanation: For $x \leq -2$, $f(x) = -4$ a horizontal line. At $x = -2$, $f(-2) = -4$. For $-2 < x < 2$, $f(x) = x - 2$ is a line with slope 1 and intercept -2. At $x = -2$, $f(-2^+) = -2 - 2 = -4$, continuous at $x=-2$. At $x=2$, $f(2^-) = 2 - 2 = 0$. For $x \geq 2$, $f(x) = -2x + 4$ is line with slope $-2$ and intercept 4. At $x=2$, $f(2) = -4 + 4 = 0$. So continuous at $x=2$. 3-8. Matching piecewise functions with graphs involves evaluating function behavior near boundaries. 9. Problem: Define the sign function $s(n)$ Solution: $$s(n) = \begin{cases}-1, & n < 0 \\ 0, & n = 0 \\ 1, & n > 0 \end{cases}$$ 10. Problem: Write the rule for the V-shaped piecewise linear function described. The graph has segments: From $-4$ to $0$: line descending from $( -4, 2 )$ to $(0, -4)$ Slope = $$\frac{-4 - 2}{0 - (-4)} = \frac{-6}{4} = -\frac{3}{2}$$ Equation left part: $$y = -\frac{3}{2}x - 4$$ shifting to pass through $(0, -4)$ From $0$ to $4$: line ascending from $(0, -4)$ to $(4, 0)$ Slope = $$\frac{0 - (-4)}{4 - 0} = 1$$ Equation right part: $$y = x - 4$$ Piecewise: $$f(x) = \begin{cases} -\frac{3}{2}x - 4, & -4 < x \leq 0 \\ x - 4, & 0 < x \leq 4 \end{cases}$$ 11. a) Graph piecewise described with peak at $(0,4)$, ascent from $(-8, 0)$ to $(0,4)$ and descent from $(0,4)$ to $(8, -8)$. Function: $$f(x) = \begin{cases} \frac{1}{2}x + 4, & -8 \leq x \leq 0 \\ -\frac{3}{2}x + 4, & 0 < x \leq 8 \end{cases}$$ Evaluate: $f(-3) = \frac{1}{2}(-3) + 4 = -1.5 + 4 = 2.5$ $f(-2) = \frac{1}{2}(-2) + 4 = -1 + 4 = 3$ $f(0) = 4$ $f(2) = -\frac{3}{2}(2) + 4 = -3 + 4 = 1$ $f(3) = -\frac{3}{2}(3) + 4 = -4.5 + 4 = -0.5$ 11. b) Evaluate $f(x)$ at points Given: $$f(x) = \begin{cases} x + 5, & x < -2 \\ -2x -1, & x \geq -2 \end{cases}$$ $f(-3) = -3 + 5 = 2$ $f(-2) = -2(-2) - 1 = 4 - 1 = 3$ $f(0) = -2(0) - 1 = -1$ $f(2) = -2(2) - 1 = -4 - 1 = -5$ $f(3) = -2(3) - 1 = -6 -1 = -7$ 12. Problem: Write piecewise salary function $S(x)$ in terms of sales $x$ Explanation: For $x \leq 500000$: $$S(x) = 35000 + 0.04x$$ If $x > 500000$: $$S(x) = 35000 + 0.04(500000) + 0.06(x - 500000) = 35000 + 20000 + 0.06(x - 500000) = 55000 + 0.06(x - 500000)$$ Piecewise: $$S(x) = \begin{cases} 35000 + 0.04x, & 0 \leq x \leq 500000 \\ 55000 + 0.06(x - 500000), & x > 500000 \end{cases}$$ 13. Problem: Find $x$ given the piecewise function graph By given graph description and function behavior, a) $f(x) = 2$ occurs where graph crosses or equals 2. From slope, find $x=-3$ approximately. b) $f(x) = 1$ occurs around $x=-4$ approximately. c) Solve $f(f(x))=2$, substitute inner $f(x)$ and solve. d) Solve $f(f(x))=1$, similarly. e) Write piecewise function with linear segments: $$f(x) = \begin{cases} \text{line from } (-5,-3) \text{ to } (0,5), \\ \text{line from } (0,5) \text{ to } (3,0), \\ \text{line from } (3,0) \text{ to } (5,4) \end{cases}$$ Calculating line equations: Segment 1: slope = $$\frac{5 - (-3)}{0 - (-5)} = \frac{8}{5}$$ Equation: $$y= \frac{8}{5}x + b$$ Using point $(-5,-3)$: $-3 = \frac{8}{5}(-5) + b \Rightarrow -3 = -8 + b \Rightarrow b=5$ So segment 1: $$f(x) = \frac{8}{5}x + 5, \quad -5 \leq x \leq 0$$ Segment 2: slope $$= \frac{0 - 5}{3 - 0} = -\frac{5}{3}$$ Equation: $$y = -\frac{5}{3}x + b$$ At $x=0$, $y=5$, so $b=5$ Segment 2: $$f(x) = -\frac{5}{3}x + 5, \quad 0 < x \leq 3$$ Segment 3: slope $$= \frac{4 - 0}{5 - 3} = 2$$ Equation: $$y = 2x + b$$ At $x=3$, $y=0$; so $0 = 2(3) + b \Rightarrow b = -6$ Segment 3: $$f(x) = 2x - 6, \quad 3 < x \leq 5$$