Answer Key 11.1

    1. No
    2. Yes
    3. No
    4. Yes
    5. Yes
    6. No
    7. Yes
    8. y^2=1+x^2
      y=\pm \sqrt{1+x^2}
      No
    9. \sqrt{y}=2-x
      y=(2-x)^2
      Yes
    10. y^2=1-x^2
      y=\pm \sqrt{1-x^2}
      No
  1. All real numbers -\infty, \infty
  2. \begin{array}{rrrrr} \\ \\ \\ \\ \\ 5&-&4x&\ge &0 \\ -5&&&&-5 \\ \midrule &&\dfrac{-4x}{-4}&\ge &\dfrac{-5}{-4} \\ \\ &&x&\le &\dfrac{5}{4} \\ \end{array}

    \left(-\infty, \dfrac{5}{4}\right]

  3. t^2\neq 0
    t\neq \sqrt{0}\text{ or }0
  4. All real or (-\infty, \infty)
  5. \begin{array}{rrrrr} \\ \\ \\ \\ \\ t^2&+&1&\neq &0 \\ &-&1&&-1 \\ \midrule &&t^2&\neq &-1 \\ &&t&\neq & i \\ \\ &&t&=&\mathbb{R} \end{array}
  6. \begin{array}{rrrrr} \\ \\ x&-&16&\ge &0 \\ &+&16&&+16 \\ \midrule &&x&\ge &16 \\ \end{array}

    [16, \infty)

  7. x^2-3x-4\neq 0
    (x-4)(x+1)\neq 0
    x\neq 4,1
  8. \begin{array}{ll} \\ \\ \\ \begin{array}{rrrrr} \\ \\ 3x&-&12&\ge &0 \\ &+&12&&+12 \\ \midrule &&\dfrac{3x}{3}&\ge &\dfrac{12}{3} \\ \\ &&x&\ge &4 \end{array} &\hspace{0.25in} \begin{array}{rrl} \\ x^2-25&\neq &0 \\ (x-5)(x+5)&\neq &0 \\ x&\neq &5, -5 \\ \\ \therefore x&\ge &4, \neq \pm 5 \end{array} \end{array}
  9. \begin{array}{rrl} \\ g(0)&=&\cancel{4(0)}-4 \\ &=&-4 \end{array}
  10. \begin{array}{rrl} \\ g(2)&=&-3\cdot 5^{-2} \\ &=&-\dfrac{3}{25} \end{array}
  11. \begin{array}{rrl} \\ f(-9)&=&(-9)^2+4 \\ &=&81+4 \\ &=&85 \end{array}
  12. \begin{array}{rrl} \\ f(10)&=&10-3 \\ &=&7 \end{array}
  13. \begin{array}{rrl} \\ \\ \\ \\ \\ \\ \\ \\ f(-2)&=&3^{-2}-2 \\ \\ &=&\dfrac{1}{9}-2 \\ \\ &=&\dfrac{1}{9}-\dfrac{18}{9} \\ \\ &=&-\dfrac{17}{9} \end{array}
  14. \begin{array}{rrl} \\ \\ f(2)&=&-3^{2-1}-3 \\ &=&-3^1-3 \\ &=&-6 \end{array}
  15. \begin{array}{rrl} \\ \\ \\ k(2)&=&-2\cdot 4^{2(2)-2} \\ &=&-2\cdot 4^{4-2} \\ &=&-2\cdot 4^2 \\ &=&-32 \end{array}
  16. \begin{array}{rrl} \\ \\ \\ \\ \\ \\ p(-2)&=&-2\cdot 4^{2(-2)+1}+1 \\ &=&-2\cdot 4^{-4+1}+1 \\ &=&-2\cdot 4^{-3}+1 \\ &=&-\dfrac{2}{64}+1 \\ \\ &=&-\dfrac{1}{32}+1 \Rightarrow \dfrac{-31}{32} \end{array}
  17. \begin{array}{rrl} \\ h(-4x)&=&(-4x)^3+2 \\ &=&-64x^3+2 \end{array}
  18. \begin{array}{rrl} \\ \\ h(n+2)&=&4(n+2)+2 \\ &=&4n+8+2 \\ &=&4n+10 \end{array}
  19. \begin{array}{rrl} \\ \\ h(-1+x)&=&3(-1+x)+2 \\ &=&-3+3x+2 \\ &=&3x-1 \end{array}
  20. \begin{array}{rrl} \\ \\ \\ h\left(\dfrac{1}{3}\right)&=&-3\cdot 2^{\frac{1}{3}+3} \\ &=& -2^3\cdot 3\sqrt[3]{2}\\ &=&-8\cdot 3\sqrt[3]{2} \\ &=&-24 \sqrt[3]{2} \end{array}
  21. \begin{array}{rrl} \\ h(x^4)&=&(x^4)^2+1 \\ &=&x^8+1 \end{array}
  22. \begin{array}{rrl} \\ h(t^2)&=&(t^2)^2+t \\ &=&t^4+t \end{array}
  23. \begin{array}{rrl} \\ f(0)&=&|\cancel{3(0)}+1|+1 \\ &=&1+1\text{ or }2 \end{array}
  24. \begin{array}{rrl} \\ \\ \\ f(-6)&=&-2 |-(-6)-2 | +1 \\ &=&-2 |6-2| + 1 \\ &=& -2(4)+1 \\ &=& -8 + 1\text{ or }-7 \end{array}
  25. \begin{array}{rrl} \\ f(10)&=&|10+3| \\ &=&13 \end{array}
  26. \begin{array}{rrl} \\ \\ p(5)&=&-|5|+1 \\ &=&-5+1 \\ &=& -4 \end{array}

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Intermediate Algebra by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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