{"id":1824,"date":"2019-08-14T16:40:11","date_gmt":"2019-08-14T20:40:11","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/?post_type=back-matter&p=1824"},"modified":"2019-08-14T18:27:28","modified_gmt":"2019-08-14T22:27:28","slug":"answer-key-11-3","status":"publish","type":"back-matter","link":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/back-matter\/answer-key-11-3\/","title":{"raw":"Answer Key 11.3","rendered":"Answer Key 11.3"},"content":{"raw":"[latexpage]\r\n
    \r\n \t
  1. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\\r\n(g\\circ f)(x)&=&-(\\sqrt[5]{-x-3})^5-3 \\\\\r\n&=&-(-x-3)-3 \\\\\r\n&=&x+3-3 \\\\\r\n&=&x\\hspace{0.75in}\\text{Inverse}\r\n\\end{array}\\)<\/li>\r\n \t
  2. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\\r\n(g\\circ f)(x)&=&4-\\left(\\dfrac{4}{x}\\right) \\\\ \\\\\r\n&=&4-\\dfrac{4}{x}\\hspace{0.5in}\\text{Not inverse}\r\n\\end{array}\\)<\/li>\r\n \t
  3. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\\r\n(g\\circ f)(x)&=&-10\\left(\\dfrac{x-5}{10}\\right)+5 \\\\ \\\\\r\n&=& -x+5+5\\\\ \\\\\r\n&=&-x+10\\hspace{0.75in}\\text{Not inverse}\r\n\\end{array}\\)<\/li>\r\n \t
  4. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n(f\\circ g)(x)&=&\\dfrac{(10x+5)-5}{10} \\\\ \\\\\r\n&=&\\dfrac{10x+5-5}{10} \\\\ \\\\\r\n&=&\\dfrac{10x}{10} \\\\ \\\\\r\n&=&x\\hspace{0.75in}\\text{Inverse}\r\n\\end{array}\\)<\/li>\r\n \t
  5. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n(f\\circ g)(x)&=&\\dfrac{-2}{\\dfrac{3x+2}{x+2}+3} \\\\ \\\\\r\n&=& \\dfrac{-2(x+2)}{3x+2+3(x+2)}\\\\ \\\\\r\n&=& \\dfrac{-2x-4}{3x+2+3x+6}\\\\ \\\\\r\n&=& \\dfrac{-2x-4}{6x+8}\\\\ \\\\\r\n&=& \\dfrac{-x-2}{3x+4}\\hspace{0.75in}\\text{Not inverse}\r\n\\end{array}\\)<\/li>\r\n \t
  6. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n(f\\circ g)&=&\\dfrac{-\\left(\\dfrac{-2x+1}{-x-1}\\right)-1}{\\dfrac{-2x+1}{-x-1}-2} \\\\ \\\\\r\n&=&\\dfrac{-(-2x+1)-1(-x-1)}{-2x+1-2(-x-1)} \\\\ \\\\\r\n&=&\\dfrac{2x-1+x+1}{-2x+1+2x+2} \\\\ \\\\\r\n&=&\\dfrac{3x}{3} \\\\ \\\\\r\n&=&x\\hspace{0.75in}\\text{Inverse}\r\n\\end{array}\\)<\/li>\r\n \t
  7. \\(\\begin{array}{rrcrr}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\ny&=&(x-2)^5&+&3 \\\\\r\nx&=&(y-2)^5&+&3 \\\\\r\n-3&&&-&3 \\\\\r\n\\midrule\r\nx-3&=&(y-2)^5&& \\\\\r\n\\sqrt[5]{x-3}&=&y-2&& \\\\ \\\\\r\ny&=&\\sqrt[5]{x-3}&+&2\r\n\\end{array}\\)<\/li>\r\n \t
  8. \\(\\begin{array}{rrcrr}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\ny&=&\\sqrt[3]{x+1}&+&2 \\\\\r\nx&=&\\sqrt[3]{y+1}&+&2 \\\\\r\n-2&&&-&2 \\\\\r\n\\midrule\r\nx-2&=&\\sqrt[3]{y+1}&& \\\\\r\n(x-2)^3&=&y+1&& \\\\ \\\\\r\ny&=&(x-2)^3&-&1\r\n\\end{array}\\)<\/li>\r\n \t
  9. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\ny&=&\\dfrac{4}{x+2} \\\\ \\\\\r\nx&=&\\dfrac{4}{y+2} \\\\ \\\\\r\ny+2&=&\\dfrac{4}{x} \\\\ \\\\\r\ny&=&\\dfrac{4}{x}-2\r\n\\end{array}\\)<\/li>\r\n \t
  10. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\ny&=&\\dfrac{3}{x-3} \\\\ \\\\\r\nx&=&\\dfrac{3}{y-3} \\\\ \\\\\r\ny-3&=&\\dfrac{3}{x} \\\\ \\\\\r\ny&=&\\dfrac{3}{x}+3\r\n\\end{array}\\)<\/li>\r\n \t
  11. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\ny&=&\\dfrac{-2x-2}{x+2} \\\\ \\\\\r\nx&=&\\dfrac{-2y-2}{y+2} \\\\ \\\\\r\nx(y+2)&=&-2y-2 \\\\\r\nxy+2x&=&-2y-2 \\\\\r\n-xy+2&&-xy+2 \\\\\r\n\\midrule\r\n2x+2&=&-2y-xy \\\\\r\n2x+2&=&y(-2-x) \\\\ \\\\\r\ny&=&\\dfrac{2x+2}{-2-x} \\\\ \\\\\r\ny&=&-\\dfrac{2x+2}{2+x}\r\n\\end{array}\\)<\/li>\r\n \t
  12. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\\r\ny&=&\\dfrac{9+x}{3} \\\\ \\\\\r\nx&=&\\dfrac{9+y}{3} \\\\ \\\\\r\n3x&=&9+y \\\\\r\ny&=&3x-9\r\n\\end{array}\\)<\/li>\r\n \t
  13. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\\r\ny&=&\\dfrac{10-x}{5} \\\\ \\\\\r\nx&=&\\dfrac{10-y}{5} \\\\ \\\\\r\n5x&=&10-y \\\\\r\ny&=&10-5x\r\n\\end{array}\\)<\/li>\r\n \t
  14. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\ny&=&\\dfrac{5x-15}{2} \\\\ \\\\\r\nx&=&\\dfrac{5y-15}{2} \\\\ \\\\\r\n5y-15&=&2x \\\\ \\\\\r\n5y&=&2x+15 \\\\ \\\\\r\ny&=&\\dfrac{2x+15}{5}\r\n\\end{array}\\)<\/li>\r\n \t
  15. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\ny&=&-(x-1)^3 \\\\\r\nx&=&-(y-1)^3 \\\\\r\n\\sqrt[3]{x}&=&-(y-1) \\\\\r\n\\sqrt[3]{x}&=&-y+1 \\\\\r\n-y&=&\\sqrt[3]{x}-1 \\\\ \\\\\r\ny&=&1-\\sqrt[3]{x}\r\n\\end{array}\\)<\/li>\r\n \t
  16. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\ny&=&\\dfrac{12-3x}{4} \\\\ \\\\\r\nx&=&\\dfrac{12-3y}{4} \\\\ \\\\\r\n4x&=&12-3y \\\\ \\\\\r\n3y&=&12-4x \\\\ \\\\\r\ny&=&\\dfrac{12-4x}{3} \\\\ \\\\\r\ny&=&4-\\dfrac{4}{3}x\r\n\\end{array}\\)<\/li>\r\n \t
  17. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\\r\ny&=&(x-3)^3 \\\\\r\nx&=&(y-3)^3 \\\\\r\n\\sqrt[3]{x}&=&y-3 \\\\ \\\\\r\ny&=&\\sqrt[3]{x}+3\r\n\\end{array}\\)<\/li>\r\n \t
  18. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\\r\ny&=&\\sqrt[5]{-x}+2 \\\\\r\nx&=&\\sqrt[5]{-y}+2 \\\\\r\nx-2&=&\\sqrt[5]{-y} \\\\\r\n(x-2)^5&=&-y \\\\ \\\\\r\ny&=&-(x-2)^5\r\n\\end{array}\\)<\/li>\r\n \t
  19. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\ny&=&\\dfrac{x}{x-1} \\\\ \\\\\r\nx&=&\\dfrac{y}{y-1} \\\\ \\\\\r\nx(y-1)&=&y \\\\ \\\\\r\nxy-x&=&y \\\\ \\\\\r\ny-xy&=&-x \\\\ \\\\\r\ny(1-x)&=&-x \\\\ \\\\\r\ny&=&\\dfrac{-x}{1-x} \\\\ \\\\\r\ny&=&\\dfrac{x}{x-1}\r\n\\end{array}\\)<\/li>\r\n \t
  20. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\ny&=&\\dfrac{-3-2x}{x+3} \\\\ \\\\\r\nx&=&\\dfrac{-3-2y}{y+3} \\\\ \\\\\r\nx(y+3)&=&-3-2y \\\\\r\nxy+3x&=&-3-2y \\\\\r\n+2y-3x&&-3x+2y \\\\\r\n\\midrule\r\nxy+2y&=&-3-3x \\\\\r\ny(x+2)&=&-3-3x \\\\ \\\\\r\ny&=&\\dfrac{-3-3x}{x+2} \\\\ \\\\\r\ny&=&-\\dfrac{3x+3}{x+2}\r\n\\end{array}\\)<\/li>\r\n \t
  21. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\ny&=&\\dfrac{x-1}{x+1} \\\\ \\\\\r\nx&=&\\dfrac{y-1}{y+1} \\\\ \\\\\r\nx(y+1)&=&y-1 \\\\\r\nxy+x&=&y-1 \\\\\r\nxy-y&=&-x-1 \\\\\r\ny(x-1)&=&-x-1 \\\\ \\\\\r\ny&=&\\dfrac{-x-1}{x-1} \\\\ \\\\\r\ny&=&-\\dfrac{x+1}{x-1}\r\n\\end{array}\\)<\/li>\r\n \t
  22. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\ny&=&\\dfrac{x}{x+2} \\\\ \\\\\r\nx&=&\\dfrac{y}{y+2} \\\\ \\\\\r\nx(y+2)&=&y \\\\\r\nxy+2x&=&y \\\\\r\n2x&=&y-xy \\\\\r\n2x&=&y(1-x) \\\\ \\\\\r\ny&=&\\dfrac{2x}{1-x}\r\n\\end{array}\\)<\/li>\r\n<\/ol>","rendered":"
      \n
    1. \"\begin{array}{rrl}<\/li>\n
    2. \"\begin{array}{rrl}<\/li>\n
    3. \"\begin{array}{rrl}<\/li>\n
    4. \"\begin{array}{rrl}<\/li>\n
    5. \"\begin{array}{rrl}<\/li>\n
    6. \"\begin{array}{rrl}<\/li>\n
    7. \"\begin{array}{rrcrr}<\/li>\n
    8. \"\begin{array}{rrcrr}<\/li>\n
    9. \"\begin{array}{rrl}<\/li>\n
    10. \"\begin{array}{rrl}<\/li>\n
    11. \"\begin{array}{rrl}<\/li>\n
    12. \"\begin{array}{rrl}<\/li>\n
    13. \"\begin{array}{rrl}<\/li>\n
    14. \"\begin{array}{rrl}<\/li>\n
    15. \"\begin{array}{rrl}<\/li>\n
    16. \"\begin{array}{rrl}<\/li>\n
    17. \"\begin{array}{rrl}<\/li>\n
    18. \"\begin{array}{rrl}<\/li>\n
    19. \"\begin{array}{rrl}<\/li>\n
    20. \"\begin{array}{rrl}<\/li>\n
    21. \"\begin{array}{rrl}<\/li>\n
    22. \"\begin{array}{rrl}<\/li>\n<\/ol>\n","protected":false},"author":727,"menu_order":104,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"back-matter-type":[],"contributor":[],"license":[],"class_list":["post-1824","back-matter","type-back-matter","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter\/1824","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/types\/back-matter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/users\/727"}],"version-history":[{"count":3,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter\/1824\/revisions"}],"predecessor-version":[{"id":1827,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter\/1824\/revisions\/1827"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter\/1824\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/media?parent=1824"}],"wp:term":[{"taxonomy":"back-matter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter-type?post=1824"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/contributor?post=1824"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/license?post=1824"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}