document.write( "Question 1203489: Determine the exact maximum and minimum y-values and their corresponding x-values for one period where
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document.write( "x > 0.
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document.write( " (For each answer, use the first occurrence for which
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document.write( "x > 0.)
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document.write( " f(x)=5sin(x-5pi/6) \n" );
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Algebra.Com's Answer #839098 by math_tutor2020(3817)![]() ![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "The min and max of y = sin(x) are y = -1 and y = 1 respectively. \n" ); document.write( "This would apply even if the input x is replaced with any expression such as x - 5pi/6\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Sticking a 5 out front will vertically stretch the graph. \n" ); document.write( "This would mean the min and max of y = 5*sin(x-5pi/6) are y = -5 and y = 5 respectively. \n" ); document.write( "Use a graphing tool like a TI83, Desmos, or GeoGebra to confirm this claim. \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Let's replace y with -5 and solve for x. \n" ); document.write( "y = 5*sin(x-5pi/6) \n" ); document.write( "-5 = 5*sin(x-5pi/6) \n" ); document.write( "-5/5 = sin(x-5pi/6) \n" ); document.write( "-1 = sin(x-5pi/6) \n" ); document.write( "x-5pi/6 = arcsin(-1) or x-5pi/6 = pi - arcsin(-1) \n" ); document.write( "x-5pi/6 = 3pi/2 or x-5pi/6 = pi - 3pi/2 \n" ); document.write( "x-5pi/6 = 3pi/2 or x-5pi/6 = 2pi/2 - 3pi/2 \n" ); document.write( "x-5pi/6 = 3pi/2 or x-5pi/6 = -pi/2 \n" ); document.write( "x = 3pi/2+5pi/6 or x = -pi/2+5pi/6 \n" ); document.write( "x = 9pi/6+5pi/6 or x = -3pi/6+5pi/6 \n" ); document.write( "x = 14pi/6 or x = 2pi/6 \n" ); document.write( "x = 7pi/3 or x = pi/3\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Because 1 < 7, it leads to pi < 7pi and further pi/3 < 7pi/3 \n" ); document.write( "pi/3 is the smaller of the two values.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "This is the first occurrence when we reach the min y = -5 such that x > 0.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "To find when the max y = 5 first occurs, plug in y = 5 and follow similar steps as shown above. \n" ); document.write( "After doing so, you should find that x = 4pi/3 is the first x value to reach the max y = 5 when x > 0.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The graphing tools I mentioned earlier can be used to confirm the answers. \n" ); document.write( "Various online calculators can be helpful as well. \n" ); document.write( "I recommend that you do the actual scratch work yourself and use the calculators to check your answers (or else you won't be ready for the exams later).\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Summary \n" ); document.write( "Min and max are y = -5 and y = 5 \n" ); document.write( "First min occurs when x = pi/3 \n" ); document.write( "First max occurs when x = 4pi/3 \n" ); document.write( "In other words, the location of the first min is (pi/3, -5) when x > 0. \n" ); document.write( "The location of the first max is (4pi/3, 5) when x > 0.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Graph \n" ); document.write( " \n" ); document.write( "A = first local min = (pi/3, -5) when focusing on the interval x > 0 \n" ); document.write( "B = first local max = (4pi/3, 5) when focusing on the interval x > 0 \n" ); document.write( "pi/3 = 1.04719755 approximately \n" ); document.write( "4pi/3 = 4.18879020 approximately\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |