Piecewise Domain Range A290E9

1. **Problem statement:** We have a piecewise linear function $f(x)$ defined by points $(-5,5)$, $(-1,-2)$, $(2,1)$, and $(4,0)$ connected by line segments. 2. **Domain:** The domain is the set of all $x$-values for which $f(x)$ is defined. Since the graph connects points from $x=-5$ to $x=4$ continuously, the domain is $$[-5,4].$$ 3. **Range:** The range is the set of all $y$-values that $f(x)$ takes. - At $x=-5$, $f(x)=5$ (maximum value). - At $x=-1$, $f(x)=-2$ (minimum value). - At $x=2$, $f(x)=1$. - At $x=4$, $f(x)=0$. Between these points, the function is linear, so it takes all values between the minimum and maximum. Thus, the range is $$[-2,5].$$ 4. **Where is $f(x)$ positive?** - From $x=-5$ to $x=-1$, $f(x)$ goes from $5$ to $-2$, crossing zero somewhere. - Find zero crossing between $(-5,5)$ and $(-1,-2)$: The slope is $$m=\frac{-2-5}{-1+5}=\frac{-7}{4}=-\frac{7}{4}.$$ The line equation is $$y-5 = -\frac{7}{4}(x+5).$$ Set $y=0$ to find zero: $$0-5 = -\frac{7}{4}(x+5) \Rightarrow -5 = -\frac{7}{4}(x+5) \Rightarrow 5 = \frac{7}{4}(x+5) \Rightarrow x+5 = \frac{20}{7} \Rightarrow x = \frac{20}{7} - 5 = \frac{20}{7} - \frac{35}{7} = -\frac{15}{7} \approx -2.14.$$ So $f(x)>0$ for $x$ in $$[-5,-\frac{15}{7}).$$ - Between $x=-1$ and $x=2$, $f(x)$ goes from $-2$ to $1$, crossing zero again. Slope: $$m=\frac{1-(-2)}{2-(-1)}=\frac{3}{3}=1.$$ Equation: $$y+2 = 1(x+1) \Rightarrow y = x -1.$$ Set $y=0$: $$0 = x -1 \Rightarrow x=1.$$ So $f(x)>0$ for $x$ in $$(1,2].$$ Combining positive intervals: $$[-5,-\frac{15}{7}) \cup (1,2].$$ 5. **Where is $f(x)$ negative?** - Between $x=-\frac{15}{7}$ and $x=1$, $f(x)$ is negative. - Also check between $x=2$ and $x=4$: At $x=2$, $f(x)=1$ (positive), at $x=4$, $f(x)=0$. Since it goes from positive to zero, $f(x)$ is not negative here. Therefore, negative interval is: $$(-\frac{15}{7},1).$$