Math 2 Exam II Practice Answers

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These are the answers to the Exam II Practice.

P-Level

1. Domain is \((2,\infty)\) and the range is \((-2,9]\)

2. \((5,-1)\)

3. \(\left(\frac{1-3\sqrt{3}}{2},\frac{3+\sqrt{3}}{2} \right)\)

4. \(p^{-1}=\left(\frac{1}{5},\frac{\sqrt{24}}{5}\right)\)

5. \(\frac{1}{2}\) of a circle

6. \(\sin(\theta)=-\frac{\sqrt{24}}{5}\), \(\cos(\theta)=\frac{1}{5}\) and \(\tan(\theta)=-\sqrt{24}\)

7. Vertex is \((1,4)\), maximum value \(4\) at the vertex, and no minimum value

8. The zeros are \(-3\), \(3\) and \(-1\) and the respective orders are \(2\), \(5\) and \(8\)

9. Zeros are \(2\) and \(-5\) with respective orders \(2\) and \(1\); the poles are at \(0\) and \(9\) with respective orders \(1\) and \(3\)

10. Asymptotic behavior is it behaves like \(\frac{1}{x}\), so it has a horizontal asymptote at \(y=0\)

11. \(\frac{11}{5}\)

12. \(f(-2)=-5\) and \(f(1)=-8\)

13. \(f(-5)=-1\) and \(f(3)=2\)

14. 1. \(2\pi\)
    2. \(\frac{2\pi}{3}\)
    3. \(\frac{2\pi}{3}\)
15. 1. horizontal asymptote is \(y=4\); no vertical asymptote
    2. horizontal asymptote is \(y=0\); no vertical asymptote
    3. horizontal asymptote is \(y=-2\); no vertical asymptote
    4. no horizontal asymptote; vertical asymptote is \(x=0\)
    5. no horizontal asymptote; vertical asymptote is \(x=2\)
    6. no horizontal asymptote; vertical asymptote is \(x=5\)

16. \(x_0=-2\) and \(y_0=5\)

17. \(\frac{3}{7}\)

18. \(y=\frac{11}{2}(x-7)+1.\)

19. \(y=4x+4.\)

C-Level

1. \(f^{-1}(x)=\sqrt[3]{x-1}-2\), Domain of the inverse and range of the inverse is \((-\infty,\infty)\)

2. \(\left(\frac{1-\sqrt{72}}{10},\frac{-\sqrt{24}-\sqrt{3}}{10}\right)\)

3. \(\left(\frac{4+2\sqrt{24}}{5}-3,\frac{2-4\sqrt{24}}{5}+1\right)\)

4. \(\sin(\theta)=-\frac{\sqrt{24}}{7}\), \(\cos(\theta)=\frac{5}{7}\), \(\tan(\theta)=-\frac{\sqrt{24}}{5}\), \(\csc(\theta)=-\frac{7}{\sqrt{24}}\), \(\sec(\theta)=\frac{7}{5}\) and \(\cot(\theta)=-\frac{5}{\sqrt{24}}\)

8. 1. \(\sin(A+B)=\frac{3+\sqrt{960}}{35}\)
   2. \(\cos(A-B)=\frac{3\sqrt{24}+\sqrt{40}}{35}\)
9. 1. Translate \(g\) up by \(4\). Domain is \((-\infty,\infty)\). Range is \((4,\infty)\). Horizontal asymptote at \(y=4\)
      
   2. Translate \(g\) right by \(2\), reflect over \(x\)-axis. Domain is \((-\infty,\infty)\). Range is \((-\infty,0)\). Horizontal asymptote at \(y=0\)
      
   3. Reflect \(g\) over \(x\)-axis, translate down by \(2\). Domain is \((-\infty,\infty)\). Range is \((-\infty,-2)\). Horizontal asymptote at \(y=-2\)
      
   4. Reflect over \(x\)-axis, translate \(g\) up by \(2\). Domain is \((0,\infty)\). Range is \((-\infty,\infty)\). Vertical asymptote at \(x=0\)
      
   5. Translate \(g\) right by \(2\). Domain is \((2,\infty)\). Range is \((-\infty,\infty)\). Vertical asymptote at \(x=2\)
      
   6. Translate \(g\) left by \(5\), reflect over \(y\)-axis, scale \(y\) by \(2\). Domain is \((-\infty,5)\). Range is \((-\infty,\infty)\). Vertical asymptote at \(x=5\)
      
10. 1. \(6\)
    2. \(-24\)

11. \(A(t)=4\left(\frac{7}{2}\right)^\frac{t-2}{3}\)

12. 1. \(m^2-16m+64=0\)
    2. \(y=8(x-2)+13\)

13. \(y=-\frac{1}{8}(x-2)+5\)

14. \(y=\frac{4}{3}\left(x-\frac{1}{3}\right)+4.\)

B-Level

1. 1. the graph is given below
   2. \(\{-4\}\cup[-3,3)\cup(3,4]\cup(5,\infty)\)
   3. \((-3,0)\cup(0,3)\cup(3,4)\cup(5,\infty)\)
   4. \((-\infty,-3]\cup\{0\}\cup[4,5)\)
   5. \((-\infty,-4)\cup(-4,-3)\cup(4,5)\)

2. 1. \(\frac{8\ln(2)}{\ln\left(\frac{9}{4}\right)}\)
   2. \(\frac{\ln\left(\frac{9}{4}\right)}{8}\)
3. 1. \(A(t)=t\left\langle \tfrac{4}{\sqrt{65}},\tfrac{7}{\sqrt{65}}\right\rangle+(1,2)\) and \(B(t)=\langle 3\cos\left(2t\right),3\sin\left(2t\right) \rangle+A(t)\)
   2. \(\sqrt{65}\).
   3. \(2\)

A-Level

1. 1. The sketches \(f\) and \(f^{-1}\) are below
      
   2. \(f^{-1}(x)=\begin{cases}\sqrt{x}&\text{ if }0\leq x\leq 1\\-\sqrt{x}&\text{ if }1<x<4\\-\sqrt{x}&\text{ if }9\leq x<16\end{cases}\)

2. \(y=-(x-2)+5\)

3. Graph of \(g\) is given below.

4. \(y=-4(x-5)+3\)

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