document.write( "Question 762384: Hi, Please help..\r
\n" ); document.write( "\n" ); document.write( "a: On the one set of axes sketch the graph of {{y = f (x)}} and
\n" ); document.write( "\"+f%28x%29=+f%5E-1%28x%29\", where domain is from 0 to infinity \"f%28x%29=+x%5E2+\"\r
\n" ); document.write( "\n" ); document.write( "b: find the coordinates of the points for which \"+f%28x%29+=+f%5E-1+%28x%29\"\r
\n" ); document.write( "\n" ); document.write( "I solve part a but I have difficulty solving part b: here is my working:\r
\n" ); document.write( "\n" ); document.write( "a) \"f+%28x%29+=+x%5E2\"
\n" ); document.write( " \"+y+=+x%5E2\"
\n" ); document.write( " \"+x+=+y%5E2\"
\n" ); document.write( " \"+sqrt+x+=+y\" I sketched the graph ok\r
\n" ); document.write( "\n" ); document.write( "b) \"+x%5E2+=+sqrt+x+\"\r
\n" ); document.write( "\n" ); document.write( " \"+x+=+x+\" How do I find the coordinate of the point x??
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Algebra.Com's Answer #463926 by KMST(5328)\"\" \"About 
You can put this solution on YOUR website!
The graph of \"f%5E%28-1%29%28x%29\" is the same as the graph of \"f%28x%29\", but exchanging the x- and y-axes. You could also say that one graph is the reflection of the other across the \"y=x\" diagonal line.
\n" ); document.write( "\"graph%28300%2C300%2C-1%2C9%2C-1%2C9%2Cx%5E2%2Csqrt%28x%29%2Cx%29\"
\n" ); document.write( "Of course, the graph of \"f%5E%28-1%29%28x%29=sqrt%28x%29\" is only for \"x%3E=0\".
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\n" ); document.write( "From \"x%5E2=sqrt%28x%29\", squaring both sides we get
\n" ); document.write( "\"x%5E4=x\" for \"x%3E=0\"
\n" ); document.write( "(You do not get \"x=x\" in any way).
\n" ); document.write( "\"x%5E4=x\" --> \"x%5E4-x=0\" --> \"%28x%5E3-1%29x=0\"
\n" ); document.write( "whose solutions are \"x=0\" and \"x=1\",
\n" ); document.write( "corresponding to the origin and point (1,1).
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