Math 2 Exam I Practice Answers
These are the answers to the Exam I Practice.
P Level
- \([-2,4)\)
- \([1,6)\)
- \((1,2)\cup(3,4)\)
The domain is \((-4,-3)\cup[-2,0)\cup[2,3)\) and the range is \((-2,-1)\cup(1,4]\)
- \(-\frac{1}{3}\)
- \(2\)
- Both domains are \((-1,4)\)
- Both domains are \([5,6)\)
- \((-\infty,\infty)\)
- \((-\infty,\infty)\)
- \((-\infty,4]\)
- \((-\infty,-3)\cup(-3,\infty)\)
\((-\infty,-2)\cup(4,5)\)
\((5,-2)\)
\(\langle 2,-1\rangle\)
\(\langle 6, 2\rangle\)
Center is \((2,-9)\) and radius is \(4\)
\(\langle 1,\frac{4}{3}\rangle\) or \(\langle 3,4\rangle\)
\(\langle -3,2\rangle\)
Slope is \(-\frac{3}{4}\)
Slope of \(\frac{1}{3}\)
Domain is \((2,\infty)\) and the range is \((-2,9]\)
C Level
- \([-2,5]\)
- \(\{\}\)
- \((1,5]\)
- \(y=4\) and \(x=\frac{4}{3}\)
- \(y=11\) and \(x=-\frac{11}{2}\)
- \(y=-5\) and \(x=1\)
- \(\mathcal{D}\left(\frac{f}{g}\right)=(-1,4)\), \(\mathcal{D}\left(\frac{g}{f}\right)=\left(-1,\frac{1}{2}\right)\cup\left(\frac{1}{2},3\right)\cup(3,4)\)
- \(\mathcal{D}\left(\frac{f}{g}\right)=\left[5,\frac{11}{2}\right)\cup\left(\frac{11}{2},6\right)\), \(\mathcal{D}\left(\frac{g}{f}\right)=(5,6)\)
\(f=\frac{c\circ b}{a}+e\circ(b+d)\)
\(\left(-\frac{3}{\sqrt{34}},\frac{5}{\sqrt{34}}\right)\)
\(\left(-\frac{1}{2},\frac{5}{2}\right)\)
\(f^{-1}(x)=\sqrt[3]{x-1}-2\) Domain of the inverse and range of the inverse is \((-\infty,\infty)\)
B Level
- \([1,4]\)
- \((-4,4]\)
\(\ell(t)=\begin{cases}\frac{t}{10}\langle 2,11 \rangle+(2,-1)&\text{ if }0\leq t\leq 10\\ \frac{t-10}{2}\langle 7,-10 \rangle+(4,10) &\text{ if }10<t\leq 12\end{cases}\)
\(\left(-\frac{3}{\sqrt{26}}+3,\frac{15}{\sqrt{26}}+11 \right)\) and \(\left(\frac{3}{\sqrt{26}}+3,-\frac{15}{\sqrt{26}}+11 \right)\)
\(\left(\frac{9}{5}+2,\frac{12}{5}\right)\)
\((0,1)\cup \left(\frac{9}{2},\infty\right)\)
A Level
\((f\circ g)(x)=\begin{cases}\frac{9}{2}x+15 &\text{ if } -4\leq x<-2\\\left(\frac{3}{2}x+5\right)^2-1 &\text{ if } 0\leq x\leq 1\\ \left(-x+9\right)^2-1 &\text{ if } 2< x\leq 4\\ \end{cases}\)
- \((-2,0)\cup(2,4]\)
- \((-4,-3]\)
- \((-3,1)\cup(1,3]\)
- The sketches \(f\) and \(f^{-1}\) are below
- \(f^{-1}(x)=\begin{cases}\sqrt{x}&\text{ if }0\leq x\leq 1\\-\sqrt{x}&\text{ if }1<x<4\\-\sqrt{x}&\text{ if }9\leq x<16\end{cases}\)
- The sketches \(f\) and \(f^{-1}\) are below
- \(y=\frac{13}{9}x-\frac{19}{9}\).