document.write( "Question 1192491: Find the equation of the tangent to;\r
\n" ); document.write( "\n" ); document.write( "f(x) = e^-x at the point where x=2.\r
\n" ); document.write( "\n" ); document.write( "So I know dy/dx=-1/e^x, but don't understand how I can sub that x value in the original equation and find the y value.
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Algebra.Com's Answer #824373 by ikleyn(52787)\"\" \"About 
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\n" ); document.write( "Find the equation of the tangent to;
\n" ); document.write( "f(x) = e^-x at the point where x=2.
\n" ); document.write( "So I know dy/dx=-1/e^x, but don't understand how I can sub that x value in the original equation
\n" ); document.write( "and find the y value.
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document.write( "You correctly determined the derivative function  \"%28dy%29%2F%28dx%29\" = -\"1%2Fe%5Ex\".\r\n" );
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document.write( "To proceed further, you should substitute x= 2 into the formula. \r\n" );
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document.write( "Doing this way, you will find the value of the derivative at the point x= 2, which is the slope of the function\r\n" );
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document.write( "    f(x) = \"e%5E%28-x%29\"\r\n" );
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document.write( "at the point x= 2.\r\n" );
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document.write( "        So, the slope is  m = -\"1%2Fe%5E2\".\r\n" );
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document.write( "Now you know the point on the plot of the function: it is  (\"x%5B0%5D\",\"y%5B0%5D\") = (\"2\",\"1%2Fe%5E2\");\r\n" );
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document.write( "and you know the slope:  it is  m = -\"1%2Fe%5E2\".\r\n" );
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document.write( "So, the equation of the tangent line is  \"y+-+y%5B0%5D\" = \"m%2A%28x-x%5B0%5D%29\",  or\r\n" );
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document.write( "    y - \"1%2Fe%5E2\" = -\"%281%2Fe%5E2%29%2A%28x+-+2%29\".\r\n" );
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document.write( "It is your ANSWER.\r\n" );
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document.write( "You can also use any other EQUIVALENT form of the last equation.\r\n" );
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\n" ); document.write( "\n" ); document.write( "Solved and explained.\r
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\n" ); document.write( "\n" ); document.write( "If you have question/questions, then let me know.\r
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\n" ); document.write( "\n" ); document.write( "If you will post your questions, please refer to the problem's ID number 1192491 .\r
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