document.write( "Question 439239: Please help...
\n" ); document.write( "Sketch the graph of Arccos x if Cos x has a domain of 0 ≤ x ≤ π
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Algebra.Com's Answer #303531 by MathLover1(20850)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "here is y = cos^-1x = arccosx and its graph:\r
\n" ); document.write( "\n" ); document.write( "open this image:\r
\n" ); document.write( "\n" ); document.write( "http://img23.imageshack.us/f/arccos3.gif/\r
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\n" ); document.write( "\n" ); document.write( "Since {y = cos^-1(x) is the \"inverse\" of the function \"y+=+cos+x\", the function y = cos^-1(x)\"if\" and \"only\"\"+if\"\"+cos+y+=+x\".
\n" ); document.write( "But, since \"y+=+cos+x\" is \"not\" \"one-to-one\", its \"domain\" must be restricted in order that y = cos^-1(x) is a function.\r
\n" ); document.write( "\n" ); document.write( "To get the graph of y = cos^-1x, start with a graph of \"y+=+cos+x\".\r
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Solved by pluggable solver: PLOT any graph
Graphing function \"cos%28x%29\":
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\n" ); document.write( " \"graph%28+500%2C+500%2C+-4%2C+4%2C+-4%2C+4%2C+cos%28x%29+%29\"

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\n" ); document.write( "\n" ); document.write( "\"Restrict\" the \"domain\" of the function to a one-to-one region - typically [\"0\", \"pi\"] is used for cos^ -1x. This leaves the range of the restricted function unchanged as [\"-1\", \"1\"].\r
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\n" ); document.write( "\n" ); document.write( "Reflect the graph across the line \"y+=+x\" to get the graph of y = cos^-1 x y = arccos x, the black curve at right on the graph above.\r
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\n" ); document.write( "\n" ); document.write( "Notice that y = cos^-1x has domain [\"-1\",\"+1\"] and range [\"0\", \"pi\"]. It is strictly decreasing on its entire domain. \r
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