Chapter 1.6 Practice

Questions

For each function \(f\), determine the domain and range of \(f.\) It may be helpful to sketch the function.
1. \(f(x)=\begin{cases}-x+2&\text{if }-2\leq x\leq 2\\x&\text{if }3\leq x\end{cases}\)
2. \(f(x)=\begin{cases}4x-2&\text{if }-4\leq x< 0\\x^2&\text{if }0\leq x<6\end{cases}\)
3. \(f(x)=\begin{cases}4&\text{if }x\leq 0\\x+5&\text{if }0<x\end{cases}\)

For each function \(f\) and \(g\), determine \(\mathcal{D}(f+g)\) and \(\mathcal{D}(fg)\) and find a formula for \(f+g\) and \(fg.\)
1. \(f(x)=x-1\quad\text{and}\quad g(x)=\begin{cases}x&\text{if } -3<x<8\\ 4&\text{if }8\leq x \end{cases}\)
2. \(f(x)=\begin{cases}2&\text{if }-4\leq x\leq 2\\ x &\text{if }2< x< 8\end{cases}\quad\text{and}\quad g(x)=\begin{cases}x&\text{if } -3<x<8\\ 4&\text{if }8\leq x \end{cases}\)
3. \(f(x)=\begin{cases}x^2&\text{if }x\leq -3\\ -3x &\text{if }-3<x<1\\2x-4&\text{if }1\leq x\end{cases}\quad\text{and}\quad g(x)=\begin{cases}x+1&\text{if } -5<x<1\\ x&\text{if }1\leq x<4\\ 4&\text{if }4\leq x \end{cases}\)

For each function \(f\) and \(g\), determine \(\mathcal{D}\left(\tfrac{f}{g}\right)\) and find a formula for \(\tfrac{f}{g}.\)
1. \(f(x)=\begin{cases}5x-2&\text{if }-7<x<-3\\ x&\text{if }-1<x\end{cases}\quad\text{and}\quad g(x)=\begin{cases}x&\text{if } -4<x<2\\ 4&\text{if }3\leq x \leq 4\\x-4&\text{if }5\leq x\end{cases}\)
2. \(f(x)=\begin{cases}x&\text{if }x\leq-8\\ 5x&\text{if }-7\leq x<-6\\-x&\text{if }-5\leq x\leq 10\end{cases}\quad\text{and}\quad g(x)=\begin{cases}x+3&\text{if } -7.5<x<-4\\ -x+2&\text{if }-2\leq x \leq 12\end{cases}\)

Solve the following inequalities:
1. \(1<f(x)\leq 2\) where \[ f(x)=\begin{cases}3x-1&\text{if }-2<x<3\\ x-9&\text{if }4\leq x\end{cases}\]
2. \(f(x)>-\tfrac{3}{2}x+2\) where \[ f(x)=\begin{cases}-2x&\text{if }-9<x<-2\\ x-4&\text{if }2\leq x\leq 5\end{cases}\]
3. \(f(x)>g(x)\) where \[ f(x)=\begin{cases}-2x&\text{if }-4<x<-2\\ 3x+1&\text{if }-2\leq x\leq 5\end{cases}\quad \text{ and }\quad g(x)=\begin{cases}-x+3&\text{if }x<-1\\ x+9&\text{if }0\leq x\end{cases}\]

Solve the following inequalities:
1. \(|x-3|>4x-3\)
2. \(|2x+1|>|-x-6|\)
3. \(|-x+8|\geq|4x-6|\)

1. \(\mathcal{D}(f)=[-2,2]\cup[3,\infty)\), \(\mathcal{R}(f)=[0,\infty).\)
2. \(\mathcal{D}(f)=[-4,6)\), \(\mathcal{R}(f)=[-18,-2)\cup[0,36].\)
3. \(\mathcal{D}(f)=(-\infty,\infty)\), \(\mathcal{R}(f)=\{4\}\cup (5,\infty)\) or \(\mathcal{R}(f)=[4,4]\cup (5,\infty)\)

1. \(\mathcal{D}(f+g)=\mathcal{D}(fg)=(-3,\infty)\) \[(f+g)(x)=\begin{cases}2x-1&\text{if }-3<x<8\\x+3&\text{if }8\leq x\end{cases}\quad\text{and}\quad (fg)(x)=\begin{cases}x^2-x&\text{if }-3<x<8\\4x-4&\text{if }8\leq x\end{cases}\]
2. \(\mathcal{D}(f+g)=\mathcal{D}(fg)=(-3,8)\) \[(f+g)(x)=\begin{cases}x+2&\text{if }-3<x\leq 2\\2x&\text{if }2<x<8\end{cases}\quad\text{and}\quad (fg)(x)=\begin{cases}2x&\text{if }-3<x\leq 2\\x^2&\text{if }2<x<8\end{cases}\]
3. \(\mathcal{D}(f+g)=\mathcal{D}(fg)=(-5,\infty)\) \((f+g)(x)=\begin{cases}x^2+x+1&\text{if }-5<x\leq -3\\ -2x+1&\text{if }-3<x<1\\3x-4&\text{if }1\leq x<4\\ 2x&\text{if }4\leq x\end{cases}\quad\text{and}\quad (fg)(x)=\begin{cases}x^3+x&\text{if }-5<x\leq -3\\ -3x^2-3x&\text{if }-3<x<1\\2x^2-4x&\text{if }1\leq x<4\\ 8x-16&\text{if }4\leq x\end{cases}\)

1. \(\mathcal{D}\left(\frac{f}{g}\right)=(-4,-3)\cup(-1,0)\cup(0,2)\cup[3,4]\cup[5,\infty)\) \[\left(\frac{f}{g}\right)(x)=\begin{cases}\frac{5x-2}{x}&\text{if }-4<x<-3\\ 1 &\text{if }-1<x<0\\ 1 &\text{if }0<x<2\\\frac{x}{4}&\text{if }3\leq x\leq 4\\\frac{x}{x-4}&\text{if }5\leq x\end{cases}\]
2. \(\mathcal{D}\left(\frac{f}{g}\right)=[-7,-6]\cup[-5,-4)\cup[-2,2)\cup(2,10]\) \[\left(\frac{f}{g}\right)(x)=\begin{cases}\frac{5x}{x+3}&\text{if }-7\leq x\leq-6\\ -\frac{x}{x+3}&\text{if }-5\leq x\leq-4\\\frac{x}{x-2}&\text{if }-2\leq x< 2\\ \frac{x}{x-2}&\text{if }2< x\leq 10\end{cases}\]

1. \((\frac{2}{3},1]\cup(10,11]\)
2. \((-9-4)\cup(\tfrac{12}{5},5]\)
3. \((-4,-3)\cup(4,5]\)

1. \((-\infty,\tfrac{6}{5})\)
2. \((-\infty,-\tfrac{7}{3})\cup(5,\infty)\)
3. \([-\tfrac{2}{3},\tfrac{14}{5}]\)