document.write( "Question 597391: I am stuck on this problem. My answer is different from what the textbook answer says.\r
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document.write( "Find the domain of the function.\r\n" );
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document.write( " _____________\r\n" );
document.write( "y = √2x² + 5x - 12\r\n" );
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Algebra.Com's Answer #378163 by AnlytcPhil(1806)![]() ![]() You can put this solution on YOUR website! \r\n" ); document.write( "\r\n" ); document.write( "I think the following is more in line with what your teacher has in mind:\r\n" ); document.write( " _____________\r\n" ); document.write( "y = √2x² + 5x - 12\r\n" ); document.write( "\r\n" ); document.write( "2x² + 5x - 12 ≧ 0\r\n" ); document.write( "\r\n" ); document.write( "(x + 4)(2x - 3) ≧ 0\r\n" ); document.write( "\r\n" ); document.write( "Now we find critical numbers, which are the \r\n" ); document.write( "zeros of the left side, by setting each factor\r\n" ); document.write( "equal to 0:\r\n" ); document.write( "\r\n" ); document.write( "x + 4 = 0; 2x - 3 = 0\r\n" ); document.write( " x = -4; x = 3/2 = 1 1/2\r\n" ); document.write( "\r\n" ); document.write( "We place these critical number on a number line: \r\n" ); document.write( "\r\n" ); document.write( "<========⚫--------------------⚫--------------\r\n" ); document.write( "-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 \r\n" ); document.write( "\r\n" ); document.write( "We use darkened circles instead of open circles because the \r\n" ); document.write( "critical numbers are themselves solutions in this case since \r\n" ); document.write( "the inequality is ≧ rather that >.\r\n" ); document.write( "\r\n" ); document.write( "There are three intervals to test for solutions to the\r\n" ); document.write( "inequality, the intervals between and beyond the critical\r\n" ); document.write( "numbers\r\n" ); document.write( "\r\n" ); document.write( "(-oo,-4), (-4, 3/2), and (3/2,oo). Or if you haven't had\r\n" ); document.write( "interval notation, then they are these inequalities:\r\n" ); document.write( "\r\n" ); document.write( "x ≦ -4, -4 ≦ x ≦ 3/2, x ≧ 3/2\r\n" ); document.write( "\r\n" ); document.write( "Choose a test point in the interval (-oo,-4), say -5.\r\n" ); document.write( "Substitute 5 into the inequality:\r\n" ); document.write( "\r\n" ); document.write( " (x + 4)(2x - 3) ≧ 0\r\n" ); document.write( "(-5 + 4)(2(-5) - 3) ≧ 0 \r\n" ); document.write( " (-1)(-10 - 3) ≧ 0\r\n" ); document.write( " (-1)(-13) ≧ 0\r\n" ); document.write( " 13 ≧ 0\r\n" ); document.write( "\r\n" ); document.write( "That is true, so (-oo,-4) is part of the solution, so we\r\n" ); document.write( "shade that interval on the number line:\r\n" ); document.write( "\r\n" ); document.write( " Choose a test point in the interval (-4,3/2), say 0.\r\n" ); document.write( "Substitute 0 into the inequality:\r\n" ); document.write( "\r\n" ); document.write( " (x + 4)(2x - 3) ≧ 0\r\n" ); document.write( " (0 + 4)(2(0) - 3) ≧ 0 \r\n" ); document.write( " (4)(0 - 3) ≧ 0\r\n" ); document.write( " (4)(-3) ≧ 0\r\n" ); document.write( " -12 ≧ 0\r\n" ); document.write( "\r\n" ); document.write( "That is false, so (-4,3/2) is NOT part of the solution, so we\r\n" ); document.write( "DO NOT shade that interval on the number line, so we still have \r\n" ); document.write( "jus:t\r\n" ); document.write( "\r\n" ); document.write( "<========⚫--------------------⚫--------------\r\n" ); document.write( "-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 \r\n" ); document.write( "\r\n" ); document.write( "Choose a test point in the interval (3/2,-oo), say 2.\r\n" ); document.write( "Substitute 2 into the inequality:\r\n" ); document.write( "\r\n" ); document.write( " (x + 4)(2x - 3) ≧ 0\r\n" ); document.write( " (2 + 4)(2(2) - 3) ≧ 0 \r\n" ); document.write( " (6)(4 - 3) ≧ 0\r\n" ); document.write( " (6)(1) ≧ 0\r\n" ); document.write( " 6 ≧ 0\r\n" ); document.write( "\r\n" ); document.write( "That is true, so (3/2,oo) is part of the solution, so we\r\n" ); document.write( "shade that interval on the number line:\r\n" ); document.write( "\r\n" ); document.write( "<========⚫--------------------⚫==============>\r\n" ); document.write( "-6 -5 -4 -3 -2 -1 0 1 2 3 4 5\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "So the domain written in interval notation is\r\n" ); document.write( "\r\n" ); document.write( "(-oo,-4] U [3/2, oo)\r\n" ); document.write( "\r\n" ); document.write( "or as an inequality the domain is written this way:\r\n" ); document.write( "\r\n" ); document.write( "x ≦ -4 OR x ≧ 3/2\r\n" ); document.write( "\r\n" ); document.write( "Edwin\n" ); document.write( " |