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Mathematics LibreTexts

11.20: A.2.3- Section 2.3 Answers

  • Page ID
    121418
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    1. (b) \(x_{0}\neq k\pi \;\;\;\;(k=\text{integer})\)

    2. (b) \( (x_{0},y_{0})\neq (0,0)\)

    3. (b) \(x_{0}y_{0}\neq (2k+1)\frac{\pi }{2} \;\;\;\; (k=\text{integer})\)

    4. (b) \(x_{0}y_{0}>0\text{ and }x_{0}y_{0}\neq 1\)

    5.

    1. all \((x_{0},y_{0})\)
    2. \((x_{0},y_{0})\) woth \(y_{0}\neq 0\)

    6. (b) all \((x_{0}, y_{0})\)

    7. (b) all \((x_{0}, y_{0})\)

    8. (b) \((x_{0}, y_{0})\) such that \(x_{0}\neq 4y_{0}\)

    9.

    1. all \((x_{0}, y_{0})\)
    2. all \((x_{0}, y_{0})\neq (0,0)\)

    10.

    1. all \((x_{0}, y_{0})\)
    2. all \((x_{0}, y_{0})\) with \(y_{0}\neq\pm 1\)

    11. (b) all \((x_{0}, y_{0})\)

    12. (b) all \((x_{0}, y_{0})\) such that \((x_{0}, y_{0}) >0\)

    13. (b) all \((x_{0}, y_{0})\) with \(x_{0}\neq 1,\quad y_{0}\neq (2k+1)\frac{\pi }{2}(k=\text{integer})\)

    16. \(y=\left(\frac{3}{5}x+1 \right)^{5/3},\quad -\infty <x<\infty\), is a solution.

    Also, \[y=\left\{\begin{array}{cc}{0,}&{-\infty <x\leq -\frac{5}{3}}\\[4pt]{(\frac{3}{5}x+1)^{5/3},}&{-\frac{5}{3}<x<\infty }\end{array} \right.\nonumber \] is a solution. For every \(a\geq \frac{5}{3}\), the following function is also a solution: \[y=\left\{\begin{array}{cc}{(\frac{3}{5}(x+a))^{5/3},}&{-\infty <x<-a,}\\[4pt]{0,}&{-a\leq x\leq -\frac{5}{3}}\\[4pt]{(\frac{3}{5}x+1)^{5/3},}&{-\frac{5}{3}<x<\infty }\end{array} \right.\nonumber \]

    17.

    1. all \((x_{0}, y_{0})\)
    2. all \((x_{0}, y_{0})\) with \(y_{0}\neq 1\)

    18. \(y_{1}=1; y_{2}=1+|x|^{3};y_{3}=1-|x|^{3};y_{4}=1+x^{3};y_{5}=1-x^{3}\)

    \[y_{6}=\left\{\begin{array}{cc}{1+x^{3},}&{x\geq 0,}\\[4pt]{1,}&{x<0}\end{array} \right.;\quad y_{7}=\left\{\begin{array}{cc}{1-x^{3},}&{x\geq 0,}\\[4pt]{1,}&{x<0}\end{array} \right.;\nonumber \]

    \[y_{8}=\left\{\begin{array}{cc}{1,}&{x\geq 0,}\\[4pt]{1+x^{3},}&{x<0}\end{array} \right.;\quad y_{9}=\left\{\begin{array}{cc}{1,}&{x\geq 0,}\\[4pt]{1-x^{3},}&{x<0}\end{array} \right.\nonumber \]

    19. \(y=1+(x^{2}+4)^{3/2},\quad -\infty <x<\infty \)

    20.

    1. The solution is unique on \((0,\infty )\). It is given by
      \[y=\left\{\begin{array}{cc}{1,}&{0<x\leq \sqrt{5}}\\[4pt]{1-(x^{2}-5)^{3/2},}&{\sqrt{5}<x<\infty }\end{array} \right.\nonumber \]
    2. \[y=\left\{\begin{array}{cc}{1,}&{-\infty <x\leq\sqrt{5},}\\[4pt]{1-(x^{2}-5)^{3/2},}&{\sqrt{5}<x<\infty }\end{array} \right.\nonumber \] is a solution of (A) on \((-\infty ,\infty )\). If \(\alpha\geq 0\), then \[y=\left\{\begin{array}{cc}{1+(x^{2}-\alpha ^{2})^{3/2},}&{-\infty <x<-\alpha ,}\\[4pt]{1,}&{-\alpha\leq x\leq\sqrt{5},}\\[4pt]{1-(x^{2}-5)^{3/2},}&{\sqrt{5}<x<\infty ,}\end{array}\right.\nonumber \] and \[y=\left\{\begin{array}{cc}{1-(x^{2}-\alpha ^{2})^{3/2},}&{-\infty <x<-\alpha ,}\\[4pt]{1,}&{-\alpha\leq x\leq\sqrt{5},}\\[4pt]{1-(x^{2}-5)^{3/2},}&{\sqrt{5}<x<\infty ,}\end{array} \right.\nonumber \] are also solutions of (A) on \((-\infty ,\infty)\).

    This page titled 11.20: A.2.3- Section 2.3 Answers is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.

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