document.write( "Question 64228: Help please!\r
\n" ); document.write( "\n" ); document.write( "1.) The solution set of the inequality x+5/x-2<=6 in interval notation\r
\n" ); document.write( "\n" ); document.write( "2.) If x+243, then x^-1/3 is ?\r
\n" ); document.write( "\n" ); document.write( "3.) The inequality -4x+7<=2x+4 is equivalent to ?\r
\n" ); document.write( "\n" ); document.write( "4.) What is the largest zero of the function f(x)=x^3+5x^2+5x-2 \n" ); document.write( "
Algebra.Com's Answer #44898 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
Help please!\r\n" );
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document.write( "1.) The solution set of the inequality x+5/x-2<=6 in interval notation\r\n" );
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document.write( "                x + 5\r\n" );
document.write( "               ------- £ 6\r\n" );
document.write( "                x - 2 \r\n" );
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document.write( "Get 0 on the right by subtracting 6 from both sides:\r\n" );
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document.write( "                x + 5\r\n" );
document.write( "               ------- - 6 £ 0\r\n" );
document.write( "                x - 2\r\n" );
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document.write( "Write 6 over 1\r\n" );
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document.write( "                x + 5     6\r\n" );
document.write( "               ------- - --- £ 0\r\n" );
document.write( "                x - 2     1\r\n" );
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document.write( "LCD = x - 2.  The first fraction already has that\r\n" );
document.write( "LCD.  The make the second fraction have it too, we\r\n" );
document.write( "multiply top and bottom by it:\r\n" );
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document.write( "                x + 5     6(x - 2)\r\n" );
document.write( "               ------- - ---------- £ 0\r\n" );
document.write( "                x - 2     1(x - 2)\r\n" );
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document.write( "                x + 5     6x - 12\r\n" );
document.write( "               ------- - ---------- £ 0\r\n" );
document.write( "                x - 2      x - 2\r\n" );
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document.write( "Combine the two fractions:\r\n" );
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document.write( "                x + 5 - (6x - 12)\r\n" );
document.write( "               ------------------- £ 0\r\n" );
document.write( "                     x - 2     \r\n" );
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document.write( "                x + 5 - 6x + 12\r\n" );
document.write( "               ------------------ £ 0\r\n" );
document.write( "                     x - 2\r\n" );
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document.write( "                -5x + 17\r\n" );
document.write( "               ---------- £ 0\r\n" );
document.write( "                  x - 2\r\n" );
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document.write( "The zero of the numerator is 17/5 or 3.4\r\n" );
document.write( "The zero of the denominator is 2\r\n" );
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document.write( "So the critical values are 2 and 3.4\r\n" );
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document.write( "We draw a number line and mark those two \r\n" );
document.write( "values:\r\n" );
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document.write( "-----------o----------o-------------\r\n" );
document.write( "           2         3.4\r\n" );
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document.write( "Select a test value less than 2, say 0,\r\n" );
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document.write( "Substitute it into\r\n" );
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document.write( "                -5x + 17\r\n" );
document.write( "               ---------- £ 0\r\n" );
document.write( "                  x - 2\r\n" );
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document.write( "                -5(0) + 17\r\n" );
document.write( "               ------------ £ 0\r\n" );
document.write( "                  (0) - 2\r\n" );
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document.write( "                    -17/2 £ 0 \r\n" );
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document.write( "This is true, so we shade that part of \r\n" );
document.write( "the number line\r\n" );
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document.write( "< ===========o----------o-------------\r\n" );
document.write( "             2         3.4\r\n" );
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document.write( "Now choose a test value between 2 and 3.4,\r\n" );
document.write( "say 3\r\n" );
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document.write( "Substitute it into\r\n" );
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document.write( "                -5x + 17\r\n" );
document.write( "               ---------- £ 0\r\n" );
document.write( "                  x - 2\r\n" );
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document.write( "                -5(3) + 17\r\n" );
document.write( "               ------------ £ 0\r\n" );
document.write( "                  (3) - 2\r\n" );
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document.write( "                    2 £ 0 \r\n" );
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document.write( "This is false, so we do not shade that part of \r\n" );
document.write( "the number line.\r\n" );
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document.write( "< ===========o----------o-------------\r\n" );
document.write( "             2         3.4\r\n" );
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document.write( "Now choose a test value greater than 3.4,\r\n" );
document.write( "say 4\r\n" );
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document.write( "Substitute it into\r\n" );
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document.write( "                -5x + 17\r\n" );
document.write( "               ---------- £ 0\r\n" );
document.write( "                  x - 2\r\n" );
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document.write( "                -5(4) + 17\r\n" );
document.write( "               ------------ £ 0\r\n" );
document.write( "                  (4) - 2\r\n" );
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document.write( "                  -3/2 £ 0 \r\n" );
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document.write( "This is true, so we shade that part of \r\n" );
document.write( "the number line\r\n" );
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document.write( "< ===========o----------o============ >\r\n" );
document.write( "             2         3.4\r\n" );
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document.write( "Now we have to test the critical points themselves.\r\n" );
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document.write( "x cannot be 2 because that causes the denominator \r\n" );
document.write( "to be 0, which is undefined. However x can be 3.4\r\n" );
document.write( "because that makes only the numerator 0 and since\r\n" );
document.write( "the inequality is \" £ \" and not \" < \", we can\r\n" );
document.write( "include 0.  So we darken the circle at 3.4.\r\n" );
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document.write( "< ===========o----------·============ >\r\n" );
document.write( "             2         3.4\r\n" );
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document.write( "So the interval notation for the solution is\r\n" );
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document.write( "  (-¥, 2) È [3.4, ¥) \r\n" );
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document.write( "2.) If x=243, then x-1/3 is ?\r\n" );
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document.write( "    Notice that 243 = 3·3·3·3·3 = 35\r\n" );
document.write( "                                   1\r\n" );
document.write( "    x-1/3 = 243-1/3 = (35)-1/3 = ----------- =\r\n" );
document.write( "                                 (35)1/3\r\n" );
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document.write( "        1         1          1            1          1 \r\n" );
document.write( "    -------- = -------- = --------- = --------- = -------- =\r\n" );
document.write( "      35/3        31+2/3     31·32/3       3·32/3     3(³Ö3²)\r\n" );
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document.write( "        1     ³Ö3        ³Ö3       ³Ö3     ³Ö3\r\n" );
document.write( "    --------·----- =  --------- = ----- = ----  \r\n" );
document.write( "     3(³Ö3²)  ³Ö3      3(³Ö3³)     3·3      9\r\n" );
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document.write( "3.) The inequality -4x+7<=2x+4 is equivalent to ?\r\n" );
document.write( "    \r\n" );
document.write( "    Solve it like an equation until the last step:\r\n" );
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document.write( "                  -4x + 7 < 2x + 4\r\n" );
document.write( "                      - 7      - 7\r\n" );
document.write( "               -----------------------\r\n" );
document.write( "                  -4x     < 2x - 3\r\n" );
document.write( "                  -2x      -2x   \r\n" );
document.write( "               -----------------------\r\n" );
document.write( "                  -6x     <     -3\r\n" );
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document.write( "But when you get to the last step, where you divide\r\n" );
document.write( "by the coefficient of x, if you divide by a POSITIVE\r\n" );
document.write( "number you KEEP the inequality sign as is, but if\r\n" );
document.write( "you divide by a NEGATIVE number you must REVERSE the\r\n" );
document.write( "inequality sign.  Here we must divide by -6, which is\r\n" );
document.write( "NEGATIVE, so we must change the < to >.\r\n" );
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document.write( "                -6x/(-6)  > -3/(-6\r\n" );
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document.write( "                        x > 1/2\r\n" );
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document.write( "4.) What is the largest zero of the function f(x)=x^3+5x^2+5x-2\r\n" );
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document.write( "The only way to find it is to approximate it by using a graphing \r\n" );
document.write( "calculator. It is approximately 0.3027756377\r\n" );
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document.write( "Edwin
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